Commutation Relations and Simultaneous Eigenvalues: Difference between revisions

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(New page: ==Commutator=== The commutator of two operators A and B is defined as follows: <math>[A,B]=AB-BA\,\!.</math> When 2 operators <math>A</math> and <math>B</math> commute, then <math>\left...)
 
 
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==Commutator===
{{Quantum Mechanics A}}
==Commutator==


The commutator of two operators A and B is defined as follows:
The commutator of two operators <math>\hat{A}\!</math> and <math>\hat{B}\!</math> is defined as follows:


<math>[A,B]=AB-BA\,\!.</math>
<math>[\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}\,\!.</math>


When 2 operators <math>A</math> and <math>B</math> commute, then <math>\left[A,B\right]=0</math>. Conversely, if <math>\left[A,B\right]\neq 0</math>, the operators do not commute, and we can think of the commutator between two operators as a quantization of how badly they fail to commute.   
When 2 operators <math>\hat{A}\!</math> and <math>\hat{B}\!</math> commute, then <math>[\hat{A},\hat{B}]=0.</math>  On the other hand, if <math>[\hat{A},\hat{B}]\neq 0,</math> then the operators do ''not'' commute, and we can think of the commutator between two operators as a measure of how badly they fail to commute.  Note that any operator will commute with an ordinary complex number.


'''Some Identities:'''<br/>
'''Some Identities:'''<br/>
<math> [A,B]+[B,A]=0 \!</math><br/>
<math> [\hat{A},\hat{B}]+[\hat{B},\hat{A}]=0 \!</math><br/>
<math>[A,A]= 0 \!</math><br/>
<math>[\hat{A},\hat{A}]= 0 \!</math><br/>
<math>[A,B+C]=[A,B]+[A,C]\!</math><br/>
<math>[\hat{A},\hat{B}+\hat{C}]=[\hat{A},\hat{B}]+[\hat{A},\hat{C}]\!</math><br/>
<math>[A+B,C]=[A,C]+[B,C]\!</math><br/>
<math>[\hat{A}+\hat{B},\hat{C}]=[\hat{A},\hat{C}]+[\hat{B},\hat{C}]\!</math><br/>
<math>[AB,C]=A[B,C]+[A,C]B\!</math><br/>
<math>[\hat{A}\hat{B},\hat{C}]=\hat{A}[\hat{B},\hat{C}]+[\hat{A},\hat{C}]\hat{B}\!</math><br/>
<math>[A,BC]=[A,B]C+B[A,C]\!</math><br/>
<math>[\hat{A},\hat{B}\hat{C}]=[\hat{A},\hat{B}]\hat{C}+\hat{B}[\hat{A},\hat{C}]\!</math><br/>
<math>[A,[B,C]]+[C,[A,B]]+[B,[C,A]]=0\!</math><br/>
<math>[\hat{A},[\hat{B},\hat{C}]]+[\hat{C},[\hat{A},\hat{B}]]+[\hat{B},[\hat{C},\hat{A}]]=0\!</math><br/>
<math> [A,B]=-[B,A]\!</math><br/>
<math> [\hat{A},\hat{B}]=-[\hat{B},\hat{A}]\!</math><br/>
<math> [A,B]^{\dagger} = [\hat{B}^{\dagger},\hat{A}^{\dagger}]</math><br/>
<math> [\hat{A},\hat{B}]^{\dagger} = [\hat{B}^{\dagger},\hat{A}^{\dagger}]</math><br/>


In addition, if any two operators are Hermitian and their product is also Hermitian, then the operators commute because


In addition, if any two operators are Hermitian and their product is Hermitian, then the operators commute because <br/>
<math> (\hat{A}\hat{B})^{\dagger} = \hat{B}^{\dagger}\hat{A}^{\dagger} = \hat{B}\hat{A} </math>
<math> (\hat{A}\hat{B})^{\dagger} = \hat{B}^{\dagger}\hat{A}^{\dagger} = \hat{B}\hat{A} </math> <br/>
and <br/>


<math>(\hat{A}\hat{B})^{\dagger} = \hat{A}\hat{B}</math><br/>
and
so we have that <math>\hat{A}\hat{B} = \hat{B}\hat{A}</math>, which means the commutator is zero.


It should also be noted that any operator will commute with a constant scalar. 
<math>(\hat{A}\hat{B})^{\dagger} = \hat{A}\hat{B}</math>


Also it should be noted that:
so that <math>\hat{A}\hat{B} = \hat{B}\hat{A}.</math>
<math>[A^n,B]=nA^{n-1}[A,B]\!</math><br/>
However this is only true if: <math>[A,[A,B]]=[B,[A,B]]=0\!</math><br/>


One may also prove that <math>[\hat{A}^n,\hat{B}]=n\hat{A}^{n-1}[\hat{A},\hat{B}]\!</math> if <math>[\hat{A},[\hat{A},\hat{B}]]=0\!</math> via mathematical induction.


