Linear Vector Spaces and Operators: Difference between revisions

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Quantum Mechanics can be conveniently formulated in the language of abstract state vectors, from which the various representations (wave mechanics, matrix mechanics, Schrödinger, Heisenberg and interaction pictures, etc.) can be derived. A formulation of quantum mechanics in terms of LVS (Linear Vector Space) hinges on the fact that the [[Schrödinger equation]] is linear.An operator is a mathematical operation performed on a vector belonging to a Linear Vector Space, the result of which is another vector belonging to the same Linear Vector Space.
{{Quantum Mechanics A}}
Quantum mechanics can be conveniently formulated in the language of abstract state vectors, from which the various representations (wave mechanics, matrix mechanics, Schrödinger, Heisenberg and interaction pictures, etc.) can be derived. A formulation of quantum mechanics in terms of linear vector spaces hinges on the fact that the [[Schrödinger equation]] is linear. An operator defines a mathematical operation performed on a vector belonging to a linear vector space, the result of which is another vector belonging to the same linear vector space.


== Ket Space ==
== The Vector (Ket) Space ==


In quantum mechanics a physical state is represented by a state vector in a complex vector space and have all the properties described above. Following Dirac we call such a vector a ket denoted by <math>|\alpha ></math> and have become an essential part of quantum mechanics. This state ket is postulated to contain complete information about the physical state i.e. everything we are allowed to ask about the state is contained in the ket. The complex linear vector space that we work in quantum mechanics are usually infinite dimensional. In such case the vector space in question is known as
In quantum mechanics, a physical state is represented by a state vector in a complex linear vector space. Following Dirac, we call such a vector a "ket", denoted by <math>|\alpha\rangle.</math> This state vector is postulated to contain complete information about the physical state (i.e. everything we are allowed to ask about the state is contained in the vector). The complex linear vector spaces that we work with in quantum mechanics are usually infinite dimensional. In this case, the vector space in question is known as a [[#The Hilbert Space|Hilbert space]] after D. Hilbert, who studied vector spaces in infinite dimensions.
a Hilbert space after D. Hilbert, who studied vector spaces in infinite dimensions.


One of the postulate of quantum mechanics is that <math>|\alpha ></math> and <math>c|\alpha ></math>, with c = 0 represent the same physical state. In other words, only the ”direction” in vector space is of significance.
One of the postulates of quantum mechanics is that <math>|\alpha\rangle</math> and <math>c|\alpha\rangle,</math> with <math>c\neq 0,</math> represent the same physical state. In other words, only the ”direction” in vector space is of significance.


== Bra Space ==
Since we assume that these vectors belong to a linear vector space, we may, given any set of state vectors <math>|\alpha\rangle,|\beta\rangle,\ldots,</math> form a superposition of the states, given by a linear combination of the vectors <math>c_{\alpha}|\alpha\rangle+c_{\beta}|\beta\rangle+\cdots.</math>


The vector space we have been dealing with is a ket space. We now introduce the notion of a bra space, a vector space ”dual to” the ket space. We postulate that corresponding to every ket <math>|\alpha ></math> there exists a bra denoted by <math><\alpha |</math> in this dual or bra space. The bra space is spanned by a set of bra vectors {<math><\alpha |</math>} which correspond to the set of kets {<math>|\alpha ></math>}.
== The Dual (Bra) Space ==
Mathematically, the dual space is a set of linear functions <math> <\beta|: V \rightarrow C </math>
which act on the member of vector space (kets) where
V is the vector space and C is the set of complex numbers.
There is one to one correspondence between a ket space and a bra space:


<math>|\alpha >\leftrightarrow <\alpha |</math>
The vector space we have been dealing with is a ket space. We now introduce the notion of a "bra" space, a vector space ”dual to” the ket space. We postulate that corresponding to every ket <math>|\alpha\rangle</math> there exists a bra denoted by <math>\langle\alpha |</math> in this dual or bra space. The dual space is spanned by a set of bra vectors <math>\left \{\langle\alpha |\right \}</math> which correspond to the set of kets <math>\left \{|\alpha\rangle\right \}.</math>  Mathematically, the dual space is a set of linear functions <math> \langle\beta|: V \rightarrow C </math> that act on the members of the corresponding vector space where <math>V\!</math> is the vector space and <math>C\!</math> is the set of complex numbers.


<math>|a_{1}>,|a_{2}>...,...\leftrightarrow <a_{1}|<a_{2}...,...</math>
There is a one-to-one correspondence between the members of a ket space and those of the corresponding bra space, <math>|\alpha\rangle\leftrightarrow\langle\alpha |,</math> where <math>\leftrightarrow</math> stands for dual correspondence. Roughly speaking, we can regard the bra space as some kind of "mirror image" of the ket space.


<math>|\alpha >+|\beta >\leftrightarrow <\alpha |+<\beta |,</math>
The bra dual to <math>c|\alpha\rangle</math> is postulated to be <math>c^{\ast}\langle\alpha |</math>, not <math>c\langle\alpha |,</math> which is a very important point to note. More generally, we have


where <math>\leftrightarrow</math> stands for dual correspondence. Roughly speaking, we can regard the bra space as some
<math>c_{\alpha}|\alpha\rangle+c_{\beta}|\beta\rangle+\cdots\leftrightarrow c^{\ast}_{\alpha}\langle\alpha |+c^{\ast}_{\beta}\langle\beta |+\cdots</math>
kind of mirror image of the ket space.


The bra dual to <math>c|\alpha ></math> is postulated to be <math>c\ast <\alpha |</math> not <math>c<\alpha |</math>, which is a very important point
== The Hilbert Space ==
to note. More generally we have
 
A Hilbert space <math>H\!</math> consisting of a set of vectors  <math>\psi,\phi,\chi,\ldots\!</math>  and a set of scalars <math>a,b,c,\ldots\!</math> obeys the following properties.
 
(a) '''<math>H\!</math> is a linear vector space.'''  It obeys all the properties of a linear vector space as mentioned in the previous section.
 