'''Problem on communtators''' : [http://wiki.physics.fsu.edu/wiki/index.php/User_talk:ChorChan]
==Compatible Observables==


==Compatible observables==
An operator which corresponds to some physically measurable property of a system is called an observable. The following is a list of common physical observables and their corresponding operators given in the coordinate representation: <br/>
Position: <math>\mathbf{r}\rightarrow\hat{\mathbf{r}}  </math><br/>
Momentum: <math>\mathbf{p}\rightarrow\frac{\hbar}{i}\nabla</math><br/>
Kinetic energy: <math>T=\frac{p^2}{2m}\rightarrow -\frac{\hbar^2}{2m}\nabla^2</math><br/>
Potential energy: <math>V(\mathbf{r})\rightarrow V(\hat{\mathbf{r}})</math><br/>
Total energy: <math>E=T+V\rightarrow -\frac{\hbar^2}{2m}\nabla^2+V(\hat{\mathbf{r}})</math><br/>


An operator which corresponds to some physically measurable property of a system is called an observable. The following is a list of common physical observables and their corresponding operators given in coordinate representation: <br/>
All observables are Hermitian.
<math> \text{Position, } \textbf{r}:  \textbf{r}  </math><br/>
<math>\text{Momentum, } \textbf{p}:  \frac{\hbar}{i}\nabla </math><br/>
<math>\text{Kinetic Energy, T}=\frac{p^2}{2m}\text{: } - \frac{\hbar^2}{2m}\nabla^2 </math><br/>
<math>\text{Potential Energy, V: } V(\textbf{r}) </math><br/>
<math>\text{Total Energy, E = T+V: } - \frac{\hbar^2}{2m}\nabla^2 + V(\textbf{r})</math><br/>


All observables are Hermitian.  If two Obersvables have simultaneous eigenkets, meaning for two obervables A and B,     
Two observables <math>\hat{A}</math> and <math>\hat{B}</math> are said to be compatible if it is possible to exactly measure both simultaneously; i.e., they possess a common set of eigenstates:      


<math>\hat{A}|\Psi_{AB}\rangle=a|\Psi_{AB}\rangle\!</math><br/>
<math>\hat{A}|\Psi_{AB}\rangle=a|\Psi_{AB}\rangle\!</math><br/>
<math>\hat{B}|\Psi_{AB}\rangle=b|\Psi_{AB}\rangle\!</math><br/>     
<math>\hat{B}|\Psi_{AB}\rangle=b|\Psi_{AB}\rangle\!</math><br/>     


Then we have that
It follows from this that
<math>\hat{A}\hat{B}|\Psi_{AB}\rangle=
\hat{A}b|\Psi_{AB}\rangle=b\hat{A}|\Psi_{AB}\rangle=ba|\Psi_{AB}\rangle\!</math><br/>
Similarly,
<math>\hat{B}\hat{A}|\Psi_{AB}\rangle=
\hat{B}a|\Psi_{AB}\rangle=a\hat{B}|\Psi_{AB}\rangle=ba|\Psi_{AB}\rangle\!</math><br/>


So we can see that,
<math>\hat{A}\hat{B}|\Psi_{AB}\rangle=\hat{A}b|\Psi_{AB}\rangle=b\hat{A}|\Psi_{AB}\rangle=ba|\Psi_{AB}\rangle\!</math><br/>
<math>\hat{A}\hat{B}-\hat{B}\hat{A}=\left[\hat{A},\hat{B}\right]=0\!</math><br/>.


The same logic works in reverse.  So if two operators, A & B commute, i.e. if , then they have simultaneous eigenkets, and they are said to correspond to compatible observables.  Conversely, if <math>\left[A,B\right]\neq 0</math>, we say that the operators  and do not commute and correspond to incompatible observables.
and
 
<math>\hat{B}\hat{A}|\Psi_{AB}\rangle=\hat{B}a|\Psi_{AB}\rangle=a\hat{B}|\Psi_{AB}\rangle=ba|\Psi_{AB}\rangle.\!</math><br/>
 
Therefore,
 
<math>\hat{A}\hat{B}-\hat{B}\hat{A}=[\hat{A},\hat{B}]=0.\!</math><br/>
 
The same logic works in reverse - if two operators commute, then they have simultaneous eigenkets, and thus they are compatible observables.  Let us consider two commuting operators <math>\hat{A}</math> and <math>\hat{B}.</math>  Suppose that we know the eigenstates of <math>\hat{A};</math> let us call them <math>|\Psi_{A}\rangle.</math>  We know that
 
<math>\hat{A}|\Psi_{A}\rangle=a|\Psi_{A}\rangle.</math>
 
We now act on both sides of this equation with <math>\hat{B}:</math>
 
<math>\hat{B}\hat{A}|\Psi_{A}\rangle=a\hat{B}|\Psi_{A}\rangle,</math>
 
or, because <math>\hat{A}</math> and <math>\hat{B}</math> commute,
 
<math>\hat{A}\hat{B}|\Psi_{A}\rangle=a\hat{B}|\Psi_{A}\rangle.</math>
 
We see that <math>\hat{B}|\Psi_{A}\rangle</math> is therefore also an eigenstate of <math>\hat{A}</math> with the same eigenvalue.  This means that we have "block diagonalized" <math>\hat{B}</math> in the sense that it can only mix eigenstates of <math>\hat{A}</math> with other eigenstates with the same eigenvalue.  All of the eigenstates of <math>\hat{B}</math> are thus also eigenstates of <math>\hat{A},</math> meaning that the two observables are compatible.
 
These observations are especially important when we consider the Hamiltonian of a system. Symmetries of the Hamiltonian are represented by operators that commute with it, and, as we will see later, identification of these symmetries often helps one diagonalize the Hamiltonian and classify its eigenstates.
 
If <math>[\hat{A},\hat{B}]\neq 0</math>, or, equivalently, if it is ''not'' possible to simultaneously diagonalize <math>\hat{A}</math> and <math>\hat{B},</math> then the two operators are said to be incompatible observables.
 
==Generalized Uncertainty Relation==
 
If two observables are incompatible, then one cannot necessarily simultaneously diagonalize them, so that an eigenstate of one may be a non-trivial linear combination of eigenstates of the other.  This fact is embodied in the existence of an uncertainty relation for the two observables, much like that between position and momentum.  We will now derive this general uncertainty relation.
 
For any observable <math>A</math>, let us define the uncertainty <math>\Delta A</math> as the standard deviation of said observable from its expectation value:
:<math>(\Delta A)^2 = \langle(A-\langle A\rangle)\Psi|(A-\langle A\rangle)\Psi\rangle = \langle f|f\rangle </math>
 
where <math> f \equiv (A-\langle A\rangle)\Psi </math>. Likewise, for any other observable <math>B</math>,
:<math>(\Delta B)^2 = \langle g|g\rangle </math> where <math> g \equiv (B-\langle B\rangle)\Psi </math>.
 
Now we invoke the [[Linear Vector Space and Operators#Schwartz Inequality|Schwartz inequality]]. Recall that this is just
:<math> \langle f|f\rangle\langle g|g\rangle\geq |\langle f|g\rangle |^2. </math>
 
Invoking this expression,
:<math> (\Delta A)^2(\Delta B)^2 \geq |\langle f|g\rangle|^2. </math>
 
Now, for any complex number <math>z</math>,
:<math> |z|^2 = [\text{Re}(z)]^2 + [\text{Im}(z)]^2 \geq [\text{Im}(z)]^2 = \left (\frac{z-z^*}{2i}\right )^2 </math>.
 
Letting <math> z = \langle f|g\rangle </math>,
:<math> (\Delta A)^2(\Delta B)^2 \geq \left (\frac{\langle f|g\rangle-\langle g|f\rangle}{2i}\right )^2 </math>.
 
The inner products are
:<math> \langle f|g\rangle = \langle(A-\langle A\rangle)\Psi|(B-\langle B\rangle)\Psi\rangle = \langle(A-\langle A\rangle)(B-\langle B\rangle)\rangle </math>
:<math> = \langle AB - A\langle B\rangle - B\langle A\rangle + \langle A\rangle\langle B\rangle\rangle </math>
:<math> = \langle AB\rangle - \langle B\rangle\langle A\rangle - \langle A\rangle\langle B\rangle + \langle A\rangle\langle B\rangle = \langle AB\rangle - \langle A\rangle\langle B\rangle </math>
 
and
:<math> \langle g|f\rangle = \langle BA\rangle - \langle B\rangle\langle A\rangle. </math>
 
Therefore,
:<math> \langle f|g\rangle - \langle g|f\rangle = \langle AB\rangle - \langle BA\rangle = \langle[A,B]\rangle </math>
 
and
:<math>(\Delta A)^2(\Delta B)^2 \geq \left (\frac{1}{2i}\langle[A,B]\rangle\right )^2,</math>
 
or
:<math>\Delta A\,\Delta B \geq \left |\frac{1}{2i}\langle[A,B]\rangle\right |.</math>
 
This is the general uncertainty relation that we sought.
 
As a simple demonstration of this relation, suppose that the first observable is position, <math> A=x</math>, and the second is momentum, <math>B=p=\frac{\hbar}{i}\frac{d}{dx}</math>.
 
The commutation relation between these two observables is just
:<math>[x,p]=i\hbar,</math>
 
so
:<math>\Delta x\,\Delta p\geq\frac{\hbar}{2}.</math>
 
We have thus obtained a formal proof of the [[Heisenberg Uncertainty Principle]].
 
'''Question:'''
What about the energy-time uncertainty relation?
 
'''Answer:''' We should note that time is not an operator in quantum mechanics, and thus we cannot apply the general uncertainty relation derived above to it. The energy-time uncertainty relation tells us that, for a short time interval, we get broadening in the energy spectrum which we observe. In other words, to get a precise energy value, we need to wait for a long time. The relation between the energy and time intervals is given by <math> \Delta E\,\Delta t\geq \hbar.</math>


==Functions of Operators==
==Functions of Operators==


It is often useful to consider functions of operators, as the Hamiltonian can be written as such.  
It is often useful to consider functions of operators, as the Hamiltonian can be written as such. For example, the exponential function of an operator is represented as a power series,
The exponential function of an operator is represented as a power series:
 
<math> e^\hat{A} = \hat{I} + \frac{\hat{A}}{1!} + \frac{\hat{A}^2}{2!}+ \frac{\hat{A}^3}{3!} + \ldots </math>
 
This leads to the identity that for some <math>f(\lambda) = e^{\lambda\hat{A}}\hat{B}e^{-\lambda\hat{A}}</math>, where <math>\hat{A}</math> and <math>\hat{B}</math> are operators:


<math> e^\hat{A} = \hat{I} + \frac{\hat{A}}{1!} + \frac{\hat{A}^2}{2!}+ \frac{\hat{A}^3}{3!} + ... </math>  
<math>e^{\lambda\hat{A}}\hat{B}e^{-\lambda\hat{A}} = \hat{B} + \frac{\lambda}{1!}[\hat{A},\hat{B}] + \frac{\lambda^2}{2!}[\hat{A},[\hat{A},\hat{B}]] + \frac{\lambda^3}{3!}[\hat{A},[\hat{A},[\hat{A},\hat{B}]]] + \ldots </math>


Which leads to the identity that for some <math>f(\lambda) = e^{\lambda\hat{A}}\hat{B}e^{-\lambda\hat{A}}</math>, where <math>\hat{A}</math> and <math>\hat{B}</math> are operators:
This follows easily when one considers a Taylor series expansion of <math>f(\lambda).\!</math> We note that


<math> e^{\lambda\hat{A}}\hat{B}e^{-\lambda\hat{A}} = \hat{B} + \frac{\lambda}{1!}[\hat{A},\hat{B}] + \frac{\lambda^2}{2!}[\hat{A},[\hat{A},\hat{B}] + \frac{\lambda^3}{3!}[,\hat{A},[\hat{A},[\hat{A},\hat{B}] + ... </math>
<math>\frac{d}{d\lambda}f(\lambda)=e^{\lambda\hat{A}}[\hat{A},\hat{B}]e^{-\lambda\hat{A}}.</math>


This follows easily when one considers a Taylor series expansion of <math>f(\lambda)</math>.  
We may then apply this relation recursively to generate the higher-order derivatives required in the Taylor expansion.