 
(b) '''The scalar product defined in <math>H\!</math> is strictly positive.'''  The scalar product of one element <math>\phi\!</math> with another element <math>\psi\!</math> is a complex number, denoted by <math>\langle\phi|\psi\rangle</math>.  This scalar satisfies the following properties.
 
'''(1)''' The scalar product of <math>\phi\!</math> with <math>\psi\!</math> is same as the complex conjugate of the scalar product of <math>\psi\!</math> with <math>\phi.\!</math>
 
<math>\langle\phi|\psi\rangle=\langle\psi|\phi\rangle^{\ast}</math>
 
'''(2)''' The scalar product of <math>\psi\!</math> with <math>\phi\!</math> is linear with respect to <math>\phi.\!</math>
 
<math>\langle\psi|a\psi_{1} + b\psi_{2}\rangle=a\langle\psi|\phi_{1}\rangle + b\langle\psi|\phi_{2}\rangle</math>
 
'''(3)''' The scalar product of a state vector <math>\psi\!</math> with itself is a positive real number.
 
<math>\langle\psi|\psi\rangle=|\psi|^2\geq 0</math>
 
In terms of this scalar product, we may express the normalization of a wave function as <math>\langle\psi|\psi\rangle=1.</math>
 
==Schwartz Inequality==
 
For two states <math>|\psi\rangle</math> and <math>|\phi\rangle</math> belonging to a linear vector space, the following theorem, known as the Schwartz inequality, holds:
 
<math>|\langle\psi|\phi\rangle|^{2}\leq\langle\psi|\psi\rangle\langle\phi|\phi\rangle</math>
 
If the vectors <math>|\psi\rangle</math>and <math>|\phi\rangle</math> are linearly dependent, i.e. <math>|\psi\rangle=\alpha|\phi\rangle,</math> then the above relation becomes an equality.


<math>c_{\alpha }|\alpha >+c_{\beta }|\beta >\leftrightarrow c\ast _{\alpha }<\alpha |+c\ast _{\beta  }<\beta </math>
''Proof:''
Let <math>|\psi\rangle</math> and <math>|\phi\rangle</math> be arbitrary vectors in the vector space <math>V.</math>  The inequality is trivial in the case that at least one of <math>|\psi\rangle</math> and <math>|\phi\rangle=0,</math> so we will consider the case that both are nonzero. Let <math>\lambda</math> be a complex number.  Then


==Linear Independence, Basis and Orthonormal Basis==
:<math> 0 \leq \langle\psi-\lambda\phi|\psi-\lambda\phi\rangle=\langle\psi|\psi\rangle-\lambda^{\ast}\langle\psi|\phi\rangle-\lambda\langle\phi|\psi\rangle+|\lambda|^2\langle\phi|\phi\rangle.</math>
Consider <math> \mathit{N}\!</math> vectors(bra states) <math> |1\rangle </math>,<math> |2\rangle </math>,...<math> |N\rangle </math>. They are linearly independent if the relation
<math> \sum_{i=1}^{N}\mathit{e}_i|1\rangle=0 </math>
necessarily implies <math> \mathit{e}_i=0 \!</math> for <math>i = 1,2,...,N\!</math>. They can be used as basis in a vector space, and decomposition of any vector in terms of basis vectors in unique.


While any set of Linearly independent vectors can be used as a basis, normally the discussion is greatly simplified if the basis is an orthonormal one, i.e, <math> \langle \mathit{e_i}|\mathit{e_j} \rangle=\delta_{ij}</math> ,<math>1\leq i\leq j\leq n </math> (<math>\mathit{n}\to \infty </math> for an infinite-dimensional vector space). The action of an operator is completely known once its action on each of the basis vectors of <math> \mathit{V}</math> is given.The Linear Vector Space used in quantum mechanics is known by the name of Hilbert Space.
The above expression is valid for any value of <math>\lambda</math>. The right-hand side of the above expression is minimized if we choose


== The Hilbert Space ==
:<math> \lambda = \frac{\langle\psi|\phi\rangle}{\langle\phi|\phi\rangle}.</math>
A Hilbert Space <math>H\!</math> consisting of a set of vectors  <math>\psi,\phi,\chi\!</math> and a set of scalars <math>a,b,c\!</math> obeys the following properties.
 
Using this value of <math>\lambda,</math> we obtain
 
:<math> 0 \leq \langle\psi|\psi\rangle-\frac{|\langle\psi|\phi\rangle|^2}{\langle\phi|\phi\rangle},</math>
 
or
 
:<math> |\langle\psi|\phi\rangle|^2 \leq \langle\psi|\psi\rangle\langle\phi|\phi\rangle.</math>
 
== Linear Operators ==
 
Let <math>V</math> be a linear vector space.  A linear operator <math>\hat{A}</math> is an operation, denoted by <math>\hat{A}|\psi\rangle,</math> that maps a given ket vector <math>|\psi\rangle</math> in <math>V</math> to a different vector <math>|\hat{A}\psi\rangle</math> in the same space, and has the property that <math>\hat{A}\left (c_{\psi}|\psi\rangle+c_{\phi}|\phi\rangle\right )=c_{\psi}\hat{A}|\psi\rangle+c_{\phi}\hat{A}|\phi\rangle.</math>
 
In addition, linear operators obey the following properties.
 
(1) If <math>\hat{A}|\psi\rangle=\hat{B}|\psi\rangle</math> for every <math>|\psi\rangle,</math> then <math>\hat{A}</math> is equal to <math>\hat{B}</math>.
 