If <math>[\hat{A},[ \hat{A}, \hat{B}]]=\beta \hat{B} </math>, we can simplify the identity to:
If <math>[\hat{A},[\hat{A},\hat{B}]]=\beta\hat{B},</math> then we can simplify the identity to


<math>e^{\lambda\hat{A}}\hat{B}e^{-\lambda\hat{A}} =\hat{B}\cosh\lambda\sqrt{\beta} +\frac{   [ \hat{A}, \hat{B} ]   }{ \sqrt{ \beta } }\sinh\lambda\sqrt{\beta}  
<math>e^{\lambda\hat{A}}\hat{B}e^{-\lambda\hat{A}}=\hat{B}\cosh\lambda\sqrt{\beta}+\frac{[\hat{A},\hat{B}]}{\sqrt{\beta}}\sinh\lambda\sqrt{\beta}.</math>
  </math>


It is also useful to consider the commutator of an operator function with an operator. Given operators <math>\hat{l}</math> and <math>\hat{m}</math> where  
It is also useful to consider the commutator of an operator function with an operator. Given operators <math>\hat{l}</math> and <math>\hat{m}</math> where  
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by use of the power series expansion of <math>f(\hat{l})</math> and the commutator identities in the above section.
by use of the power series expansion of <math>f(\hat{l})</math> and the commutator identities in the above section.


==Physical meaning of various representations==
== Physical Meaning of Various Representations ==


In a given Hilbert basis, it is obvious that the state vector <math>|\psi \rangle</math> is completely determined by the set of its components {Cn },
In a given Hilbert basis, it is obvious that the state vector <math>|\psi \rangle</math> is completely determined by the set of its components <math>\{c_n\},\!</math>


          <math>|\psi \rangle\leftrightarrow \left \{ c_{n}=\langle\phi _{n}| \psi _{n}\right \}</math>
<math>|\psi\rangle\leftrightarrow\left \{c_{n}=\langle\phi_{n}|\psi\rangle\right \}</math>


which we can write as a column vector, the corresponding bra being the conjugate line vector.
which we can write as a column vector; the corresponding bra is then represented by the conjugate line vector. This representation of the state vector is completely equivalent to the wave function <math>\Psi(\mathbf{r},t).</math>
This representation of the state vector is completely equivalent to the wave function <math>\psi \left ( r,t \right )</math>.


Therefore, there are not only two, but an infinite number of equivalent representations of the state of the system. What is their physical meaning?
Therefore, there are not only two, but an infinite number of equivalent representations of the state of the system. What is their physical meaning?


In the basis of the eigenstates of the Hamiltonian, the interpretation of this representation is simple and crystal clear: the <math>C_{n}</math>’s are the amplitudes to find <math>E_{n}</math> in an energy measurement. Therefore,
In the basis of the eigenstates of the Hamiltonian, the interpretation of this representation is simple and crystal clear: the coefficients <math>c_{n}\!</math> are the probability amplitudes for obtaining <math>E_{n}\!</math> in an energy measurement.


• The representation <math>\psi \left ( r,t \right )</math> is more convenient if we are interested in the properties of the particle in space,
Therefore,


• Its Fourier transform <math>\phi \left ( p,t \right )</math> is more convenient if we are interested in its momentum properties,
• The position representation, <math>\Psi(\mathbf{r},t),</math> is more convenient if we are interested in the properties of the particle in space.


• And the components {Cn} in the basis of energy eigenstates are more convenient if we are interested in the energy of the particle.
• Its Fourier transform, <math>\Phi(\mathbf{p},t),</math> is more convenient if we are interested in its momentum properties.


But, owing to Riesz’s theorem, this can be done with any physical quantity, for instance, the angular momentum, which we examine later on and which
• The coefficients <math>c_n\!</math> in the basis of energy eigenstates are more convenient if we are interested in the energy of the particle.
also has discrete eigenvalues. This can be thought of as a “generalization” of the properties of the Fourier transform.    


Owing to Riesz’s theorem, this can be done with any observable, such as the angular momentum, which we examine later on and also has discrete eigenvalues.  This can be thought of as a “generalization” of the Fourier transform.


==Position and momentum operators==
== Position and Momentum Operators ==
An extremely useful example is the commutation relation of the position operator <math>\hat{x}</math> and momentum <math>\hat{p}</math>.   
An extremely useful example is the commutation relation of the position <math>\hat{x}</math> and momentum <math>\hat{p}</math> operators.  In the position representation, the position operator is <math>\hat{x}\rightarrow x</math> and the momentum operator is <math>\hat{p}\rightarrow \frac{\hbar}{i}\frac{\partial}{\partial x}</math>. Similarly, in the momentum representation, the momentum operator is <math>\hat{p}\rightarrow p</math> and position operator <math>\hat{x}\rightarrow i\hbar\frac{\partial}{\partial p}</math>.
In the position representation, position operator <math>\hat{x}= x</math> and momentum operator <math>\hat{p}= \frac{\hbar}{i}\frac{\partial}{\partial x}</math>. On the other hand, in the momentum representation, momentum operator <math>\hat{p}= p</math> and position operator <math>\hat{x}= i\hbar\frac{\partial}{\partial p}</math>.


Applying <math>\hat{x}</math> and <math>\hat{p}</math> to an arbitrary state ket we can see that:
Applying <math>\hat{x}</math> and <math>\hat{p}</math> to an arbitrary state vector, we can see that
<math>\left[\hat{x},\hat{p}\right]= i\hbar.</math>


The position and momentum operators are incompatible.  This provides a fundamental contrast to classical mechanics in which x and p obviously commute.  
<math>\left [\hat{x},\hat{p}\right ]= i\hbar.</math>


In three dimensions the canonical commutation relations are:
The position and momentum operators are thus incompatible.  This provides a fundamental contrast to classical mechanics in which <math>x</math> and <math>p</math> obviously commute.  Such a commutation relation holds for any coordinate and its canonically conjugate momentum; the commutation relations for all coordinates and their corresponding momenta are known as the canonical commutation relations.
<math>\left[\hat{r}_i,\hat{p}_j\right]= i\hbar\delta_{ij}</math><br/>
<math>\left[\hat{r}_i,\hat{r}_j\right]=
\left[\hat{p}_i,\hat{p}_j\right]=0,</math><br/>
where the indices stand for x,y, or z components of the 3-vectors.


It is again interesting to consider functions of these operators. (Note: <math>\hat{q}</math> and <math>\hat{p}</math> are the vector canonical position and momentum operators, respectively.)
In three dimensions, the canonical commutation relations are
For some function <math>f(\hat{q})</math>, it is easily shown that


<math>[\hat{p},f(\hat{q})] = -i\hbar\nabla f(\hat{q})</math>.
<math>\left[\hat{x}_i,\hat{p}_j\right]= i\hbar\delta_{ij},</math>


Additionally, for some <math>F(\hat{q})</math>, the expression <math>e^{\frac{i\lambda}{\hbar}\hat{p}}F(\hat{q})e^{\frac{-i\lambda}{\hbar}\hat{p}}</math> is equivalent to <math>F(\hat{q} + \lambda)</math> where <math>\lambda</math> is some c-number.
<math>\left[\hat{x}_i,\hat{x}_j\right]=0,</math>


[[Worked problem]]
and


==Connection between classical and quantum mechanics==
<math>\left[\hat{p}_i,\hat{p}_j\right]=0,</math>


There is a wonderful connection between Classical mechanics and Quantum Mechanics. The Hamiltonian is a concept in the frame of classical mechanics. In this frame, the Hamiltonian is defined as:
where the indices stand for the <math>x,</math> <math>y,</math> and <math>z</math> components of the 3-vectors.
<math>H=\sum_kp_{k}\dot{q}_{k}-L(q,\dot{q},t).</math>


There are two possibilities.  
It is again interesting to consider functions of these operators. For some function of a coordinate <math>\hat{q},</math> <math>f(\hat{q}),</math> it is easily shown that


1. If the Lagrangian <math>L(q,\dot{q},t)</math> does not depend explicitly on time the quantity H is conserved.<br/>
<math>[\hat{p},f(\hat{q})] = -i\hbar\frac{\partial}{\partial q}f(\hat{q}).</math>  
2. If the Potential and the constraints of the system are time independent, then H is conserved and H is the energy of the system.<br/>


It is clear from the above equation that:<br/>
Additionally, the expression <math>e^{i\lambda\hat{p}/\hbar}f(\hat{q})e^{-i\lambda\hat{p}/\hbar}=f(\hat{q}+\lambda);</math> i.e., the operator <math>\hat{T}(\lambda)=e^{i\lambda\hat{p}/\hbar}</math> is a translation operator that shifts the coordinate <math>\hat{q}</math> by an amount <math>\lambda.\!</math>
<math>\dot{q}_k=\frac{\partial H}{\partial p_{k}}</math><br/>
<math>\dot{p}_k=-\frac{\partial H}{\partial q_{k}}</math><br/>


This pair of the equations is called Hamilton's equations of motions. The following object
== Connection between Classical and Quantum Mechanics ==


<math>[A,B]=\sum_{k}\left(\frac{\partial A}{\partial q_{k}}\frac{\partial B}{\partial p_{k}}-
There is a wonderful connection between classical and quantum mechanics.  The Hamiltonian is a function that appears in classical mechanics and is "promoted" to an operator in quantum mechanics.  Classically, the Hamiltonian is defined as
\frac{\partial A}{\partial p_{k}}\frac{\partial B}{\partial q_{k}}\right)
 
</math>
<math>H(q,p;t)=\sum_kp_{k}\dot{q}_{k}-L(q,\dot{q};t),</math>
 
where <math>L(q,\dot{q};t)</math> is the Lagrangian.  There are two properties of this quantity.
 
1. If <math>L</math> does not depend explicitly on time, then <math>H</math> is conserved.<br/>
2. Furthermore, if the potential energy and the constraints of the system are time-independent, then <math>H</math> is not only conserved, but <math>H</math> is the energy of the system.<br/>
 
The equations of motion for the system that we may derive from the Hamiltonian are
 
<math>\dot{q}_k=\frac{\partial H}{\partial p_{k}}</math>
 
and
 
<math>\dot{p}_k=-\frac{\partial H}{\partial q_{k}}.</math>
 
These equations are called Hamilton's equations of motion.  We also define the quantity,
 
<math>[A,B]=\sum_{k}\left (\frac{\partial A}{\partial q_{k}}\frac{\partial B}{\partial p_{k}}-\frac{\partial A}{\partial p_{k}}\frac{\partial B}{\partial q_{k}}\right ),</math>


is called Poisson Bracket, and it has interesting properties. To see, let's calculate commutation relationships between coordinates and momenta.
known as the Poisson bracket.  We may observe an interesting connection between this purely classical concept and the commutators that we encounter in quantum mechanics as follows. Let us calculate this quantity with <math>A</math> and <math>B</math> different coordinates and momenta.