(2) Commutative law: <math>\hat{A}+\hat{B}=\hat{B}+\hat{A} \!</math>
 
(3) Associative law: <math>(\hat{A}+\hat{B})+\hat{C}=\hat{A}+(\hat{B}+\hat{C}) \!</math>
 
(4) Multiplication of operators: <math>\hat{A}\hat{B}|\psi\rangle=\hat{A}(\hat{B}|\psi\rangle) </math>
 
(5) There exists an identity operator <math>\hat{I}</math> such that <math>\hat{I}|\psi\rangle=|\psi\rangle.</math>
 
For some, but not all, operators <math>\hat{A},</math> there exists an inverse operator <math>\hat{A}^{-1}</math> such that <math>\hat{A}^{-1}\hat{A}=\hat{A}\hat{A}^{-1}=\hat{I}.</math>
 
== The Hermitian Adjoint ==
 
The dual vector to <math>\hat{A}|\psi\rangle</math>, where <math>\hat{A}</math> is a linear operator, is <math>\langle\hat{A}\psi|=\langle\psi|\hat{A}^{\dagger},</math> where <math>\hat{A}^{\dagger}</math> is known as the Hermitian adjoint of <math>\hat{A},</math> and is itself a linear operator.  The properties obeyed by Hermitian adjoints are as follows.
 
(1) <math> (\hat{A}^{\dagger})^{\dagger}=\hat{A}</math><br/>
(2) For any complex number <math>c,\,(c\hat{A})^{\dagger}=c^{\ast}\hat{A}^{\dagger}</math><br/>
(3) <math> (\hat{A}+\hat{B})^{\dagger}=\hat{A}^{\dagger}+\hat{B}^{\dagger} </math><br/>
(4) <math> (\hat{A}\hat{B})^{\dagger}=\hat{B}^{\dagger}\hat{A}^{\dagger}</math><br/>
(5) For any complex number <math>c,\,c^{\dagger}=c^{\ast}</math><br/>
(6) <math> |a\rangle ^{\dagger}=\langle a| </math><br/>
(7) <math> (|a\rangle\langle b|)^{\dagger}=|b\rangle\langle a|</math>
 
 
== Linear Independence and Bases ==
 
Consider <math>N\!</math> vectors (ket states) <math> |1\rangle </math>,<math> |2\rangle,\ldots,|N\rangle</math> belonging to a linear vector space <math>V.</math>  They are linearly independent if the relation
<math> \sum_{i=1}^{N}c_i|i\rangle=0 </math>
necessarily implies <math> c_i=0 \!</math> for <math>i = 1,2,...,N\!</math>. They can be used as a basis in a vector space, and decomposition of any vector in terms of basis vectors in unique.  Any such set of basis vectors must be complete; i.e., ''any'' vector can be written as a linear combination of vectors from the set.


While any set of linearly independent vectors can be used as a basis, it is usually easier to work in an orthonormal basis; i.e., <math> \langle i|j \rangle=\delta_{ij}.</math>  For such a case, we may write a completeness relation for the basis.  Consider an arbitrary vector <math>|\psi\rangle</math> expanded in terms of a given orthonormal basis:


(a) <math>H\!</math> is a linear vector space.
<math>|\psi\rangle=\sum_{i=1}^{N}c_i|i\rangle</math>


(It obeys all the properties of a linear vector space as mentioned in the previous section.)
We may find the coefficients <math>c_i\!</math> by simply taking the scalar product of <math>|\psi\rangle</math> with each of the basis vectors, obtaining


<math>c_i=\langle i|\psi\rangle.</math>


(b) The scalar product defined in <math>H\!</math> is strictly positive.
Substituting back into the expansion for <math>|\psi\rangle,</math> we find that


(The scalar product of one element <math>\phi\!</math> with another element <math>\psi\!</math> is a complex number, denoted by <math>\langle\phi|\psi\rangle</math>.This scalar satisfies the following properties
<math>\sum_{i=1}^{N}|i\rangle\langle i|\psi\rangle=|\psi\rangle.</math>


1. The scalar product of <math>\phi\!</math> with that of <math>\psi\!</math> is same as the complex conjugate of the scalar product of <math>\psi\!</math> with <math>\phi\!</math>.
Since <math>|\psi\rangle</math> is arbitrary, we conclude that
              <math>\langle\phi|\psi\rangle</math>=<math>\langle\psi|\phi\rangle^*</math>


2. The scalar product of <math>\psi\!</math> with <math>\phi\!</math> is linear with respect to <math>\phi\!</math>.
<math>\sum_{i=1}^{N}|i\rangle\langle i|=\hat{I},</math>
            <math>\langle\psi|a\psi_{1} + b\psi_{2}\rangle</math> = <math>a\langle\psi|\phi_{1}\rangle + b\langle\psi|\phi_{2}\rangle</math>


3. The scalar product of a state vector <math>\psi\!</math> with itself is a positive real number.
where <math>\hat{I}</math> is the identity operator.  This is the completeness relation that we sought.  It is possible to derive similar relations for more general bases through a similar line of reasoning.
              <math>\langle\psi|\psi\rangle</math> = <math>|\psi|^{2}  \geq 0 </math>


== Schwartz Inequality==
== Matrix Elements of a Linear Operator ==


For two states <math>|\psi\rangle</math> and <math>|\phi\rangle</math> belonging to the Hilbert Space,the Schwartz Inequality:
The action of a linear operator <math>\hat{A}</math> is completely known once its action on each of the basis vectors of <math>V</math> is given.  To see this, let us consider the action of such an operator on an arbitrary vector:
                          <math>|\langle\psi|\phi\rangle|^{2}</math> <math>\leq</math> <math>\langle\psi|\psi\rangle</math><math>\langle\phi|\phi\rangle</math>


If the vectors <math>|\psi\rangle</math>and <math>|\phi\rangle</math>are linearly dependent such that <math>|\psi\rangle</math>=<math>\alpha|\phi\rangle</math>, then the above relation reduces to an equality.
<math>\hat{A}|\psi\rangle=\sum_{i=1}^{N}c_i\hat{A}|i\rangle=\sum_{i=1}^{N}\sum_{j=1}^{N}c_i A_{ji}|j\rangle,</math>


''Proof:''
where <math>A_{ji}=\langle j|\hat{A}|i\rangle.</math> Substituting in the expression for the <math>c_i\!</math> derived earlier, we obtain
Let ''u'',&nbsp;''v'' be arbitrary vectors in the vector space ''V''.  The inequality is trivial in the case ''v''&nbsp;=&nbsp;0, so we will start with <''v'',&nbsp;''v''> is nonzero. Let <math> \delta</math> be any number in the field&nbsp;''F''. Then,


:<math> 0 \leq \left\| u-\delta v \right\|^2
<math>\hat{A}|\psi\rangle=\sum_{i=1}^{N}\sum_{j=1}^{N}A_{ji}|j\rangle\langle i|\psi\rangle.</math>
= \langle u-\delta v,u-\delta v \rangle = \langle u,u \rangle - \bar{\delta} \langle u,v \rangle - \delta \langle v,u \rangle + |\delta|^2 \langle v,v\rangle. \, </math>


The above expression is valid for any number <math> \delta</math>. Choosing
Again, since <math>|\psi\rangle</math> is arbitrary, we conclude that


:<math> \delta = \langle u,v \rangle \cdot \langle v,v \rangle^{-1}, \, </math>  
<math>\hat{A}=\sum_{i=1}^{N}\sum_{j=1}^{N}A_{ji}|j\rangle\langle i|.</math>


we obtain
The coefficients <math>A_{ji}\!</math> appearing in the above expression are known as the matrix elements of the operator <math>\hat{A}.</math>  This is because the expansion coefficients for the "input" and "output" vectors are related by a matrix equation, with the <math>A_{ji}\!</math> being the elements of the matrix appearing in said equation.  Let us write the "output" <math>\hat{A}|\psi\rangle</math> as


:<math> 0 \leq \langle u,u \rangle - |\langle u,v \rangle|^2 \cdot \langle v,v \rangle^{-1} \, </math>
<math>\hat{A}|\psi\rangle=\sum_{j=1}^{N}c'_j|j\rangle.</math>


which is true if and only if
We see, however, from our earlier derivation that


:<math> |\langle u,v \rangle|^2 \leq \langle u,u \rangle \cdot \langle v,v \rangle, \, </math>
<math>c'_j=\sum_{i=1}^{N}A_{ji}c_i,</math>


or equivalently:
or <math>\mathbf{c}'=\mathbf{A}\mathbf{c},</math> where <math>\mathbf{A}</math> is the matrix whose elements are given by the <math>A_{ji}\!</math> and <math>\mathbf{c}</math> and <math>\mathbf{c}'</math> are column vectors of the expansion coefficients of <math>|\psi\rangle</math> and <math>\hat{A}|\psi\rangle,</math> respectively.


:<math> \big| \langle u,v \rangle \big|
If we follow a similar line of reasoning for vectors in the dual vector space, we obtain <math>\mathbf{c}'^{\dagger}=\mathbf{c}^{\dagger}\mathbf{A}^{\dagger},</math> where <math>\mathbf{A}^{\dagger}</math> is the transposed complex conjugate, or Hermitian adjoint, of <math>\mathbf{A}.</math> We have thus justified our use of the term "Hermitian adjoint" in describing the action of an operator in the dual space.
\leq \left\|u\right\| \left\|v\right\|, \, </math>


which completes the proof.
== Special Linear Operators in Quantum Mechanics ==


==Dual Spaces and Adjoint Operators==
'''Hermitian Operator:''' An operator <math>\hat{H}</math> is called Hermitian if <math>\hat{H}^{\dagger}=\hat{H}.</math> All physical observables in quantum mechanics are represented by Hermitian operators.
Let <math> \mathit{V}</math> be a Linear Vector Space. From <math> \mathit{V}\!</math>, we can construct another Linear Vector Space <math> \mathit{V^'}</math> by the following rule:
For each element in <math> \mathit{V}</math> of the form <math>|z\rangle \equiv [\alpha |x\rangle + \beta |y\rangle] \in \mathit{V}</math>, we associate it with <math>\langle z| \in [\alpha^* \langle x| + \beta^* \langle y| \in \mathit{V}</math>. <math> \mathit{V^*}\!</math> is called the dual of <math> \mathit{V}\!</math> (the bra space). together, they form a bracket.
Let <math> \mathit{A}</math> be an operator.It is completely known if <math> \langle y|A|x\rangle</math> is known for all <math>|x\rangle</math> and <math>|y\rangle</math> <math> \in V</math>.
We define a new operator <math>A^\dagger</math> (the adjoint) such that <math> \langle x|A^\dagger|y\rangle=\langle y|A|x\rangle</math> for all <math>|x\rangle</math> and <math>|y\rangle</math> <math> \in V</math>.


==Special Linear Operators in Quantum Mechanics==
'''Anti-Hermitian operator:''' An operator <math>\hat{A}</math> is called anti-Hermitian or skew-Hermitian if <math>\hat{A}^{\dagger}=-\hat{A}.</math>


'''Hermitian Operator:''' An operator <math>H\!</math>is called hermitian if <math>H = H^\dagger</math>. all physical observables in quantum mechanics are represented by hermitian operators.
Every operator <math>\hat{C}</math> can be decomposed uniquely in terms of a Hermitian and an anti-Hermitian part:
<math>\hat{C}=\frac{\hat{C}+\hat{C}^{\dagger}}{2}+\frac{\hat{C}-\hat{C}^{\dagger}}{2}\equiv H+A</math>.


'''Unitary Operator:''' An operator <math>U\!</math>is called unitary if there exits an unique <math>U^{-1}\!</math> and is equal to <math>U^\dagger</math>, i.e., <math>UU^\dagger = U^\dagger U = 1</math>. An important property of unitary operators is that it preserves the norm of a vector,which in quantum mechanics refers to the conservation of probability under physical operations.
'''Unitary Operator:''' An operator <math>\hat{U}</math> is called unitary if there exits an unique <math>\hat{U}^{-1}</math> and is equal to <math>\hat{U}^\dagger</math>, i.e., <math>\hat{U}\hat{U}^{\dagger}=\hat{U}^{\dagger}\hat{U}=1</math>. An important property of unitary operators is that they preserve the norm of a vector, which in quantum mechanics refers to the conservation of probability under physical operations.  Most transformations of importance in quantum mechanics are given by unitary operators.