<math>[p_i,p_j]=\sum_{k}\left(\frac{\partial p_i}{\partial q_{k}}\frac{\partial p_j}{\partial p_{k}}-
<math>[p_i,p_j]=\sum_{k}\left(\frac{\partial p_i}{\partial q_{k}}\frac{\partial p_j}{\partial p_{k}}-
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</math>
</math>


This relations clearly shows how close are the quantum commutators with classical world. If we perform the following identification:
These relations are very similar to the canonical commutation relations discussed earlier.  In fact, if we make the replacement,
 
<math>\delta_{ij}\rightarrow i\hbar\delta_{ij},</math>
 
and replace the Poisson brackets with commutators, then we obtain the canonical commutation relations. This identification is called canonical quantization.
 
As a final important remark, there is no classical analogue to the Heisenberg uncertainty relation for conjugate classical variables in terms of, say, their Poisson brackets.  This is because, in classical mechanics, the objects that we study are point particles (or collections thereof) with well-defined positions and momenta. In quantum mechanics, on the other hand, we study the state of a system as encoded in its wave function.  This wave function describes the ''probability'' of finding a particle at a given position or with a given momentum; the particles that make up the system do not have definite positions or momenta.


<math>\delta_{ij}\rightarrow i\hbar \delta_{ij}.</math>
== Problems ==


Then we get quantum commutators. This identifications is called canonical quantization. As a final an important remark, the fact that we have classical commutators doesn't mean that we will have Heisenberg uncertainty relation for conjugate classical variables. This is because in classical mechanics the object of study are points (or body as a collection of points). In quantum mechanics we object of study is the state of a particle or system of particles - which describes the probability of finding a particle, and not it's exact, point-like, location of momentum.
'''(1)''' Let <math> f(x) \!</math> be a differentiable function. Using the fact that <math>[\hat{x},\hat{p}_{x}]=i\hbar,</math> prove the following identities:


==Hamiltonian==
'''(a)''' <math>[\hat{x},\hat{p}^{2}_{x}f(\hat{x})  ]=2i\hbar \hat{p}_{x}f(\hat{x})</math>


In Quantum Mechanics an important property is the commutation of a given operator (let's say <math>\hat{O}</math>) and the Hamiltonian <math>\hat{H}</math>.  If  <math>\hat{O}</math> commutes with <math>\hat{H}</math>, then the eigenfunctions of <math>\hat{H}</math> can always be chosen to be simultaneous eigenfunctions of <math>\hat{O}</math>.  If <math>\hat{O}</math> commutes with the Hamiltonian and does not explicitly depend on time, then  <math>\hat{O}</math> is a constant of motion.
'''(b)''' <math>[\hat{x},\hat{p}_{x}f(\hat{x})\hat{p}_{x}]=i\hbar[f(\hat{x})\hat{p}_{x}+\hat{p}_{x}f(\hat{x})]</math>


Prove:
'''(c)''' <math>[\hat{p}_{x},\hat{p}^{2}_{x}f(\hat{x})]=-i\hbar\hat{p}^{2}_{x}\frac{df(\hat{x})}{dx}</math>
If there is an operator <math>\hat{O}</math>, which <math>[\hat{H},\hat{O}]=0 </math>,


<math>\hat{H}|\psi_i\rangle=E_i|\psi_i\rangle </math>
'''(d)''' <math>[\hat{p}_{x},\hat{p}_{x}f(\hat{x})\hat{p}_{x}]=-i\hbar\hat{p}_{x}\frac{df(\hat{x})}{dx}\hat{p}_{x}</math>


<math>\hat{O}\hat{H}|\psi_i\rangle=E_i\hat{O}|\psi_i\rangle </math>
[[Commutation Problem|Solution]]


so, <math>\hat{H}\hat{O}|\psi_i\rangle=E_i\hat{O}|\psi_i\rangle </math>, eg. <math>\hat{O}|\psi_i\rangle </math> becomes another eigenfunction of energy <math>E_i</math>.


Case I: Non-degenerate energy spectrum.  
'''(2)''' (From Sakurai, ''Modern Quantum Mechanics'', Problem 1.30)


<math>\hat{O}|\psi_i\rangle=a|\psi_i\rangle </math>, eg. <math>|\psi_i\rangle </math> is an eigenstate of <math>\hat{O}</math>.
The translation operator for a finite (spatial) displacement <math>\mathbf{l}</math> is given by <math>\hat{T}(\mathbf{l})=\exp\left (-\frac{i\hat{\mathbf{p}}\cdot\mathbf{l}}{\hbar}\right ),</math> where <math>\hat{\mathbf{p}}</math> is the momentum operator.  


Case II: degenerate energy spectrum.
'''(a)''' Evaluate <math>[\hat{x}_{i},\hat{T}(\mathbf{l})].</math>


<math>|\psi_i\rangle </math> and <math> \hat{O}|\psi_i\rangle </math> have the same energy.
'''(b)''' Using your result from (a), demonstrate how the expectation value <math>\langle\hat{\mathbf{x}}\rangle</math> changes under translation.


So, there is always a linear combination of <math>|\psi_i\rangle </math>, such that the linear combination is an eigenstate of <math> \hat{O} </math>.
[[Translation operator problem|Solution]]

Latest revision as of 12:40, 12 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

Commutator

The commutator of two operators 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}\!} 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 \hat{B}\!} is defined as follows:

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},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}\,\!.}

When 2 operators 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}\!} 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 \hat{B}\!} commute, then 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},\hat{B}]=0.} On the other hand, if 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},\hat{B}]\neq 0,} then the operators do not commute, and we can think of the commutator between two operators as a measure of how badly they fail to commute. Note that any operator will commute with an ordinary complex number.