'''Antihermitian operator:''' An operator <math>A\!</math> is called antihermitian if <math>A^\dagger = -A</math>. Every operator <math>C\!</math> can be decomposed uniquely in terms of a hermitian and an antihermitian part:
'''Antilinear operator:''' An operator <math>\hat{A}</math> is called antilinear if for any two vectors <math>|\psi\rangle</math> and <math>|\phi\rangle\in V</math> and for any two complex numbers <math>c_{\psi}</math> and <math>c_{\phi}</math>, <math>\hat{A}\left (c_{\psi}|\psi\rangle + c_{\phi}|\phi\rangle\right )=c_{\psi}^{\ast}\hat{A}|\psi\rangle+c_{\phi}^{\ast}\hat{A}|\phi\rangle.</math> All operators of importance in quantum mechanics are linear, with one important exception - the time reversal operator <math>\hat{T}</math> is an antilinear operator.
<math>B = \frac{B+B^\dagger}{2}+\frac{B-B^\dagger}{2}\equiv H+A</math>.


'''Antilinear operator:''' An operator <math>A\!</math> is called antilinear if for all <math>|x\rangle</math> and <math>|y\rangle</math> <math> \in V</math> and for every complex numbers <math>\alpha</math> and  <math>\beta</math>,
==Theorem on Eigenvalues and Eigenstates of Hermitian Operator==
<math>A[\alpha |x\rangle + \beta |y\rangle]=[\alpha^* |x\rangle + \beta^* |y\rangle]</math>.
We will now prove that the eigenvalues of Hermitian operators are real and that two eigenvectors of a Hermitian operator that correspond to different eigenvalues are orthogonal.
In quantum mechanics, we need only linear operators but for one important exception: The operator <math>T\!</math> that represents time reversal of states is an "antilinear operator".


==Theorem on Hermitian Operator==
''Proof:'' Consider an eigenstate <math>|\psi\rangle</math> of a Hermitian operator <math>\hat{H}</math>, corresponding to an eigenvalue <math>\lambda;</math> i.e.,
Eigenvalues of a hermitian operators are real and eigenvectors of a hermitian operator and corresponding to different eigenvalues are orthogonal.


''Proof:'' Consider the eigenvalue equation:
<math>\hat{H}|\psi\rangle = \lambda|\psi\rangle.</math>


<math>H|x\rangle = \lambda|x\rangle</math>
Taking the Hermitian adjoint of both sides, and using the fact that <math>H^\dagger = H</math>, we get
where <math>|x\rangle</math> is an eigenvector corresponding to an eigenvalue <math>\lambda\!</math> of <math>H\!</math>. Taking the dual of the equation, and using the fact that <math>H^\dagger = H</math>, we get


<math>\langle x|H = \lambda\langle x|</math>
<math>\langle\psi|\hat{H} = \lambda\langle\psi|.</math>


Taking the scalar product of the first equation with <math>\langle x|</math> and of the second with <math>|x\rangle</math>, we get  
Taking the scalar product of the first equation with <math>\langle\psi|</math> and of the second with <math>|\psi\rangle</math>, we get  


<math>\langle x|H|x\rangle = \lambda\langle x|x\rangle = \lambda^*\langle x|x\rangle</math>
<math>\langle\psi|\hat{H}|\psi\rangle = \lambda\langle\psi|\psi\rangle = \lambda^{\ast}\langle\psi|\psi\rangle.</math>


As <math>|x\rangle</math> is not a null vector, <math> \lambda^* = \lambda\!</math>,i.e, <math>\lambda</math> is real. To prove the second part of the theorem, consider another eigenstate <math>|y\rangle</math> with a distinct (real)eigenvalue <math>\lambda^'\neq \lambda</math>. Taking scalar product of the first equation with <math>\langle y|</math> and of the second equation withn <math>\langle x|</math>, we get
Because <math>|\psi\rangle</math> is not a null vector, we conclude that <math> \lambda^{\ast} = \lambda;</math> i.e, <math>\lambda</math> is real.


<math>\langle|H|x\rangle = \lambda\langle y|x\rangle = \lambda^'\langle y|x\rangle</math>.
To prove the second part of the theorem, consider another eigenstate <math>|\phi\rangle</math> with a different eigenvalue <math>\lambda'\neq\lambda;</math> i.e.,


As <math>\lambda \neq \lambda^'</math>, <math>\langle y|x\rangle = 0</math>, i.e., <math>|x\rangle</math> and <math>|y\rangle</math> are mutually orthogonal.
<math>\hat{H}|\phi\rangle = \lambda'|\phi\rangle.</math>


==Hermitian adjoint==
Taking the scalar product of the first equation with <math>\langle\phi|</math> and of the second equation withn <math>\langle \psi|</math>, we get
Suppose that there exists an operator <math> \hat{A}^{\dagger}</math>, for which <math>\langle \hat{A}^{\dagger} \psi_1|\psi_2\rangle=\langle\psi_1|\hat{A}|\psi_2\rangle </math>. If this holds for all wavefunction, then we construct a Hermitian adjoint.


An operator is called Hermitian if: <math> \hat{O}^{\dagger}=\hat{O}</math>.
<math>\langle\phi|\hat{H}|\psi\rangle = \lambda\langle\phi|\psi\rangle=\lambda'\langle\phi|\psi\rangle</math>.


==Relations of Hermitian adjoint==
Because <math>\lambda\neq\lambda',</math>, we conclude that <math>\langle\phi|\psi\rangle = 0;</math> i.e., <math>|\psi\rangle</math> and <math>|\phi\rangle</math> are mutually orthogonal.
<math> (\hat{F}^{\dagger})^{\dagger}=\hat{F} </math><br/>
<math> (c\hat{F})^{\dagger}=c^* \hat{F}^{\dagger} </math><br/>
<math> (\hat{F}+\hat{G})^{\dagger}=\hat{F}^{\dagger}+\hat{G}^{\dagger} </math><br/>
<math> (\hat{F}\hat{G})^{\dagger}=\hat{G}^{\dagger}\hat{F}^{\dagger}</math><br/>
<math> c^{\dagger}=c^* </math><br/>
<math> |a\rangle ^{\dagger}=\langle a| </math><br/>
<math> \langle a|b \rangle ^{\dagger}=\langle b|a \rangle </math><br/>
<math> (|a\rangle\langle b|)^{\dagger}=|b\rangle\langle a|</math>


==The Law of Calculation for Operators==
Since the eigenstates of a Hermitian operator are orthogonal (in fact, if they are normalized, then they are orthonormal), they often form a convenient basis in which to expand vectors; we will in fact often use the eigenstates of some observable as a basis.
(1) If <math> \hat{F} |\psi\rangle=\hat{G} |\psi\rangle </math> for every <math>|\psi\rangle</math>, then <math> \hat{F}</math> is equal to <math> \hat{G}</math>.


(2) Commutative Law: <math>\hat{F}+\hat{G}=\hat{G}+\hat{F} \!</math>
== Projection Operators ==


(3) Associative Law: <math>(\hat{F}+\hat{G})+\hat{H}=\hat{F}+(\hat{G}+\hat{H}) \!</math>
A projection operator is an operator that "projects" the vector that it acts on onto the direction of another given vector.  The projection operator corresponding to the vector <math>|x\rangle</math> is given by <math>\hat{P}=|x\rangle\langle x|.</math>  We may see that this is the case by acting with this operator on an arbitrary vector <math>|\psi\rangle:</math>


(4) Multiply: <math> \hat{F}\hat{G} |\psi\rangle=\hat{F}(\hat{G} |\psi\rangle) </math>
<math>\hat{P}|\psi\rangle = |x\rangle\langle x|\psi\rangle = |x\rangle v_x</math>


(5) Identity operator: <math> \hat{I} |\psi\rangle= |\psi\rangle </math>
We see that the result of the operation is a vector "parallel" to <math>|x\rangle,</math> as asserted.


(6) Zero operator: <math> \hat{0} |\psi\rangle= |0\rangle </math>
Let us now state some properties of projection operators.


==Projection Operator==
'''(1) A projection operator is Hermitian and equal to its own square; i.e., it is idempotent.'''


An operator<math>\hat{P}</math> is said to be a Projection operator , if it is Hermitian and equal to its own square.
''Proof:'' The fact that a projection operator is Hermitian follows immediately from the properties of the Hermitian adjoint stated above. The fact that it is also idempotent may be seen as follows:
Thus,<math>\hat{P}</math>= <math>\hat{P}^{\dagger}</math>    and <math>\hat{P}</math>  = <math>\hat{P}^2</math>
Properties of Projection operator:
<math>\hat{P}^2=|x\rangle\langle x|x\rangle\langle x|=|x\rangle\langle x|=\hat{P}</math>


1. The Product of two commuting projection operators <math>\hat{P}_1</math> and <math>\hat{P}_2</math> is also a projection operator.
'''(2) The product of two commuting projection operators <math>\hat{P}_1</math> and <math>\hat{P}_2</math> is also a projection operator.'''
    <math>(\hat{P}_1\hat{P}_2)^{\dagger} =\hat{P}_2^{\dagger}\hat{P}_1^{\dagger} = \hat{P}_2\hat{P}_1 =\hat{P}_1\hat{P}_2 </math>
and  <math>(\hat{P}_1\hat{P}_2)^2 = \hat{P}_1\hat{P}_2 \hat{P}_1\hat{P}_2 =\hat{P}_1^2\hat{P}_2^2= \hat{P}_1\hat{P}_2</math>


2. The sum of two or more Projection operators is not a Projection operator in general.The sum is a Projection operator iff the operators are mutually orthogonal.
''Proof:''


3. Two Projection operators are said to be orthogonal if the product of the two operators is zero.
<math>(\hat{P}_1\hat{P}_2)^{\dagger} =\hat{P}_2^{\dagger}\hat{P}_1^{\dagger} = \hat{P}_2\hat{P}_1 =\hat{P}_1\hat{P}_2 </math>


The Projection operator corresponding to the ket <math>|x \rangle </math> can be written as  <math>\hat{P} = |x\rangle\langle x|</math>.
and
The completion relation is given by  <math>\Sigma |x\rangle\langle x| = I </math>  or,  <math>\Sigma P_x \!= I</math>


For any vector V,
<math>(\hat{P}_1\hat{P}_2)^2 = \hat{P}_1\hat{P}_2 \hat{P}_1\hat{P}_2 =\hat{P}_1^2\hat{P}_2^2= \hat{P}_1\hat{P}_2.</math>
  <math>P_x|V\rangle = |x\rangle\langle x|V\rangle = |x\rangle v_x</math>


The Projection operator selects the component of <math>|V\rangle</math> in the direction of    <math>|x\rangle</math>
The sum of two projection operators is ''not'', in general, a projection operator itself; it will only be a projection operator if the two original operators are orthogonal; i.e., if <math>\hat{P}_1\hat{P}_2=0.</math>

Latest revision as of 14:11, 8 August 2013

Quantum Mechanics A
SchrodEq.png
Schrödinger Equation
The most fundamental equation of quantum mechanics; given a Hamiltonian Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}} , it describes how a state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi\rangle} evolves in time.
Basic Concepts and Theory of Motion
UV Catastrophe (Black-Body Radiation)
Photoelectric Effect
Stability of Matter
Double Slit Experiment
Stern-Gerlach Experiment
The Principle of Complementarity
The Correspondence Principle
The Philosophy of Quantum Theory
Brief Derivation of Schrödinger Equation
Relation Between the Wave Function and Probability Density
Stationary States
Heisenberg Uncertainty Principle
Some Consequences of the Uncertainty Principle
Linear Vector Spaces and Operators
Commutation Relations and Simultaneous Eigenvalues
The Schrödinger Equation in Dirac Notation
Transformations of Operators and Symmetry
Time Evolution of Expectation Values and Ehrenfest's Theorem
One-Dimensional Bound States
Oscillation Theorem
The Dirac Delta Function Potential
Scattering States, Transmission and Reflection
Motion in a Periodic Potential
Summary of One-Dimensional Systems
Harmonic Oscillator Spectrum and Eigenstates
Analytical Method for Solving the Simple Harmonic Oscillator
Coherent States
Charged Particles in an Electromagnetic Field
WKB Approximation
The Heisenberg Picture: Equations of Motion for Operators
The Interaction Picture
The Virial Theorem
Commutation Relations
Angular Momentum as a Generator of Rotations in 3D
Spherical Coordinates
Eigenvalue Quantization
Orbital Angular Momentum Eigenfunctions
General Formalism
Free Particle in Spherical Coordinates
Spherical Well
Isotropic Harmonic Oscillator
Hydrogen Atom
WKB in Spherical Coordinates
Feynman Path Integrals
The Free-Particle Propagator
Propagator for the Harmonic Oscillator
Differential Cross Section and the Green's Function Formulation of Scattering
Central Potential Scattering and Phase Shifts
Coulomb Potential Scattering