Some Identities:
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},\hat{B}]+[\hat{B},\hat{A}]=0 \!}
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},\hat{A}]= 0 \!}
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},\hat{B}+\hat{C}]=[\hat{A},\hat{B}]+[\hat{A},\hat{C}]\!}
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}+\hat{B},\hat{C}]=[\hat{A},\hat{C}]+[\hat{B},\hat{C}]\!}
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}\hat{B},\hat{C}]=\hat{A}[\hat{B},\hat{C}]+[\hat{A},\hat{C}]\hat{B}\!}
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},\hat{B}\hat{C}]=[\hat{A},\hat{B}]\hat{C}+\hat{B}[\hat{A},\hat{C}]\!}
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},[\hat{B},\hat{C}]]+[\hat{C},[\hat{A},\hat{B}]]+[\hat{B},[\hat{C},\hat{A}]]=0\!}
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},\hat{B}]=-[\hat{B},\hat{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 [\hat{A},\hat{B}]^{\dagger} = [\hat{B}^{\dagger},\hat{A}^{\dagger}]}

In addition, if any two operators are Hermitian and their product is also Hermitian, then the operators commute because

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}\hat{B})^{\dagger} = \hat{B}^{\dagger}\hat{A}^{\dagger} = \hat{B}\hat{A} }

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 (\hat{A}\hat{B})^{\dagger} = \hat{A}\hat{B}}

so 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}\hat{B} = \hat{B}\hat{A}.}

One may also prove 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}^n,\hat{B}]=n\hat{A}^{n-1}[\hat{A},\hat{B}]\!} if 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},[\hat{A},\hat{B}]]=0\!} via mathematical induction.

Compatible Observables

An operator which corresponds to some physically measurable property of a system is called an observable. The following is a list of common physical observables and their corresponding operators given in the coordinate representation:
Position: 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{r}\rightarrow\hat{\mathbf{r}} }
Momentum: 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{p}\rightarrow\frac{\hbar}{i}\nabla}
Kinetic energy: 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 T=\frac{p^2}{2m}\rightarrow -\frac{\hbar^2}{2m}\nabla^2}
Potential energy: 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(\mathbf{r})\rightarrow V(\hat{\mathbf{r}})}
Total energy: 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 E=T+V\rightarrow -\frac{\hbar^2}{2m}\nabla^2+V(\hat{\mathbf{r}})}

All observables are Hermitian.

Two observables 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}} 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 \hat{B}} are said to be compatible if it is possible to exactly measure both simultaneously; i.e., they possess a common set of eigenstates:

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_{AB}\rangle=a|\Psi_{AB}\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 \hat{B}|\Psi_{AB}\rangle=b|\Psi_{AB}\rangle\!}

It follows from this 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}\hat{B}|\Psi_{AB}\rangle=\hat{A}b|\Psi_{AB}\rangle=b\hat{A}|\Psi_{AB}\rangle=ba|\Psi_{AB}\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 \hat{B}\hat{A}|\Psi_{AB}\rangle=\hat{B}a|\Psi_{AB}\rangle=a\hat{B}|\Psi_{AB}\rangle=ba|\Psi_{AB}\rangle.\!}

Therefore,

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}\hat{B}-\hat{B}\hat{A}=[\hat{A},\hat{B}]=0.\!}

The same logic works in reverse - if two operators commute, then they have simultaneous eigenkets, and thus they are compatible observables. Let us consider two commuting operators 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}} 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 \hat{B}.} Suppose that we know the eigenstates 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 \hat{A};} let us call them 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_{A}\rangle.} We know 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}|\Psi_{A}\rangle=a|\Psi_{A}\rangle.}

We now act on both sides of this equation 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 \hat{B}:}

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{B}\hat{A}|\Psi_{A}\rangle=a\hat{B}|\Psi_{A}\rangle,}

or, because 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}} 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 \hat{B}} commute,

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}\hat{B}|\Psi_{A}\rangle=a\hat{B}|\Psi_{A}\rangle.}

We see 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{B}|\Psi_{A}\rangle} is therefore also an eigenstate 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 \hat{A}} with the same eigenvalue. This means that we have "block diagonalized" 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{B}} in the sense that it can only mix eigenstates 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 \hat{A}} with other eigenstates with the same eigenvalue. All of the eigenstates 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 \hat{B}} are thus also eigenstates of meaning that the two observables are compatible.

These observations are especially important when we consider the Hamiltonian of a system. Symmetries of the Hamiltonian are represented by operators that commute with it, and, as we will see later, identification of these symmetries often helps one diagonalize the Hamiltonian and classify its eigenstates.

If , or, equivalently, if it is not possible to simultaneously diagonalize and then the two operators are said to be incompatible observables.

Generalized Uncertainty Relation

If two observables are incompatible, then one cannot necessarily simultaneously diagonalize them, so that an eigenstate of one may be a non-trivial linear combination of eigenstates of the other. This fact is embodied in the existence of an uncertainty relation for the two observables, much like that between position and momentum. We will now derive this general uncertainty relation.

For any observable , let us define the uncertainty as the standard deviation of said observable from its expectation value:

where . Likewise, for any other observable ,

where .

Now we invoke the Schwartz inequality. Recall that this is just

Invoking this expression,

Now, for any complex number ,

.

Letting ,

.

The inner products are

and

Therefore,

and

or

This is the general uncertainty relation that we sought.

As a simple demonstration of this relation, suppose that the first observable is position, , and the second is momentum, .

The commutation relation between these two observables is just

so

We have thus obtained a formal proof of the Heisenberg Uncertainty Principle.

Question: What about the energy-time uncertainty relation?