Quantum mechanics can be conveniently formulated in the language of abstract state vectors, from which the various representations (wave mechanics, matrix mechanics, Schrödinger, Heisenberg and interaction pictures, etc.) can be derived. A formulation of quantum mechanics in terms of linear vector spaces hinges on the fact that the Schrödinger equation is linear. An operator defines a mathematical operation performed on a vector belonging to a linear vector space, the result of which is another vector belonging to the same linear vector space.

The Vector (Ket) Space

In quantum mechanics, a physical state is represented by a state vector in a complex linear vector space. Following Dirac, we call such a vector a "ket", denoted by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\alpha\rangle.} This state vector is postulated to contain complete information about the physical state (i.e. everything we are allowed to ask about the state is contained in the vector). The complex linear vector spaces that we work with in quantum mechanics are usually infinite dimensional. In this case, the vector space in question is known as a Hilbert space after D. Hilbert, who studied vector spaces in infinite dimensions.

One of the postulates of quantum mechanics is that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\alpha\rangle} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c|\alpha\rangle,} with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c\neq 0,} represent the same physical state. In other words, only the ”direction” in vector space is of significance.

Since we assume that these vectors belong to a linear vector space, we may, given any set of state vectors Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\alpha\rangle,|\beta\rangle,\ldots,} form a superposition of the states, given by a linear combination of the vectors Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{\alpha}|\alpha\rangle+c_{\beta}|\beta\rangle+\cdots.}

The Dual (Bra) Space

The vector space we have been dealing with is a ket space. We now introduce the notion of a "bra" space, a vector space ”dual to” the ket space. We postulate that corresponding to every ket Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\alpha\rangle} there exists a bra denoted by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle\alpha |} in this dual or bra space. The dual space is spanned by a set of bra vectors Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left \{\langle\alpha |\right \}} which correspond to the set of kets Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left \{|\alpha\rangle\right \}.} Mathematically, the dual space is a set of linear functions Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle\beta|: V \rightarrow C } that act on the members of the corresponding vector space where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V\!} is the vector space and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C\!} is the set of complex numbers.

There is a one-to-one correspondence between the members of a ket space and those of the corresponding bra space, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\alpha\rangle\leftrightarrow\langle\alpha |,} where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \leftrightarrow} stands for dual correspondence. Roughly speaking, we can regard the bra space as some kind of "mirror image" of the ket space.

The bra dual to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c|\alpha\rangle} is postulated to be Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c^{\ast}\langle\alpha |} , not Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c\langle\alpha |,} which is a very important point to note. More generally, we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{\alpha}|\alpha\rangle+c_{\beta}|\beta\rangle+\cdots\leftrightarrow c^{\ast}_{\alpha}\langle\alpha |+c^{\ast}_{\beta}\langle\beta |+\cdots}

The Hilbert Space

A Hilbert space Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H\!} consisting of a set of vectors Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi,\phi,\chi,\ldots\!} and a set of scalars Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a,b,c,\ldots\!} obeys the following properties.

(a) Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H\!} is a linear vector space. It obeys all the properties of a linear vector space as mentioned in the previous section.


(b) The scalar product defined in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H\!} is strictly positive. The scalar product of one element Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi\!} with another element Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi\!} is a complex number, denoted by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle\phi|\psi\rangle} . This scalar satisfies the following properties.

(1) The scalar product of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi\!} with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi\!} is same as the complex conjugate of the scalar product of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi\!} with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi.\!}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle\phi|\psi\rangle=\langle\psi|\phi\rangle^{\ast}}

(2) The scalar product of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi\!} with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi\!} is linear with respect to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi.\!}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle\psi|a\psi_{1} + b\psi_{2}\rangle=a\langle\psi|\phi_{1}\rangle + b\langle\psi|\phi_{2}\rangle}

(3) The scalar product of a state vector Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi\!} with itself is a positive real number.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle\psi|\psi\rangle=|\psi|^2\geq 0}

In terms of this scalar product, we may express the normalization of a wave function as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle\psi|\psi\rangle=1.}

Schwartz Inequality

For two states Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\psi\rangle} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\phi\rangle} belonging to a linear vector space, the following theorem, known as the Schwartz inequality, holds:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\langle\psi|\phi\rangle|^{2}\leq\langle\psi|\psi\rangle\langle\phi|\phi\rangle}

If the vectors Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\psi\rangle} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\phi\rangle} are linearly dependent, i.e. Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\psi\rangle=\alpha|\phi\rangle,} then the above relation becomes an equality.

Proof: Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\psi\rangle} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\phi\rangle} be arbitrary vectors in the vector space Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V.} The inequality is trivial in the case that at least one of and so we will consider the case that both are nonzero. Let be a complex number. Then

The above expression is valid for any value of . The right-hand side of the above expression is minimized if we choose

Using this value of we obtain

or

Linear Operators

Let be a linear vector space. A linear operator is an operation, denoted by that maps a given ket vector in to a different vector in the same space, and has the property that

In addition, linear operators obey the following properties.

(1) If for every then is equal to .

(2) Commutative law:

(3) Associative law:

(4) Multiplication of operators:

(5) There exists an identity operator such that

For some, but not all, operators there exists an inverse operator such that

The Hermitian Adjoint

The dual vector to , where is a linear operator, is where is known as the Hermitian adjoint of and is itself a linear operator. The properties obeyed by Hermitian adjoints are as follows.

(1)
(2) For any complex number
(3)
(4)
(5) For any complex number
(6)
(7) Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (|a\rangle\langle b|)^{\dagger}=|b\rangle\langle a|}


Linear Independence and Bases

Consider Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N\!} vectors (ket states) Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |1\rangle } ,Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |2\rangle,\ldots,|N\rangle} belonging to a linear vector space Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V.} They are linearly independent if the relation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{i=1}^{N}c_i|i\rangle=0 } necessarily implies Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_i=0 \!} for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i = 1,2,...,N\!} . They can be used as a basis in a vector space, and decomposition of any vector in terms of basis vectors in unique. Any such set of basis vectors must be complete; i.e., any vector can be written as a linear combination of vectors from the set.

While any set of linearly independent vectors can be used as a basis, it is usually easier to work in an orthonormal basis; i.e., Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle i|j \rangle=\delta_{ij}.} For such a case, we may write a completeness relation for the basis. Consider an arbitrary vector Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\psi\rangle} expanded in terms of a given orthonormal basis:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\psi\rangle=\sum_{i=1}^{N}c_i|i\rangle}

We may find the coefficients Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_i\!} by simply taking the scalar product of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\psi\rangle} with each of the basis vectors, obtaining

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_i=\langle i|\psi\rangle.}

Substituting back into the expansion for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\psi\rangle,} we find that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{i=1}^{N}|i\rangle\langle i|\psi\rangle=|\psi\rangle.}

Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\psi\rangle} is arbitrary, we conclude that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{i=1}^{N}|i\rangle\langle i|=\hat{I},}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{I}} is the identity operator. This is the completeness relation that we sought. It is possible to derive similar relations for more general bases through a similar line of reasoning.

Matrix Elements of a Linear Operator

The action of a linear operator Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}} is completely known once its action on each of the basis vectors of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V} is given. To see this, let us consider the action of such an operator on an arbitrary vector:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}|\psi\rangle=\sum_{i=1}^{N}c_i\hat{A}|i\rangle=\sum_{i=1}^{N}\sum_{j=1}^{N}c_i A_{ji}|j\rangle,}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_{ji}=\langle j|\hat{A}|i\rangle.} Substituting in the expression for the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_i\!} derived earlier, we obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}|\psi\rangle=\sum_{i=1}^{N}\sum_{j=1}^{N}A_{ji}|j\rangle\langle i|\psi\rangle.}

Again, since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\psi\rangle} is arbitrary, we conclude that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}=\sum_{i=1}^{N}\sum_{j=1}^{N}A_{ji}|j\rangle\langle i|.}

The coefficients Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_{ji}\!} appearing in the above expression are known as the matrix elements of the operator Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}.} This is because the expansion coefficients for the "input" and "output" vectors are related by a matrix equation, with the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_{ji}\!} being the elements of the matrix appearing in said equation. Let us write the "output" Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}|\psi\rangle} as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}|\psi\rangle=\sum_{j=1}^{N}c'_j|j\rangle.}

We see, however, from our earlier derivation that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c'_j=\sum_{i=1}^{N}A_{ji}c_i,}

or Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{c}'=\mathbf{A}\mathbf{c},} where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{A}} is the matrix whose elements are given by the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_{ji}\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{c}} and are column vectors of the expansion coefficients of and respectively.

If we follow a similar line of reasoning for vectors in the dual vector space, we obtain where is the transposed complex conjugate, or Hermitian adjoint, of We have thus justified our use of the term "Hermitian adjoint" in describing the action of an operator in the dual space.

Special Linear Operators in Quantum Mechanics

Hermitian Operator: An operator is called Hermitian if All physical observables in quantum mechanics are represented by Hermitian operators.

Anti-Hermitian operator: An operator is called anti-Hermitian or skew-Hermitian if

Every operator can be decomposed uniquely in terms of a Hermitian and an anti-Hermitian part: .

Unitary Operator: An operator is called unitary if there exits an unique and is equal to , i.e., . An important property of unitary operators is that they preserve the norm of a vector, which in quantum mechanics refers to the conservation of probability under physical operations. Most transformations of importance in quantum mechanics are given by unitary operators.

Antilinear operator: An operator is called antilinear if for any two vectors and and for any two complex numbers and , All operators of importance in quantum mechanics are linear, with one important exception - the time reversal operator is an antilinear operator.

Theorem on Eigenvalues and Eigenstates of Hermitian Operator

We will now prove that the eigenvalues of Hermitian operators are real and that two eigenvectors of a Hermitian operator that correspond to different eigenvalues are orthogonal.

Proof: Consider an eigenstate of a Hermitian operator , corresponding to an eigenvalue i.e.,

Taking the Hermitian adjoint of both sides, and using the fact that , we get

Taking the scalar product of the first equation with and of the second with , we get

Because is not a null vector, we conclude that i.e, is real.

To prove the second part of the theorem, consider another eigenstate with a different eigenvalue i.e.,

Taking the scalar product of the first equation with and of the second equation withn , we get

.

Because , we conclude that i.e., and are mutually orthogonal.

Since the eigenstates of a Hermitian operator are orthogonal (in fact, if they are normalized, then they are orthonormal), they often form a convenient basis in which to expand vectors; we will in fact often use the eigenstates of some observable as a basis.

Projection Operators

A projection operator is an operator that "projects" the vector that it acts on onto the direction of another given vector. The projection operator corresponding to the vector is given by We may see that this is the case by acting with this operator on an arbitrary vector

We see that the result of the operation is a vector "parallel" to as asserted.

Let us now state some properties of projection operators.

(1) A projection operator is Hermitian and equal to its own square; i.e., it is idempotent.

Proof: The fact that a projection operator is Hermitian follows immediately from the properties of the Hermitian adjoint stated above. The fact that it is also idempotent may be seen as follows:

(2) The product of two commuting projection operators and is also a projection operator.

Proof:

and

The sum of two projection operators is not, in general, a projection operator itself; it will only be a projection operator if the two original operators are orthogonal; i.e., if

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