Answer: We should note that time is not an operator in quantum mechanics, and thus we cannot apply the general uncertainty relation derived above to it. The energy-time uncertainty relation tells us that, for a short time interval, we get broadening in the energy spectrum which we observe. In other words, to get a precise energy value, we need to wait for a long time. The relation between the energy and time intervals is given by

Functions of Operators

It is often useful to consider functions of operators, as the Hamiltonian can be written as such. For example, the exponential function of an operator is represented as a power series,

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 e^\hat{A} = \hat{I} + \frac{\hat{A}}{1!} + \frac{\hat{A}^2}{2!}+ \frac{\hat{A}^3}{3!} + \ldots }

This leads to the identity that for some 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 f(\lambda) = e^{\lambda\hat{A}}\hat{B}e^{-\lambda\hat{A}}} , 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{A}} 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 \hat{B}} are operators:

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 e^{\lambda\hat{A}}\hat{B}e^{-\lambda\hat{A}} = \hat{B} + \frac{\lambda}{1!}[\hat{A},\hat{B}] + \frac{\lambda^2}{2!}[\hat{A},[\hat{A},\hat{B}]] + \frac{\lambda^3}{3!}[\hat{A},[\hat{A},[\hat{A},\hat{B}]]] + \ldots }

This follows easily when one considers a Taylor series expansion 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 f(\lambda).\!} We note 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 \frac{d}{d\lambda}f(\lambda)=e^{\lambda\hat{A}}[\hat{A},\hat{B}]e^{-\lambda\hat{A}}.}

We may then apply this relation recursively to generate the higher-order derivatives required in the Taylor expansion.

If 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},[\hat{A},\hat{B}]]=\beta\hat{B},} then we can simplify the identity 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 e^{\lambda\hat{A}}\hat{B}e^{-\lambda\hat{A}}=\hat{B}\cosh\lambda\sqrt{\beta}+\frac{[\hat{A},\hat{B}]}{\sqrt{\beta}}\sinh\lambda\sqrt{\beta}.}

It is also useful to consider the commutator of an operator function with an operator. Given operators 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{l}} 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 \hat{m}} 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{l},\hat{m}] = 1 }

it can be shown that, for arbitrary 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 f(\hat{l})} ,

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 [f(\hat{l}),\hat{m}] = \frac{d}{d\hat{l}}f(\hat{l})}

by use of the power series expansion 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 f(\hat{l})} and the commutator identities in the above section.

Physical Meaning of Various Representations

In a given Hilbert basis, it is obvious that the 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 \rangle} is completely determined by the set of its components 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_n\},\!}

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\leftrightarrow\left \{c_{n}=\langle\phi_{n}|\psi\rangle\right \}}

which we can write as a column vector; the corresponding bra is then represented by the conjugate line vector. This representation of the state vector is completely equivalent to the wave function 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(\mathbf{r},t).}

Therefore, there are not only two, but an infinite number of equivalent representations of the state of the system. What is their physical meaning?

In the basis of the eigenstates of the Hamiltonian, the interpretation of this representation is simple and crystal clear: 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_{n}\!} are the probability amplitudes for 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 E_{n}\!} in an energy measurement.

Therefore,

• The position representation, is more convenient if we are interested in the properties of the particle in space.

• Its Fourier transform, is more convenient if we are interested in its momentum properties.

• The coefficients in the basis of energy eigenstates are more convenient if we are interested in the energy of the particle.

Owing to Riesz’s theorem, this can be done with any observable, such as the angular momentum, which we examine later on and also has discrete eigenvalues. This can be thought of as a “generalization” of the Fourier transform.

Position and Momentum Operators

An extremely useful example is the commutation relation of the position and momentum operators. In the position representation, the position operator is and the momentum operator is . Similarly, in the momentum representation, the momentum operator is and position operator .

Applying and to an arbitrary state vector, we can see that

The position and momentum operators are thus incompatible. This provides a fundamental contrast to classical mechanics in which and obviously commute. Such a commutation relation holds for any coordinate and its canonically conjugate momentum; the commutation relations for all coordinates and their corresponding momenta are known as the canonical commutation relations.

In three dimensions, the canonical commutation relations are

and

where the indices stand for the and components of the 3-vectors.

It is again interesting to consider functions of these operators. For some function of a coordinate it is easily shown that

Additionally, the expression i.e., the operator is a translation operator that shifts the coordinate by an amount

Connection between Classical and Quantum Mechanics

There is a wonderful connection between classical and quantum mechanics. The Hamiltonian is a function that appears in classical mechanics and is "promoted" to an operator in quantum mechanics. Classically, the Hamiltonian is defined as

where is the Lagrangian. There are two properties of this quantity.

1. If does not depend explicitly on time, then is conserved.
2. Furthermore, if the potential energy and the constraints of the system are time-independent, then is not only conserved, but is the energy of the system.

The equations of motion for the system that we may derive from the Hamiltonian are

and

These equations are called Hamilton's equations of motion. We also define the quantity,

known as the Poisson bracket. We may observe an interesting connection between this purely classical concept and the commutators that we encounter in quantum mechanics as follows. Let us calculate this quantity with and different coordinates and momenta.

These relations are very similar to the canonical commutation relations discussed earlier. In fact, if we make the replacement,

and replace the Poisson brackets with commutators, then we obtain the canonical commutation relations. This identification is called canonical quantization.

As a final important remark, there is no classical analogue to the Heisenberg uncertainty relation for conjugate classical variables in terms of, say, their Poisson brackets. This is because, in classical mechanics, the objects that we study are point particles (or collections thereof) with well-defined positions and momenta. In quantum mechanics, on the other hand, we study the state of a system as encoded in its wave function. This wave function describes the probability of finding a particle at a given position or with a given momentum; the particles that make up the system do not have definite positions or momenta.

Problems

(1) Let be a differentiable function. Using the fact that prove the following identities:

(a)

(b)

(c)

(d)

Solution


(2) (From Sakurai, Modern Quantum Mechanics, Problem 1.30)

The translation operator for a finite (spatial) displacement is given by where is the momentum operator.

(a) Evaluate

(b) Using your result from (a), demonstrate how the expectation value changes under translation.

Solution

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