Transformations of Operators and Symmetry: Difference between revisions

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Symmetry of any quantum mechanical state is determined by how the state transforms under certain mathematical transformations, examples being translation and rotation. A symmetry transformation is a transformation that keeps the physical characteristics of the system unchanged (for example, a rotation of a spherical object). Of special importance are problems for which the Hamiltonian is left invariant under a symmetry transformation.
{{Quantum Mechanics A}}


In addition, in both classical and quantum mechanics, symmetry transformations become important due to their relation to conserved quantities. Moreover in quantum mechanics the importance of symmetries is further enhanced by the fact that observation of conserved quantities can be exactly predictable in spite of the probabilistic nature of quantum predictions.
==Transformations of Operators==


Let us consider an arbitrary transformation of an arbitrary state | n > to be given by the operator U such that the transformation gives | n > <math> \rightarrow </math> U | n >
In the previous section, we discussed operators as transformations of vectors.  In many cases, however, we will be interested in how operators, observables in particular, will transform under the action of another operator.  Given an operator <math>\hat{A}</math> and a transformation <math>\hat{T},</math> we define the transformed operator <math>\hat{A}'</math> as follows.  Given the relation,


If U produce a symmetry transformation, the following theorems hold.
<math>\hat{A}|\psi\rangle=|\phi\rangle,</math>


between two vectors <math>|\psi\rangle</math> and <math>|\phi\rangle</math>, the operator <math>\hat{A}'</math> is the operator giving the relation between <math>|\psi'\rangle=\hat{T}|\psi\rangle</math> and <math>|\phi'\rangle=\hat{U}|\phi\rangle;</math> i.e.,


If the operator U produces a symmetry transformation on all ket vectors, then it must commute with the hamiltonian.
<math>\hat{A}'|\psi'\rangle=|\phi'\rangle.</math>


Proof: By definition of a symmetry transformation, the operator U could transform an energy eigenstate either to itself or another eigenstate degenerate to it. Hence, if | E_i > is an eigenstate of H with eigenvalue Ei then
To find <math>\hat{A}',</math> let us first act on both sides of the original relation with <math>\hat{T}:</math>


<math>HU | E_{i} > = HU  | E^{'}_{i}> = E_{i} | E^{'}_{i}> = E_i U | E_{i}> = UE_{i} | E_{i}> = UH | E_{i}></math>
<math>\hat{T}\hat{A}|\psi\rangle=\hat{T}|\phi\rangle</math>


We now introduce the identity between <math>\hat{A}</math> and <math>|\psi\rangle</math> in the form, <math>\hat{T}^{-1}\hat{T}:</math>


<math> | E_{i}> </math> and <math> | E^{'}_{i}></math>
<math>\hat{T}\hat{A}\hat{T}^{-1}\hat{T}|\psi\rangle=\hat{T}|\phi\rangle</math>


Using the above definitions of <math>|\psi'\rangle</math> and <math>|\phi'\rangle,</math> we may write this as


Therefore we can write,
<math>\hat{T}\hat{A}\hat{T}^{-1}|\psi'\rangle=|\phi'\rangle</math>


<math>\left [ H,U \right ]|E_{i}> = 0</math>
We see then that the transformed operator <math>\hat{A}'=\hat{T}\hat{A}\hat{T}^{-1}.</math>  In matrix form, this would simply correspond to a similarity transformation of <math>\hat{A}.</math>


This is valid for all energy eigenstates <math> | E_{i}> .</math>
Of particular importance is the case in which <math>\hat{T}</math> is unitary and <math>\hat{A}</math> is an observable. This is because, in addition to preserving the normalization of the state vectors, as mentioned in the previous section, unitary transformations also preserve the Hermitian nature of <math>\hat{A}:</math>


Now From the completeness theorem any arbitrary state <math> | n></math>
<math>\hat{A}'^{\dagger}=(\hat{T}\hat{A}\hat{T}^\dagger)^\dagger=\hat{T}\hat{A}^\dagger\hat{T}^\dagger=\hat{T}\hat{A}\hat{T}^\dagger=\hat{A}'</math>


can be written as a linear combination of the eigenstates <math>| E_{i}></math> .
==Symmetry and its Role in Quantum Mechanics==


Hence, we can write,
Having discussed the transformation of operators, we will now apply our results to discuss symmetries of the Hamiltonian, a very important topic.  As alluded to in the previous section, identifying the symmetries of the Hamiltonian will allow us to greatly simplify the problem at hand.  In addition, in both classical and quantum mechanics, symmetry transformations become important due to their relation to conserved quantities via Noether's Theorem.  In quantum mechanics, the importance of symmetries is further enhanced by the fact that measurements of conserved quantities can be exact in spite of the probabilistic nature of quantum predictions.


<math>\left [ H,U \right ]|n> = 0</math>
Given a unitary transformation <math>\hat{U},</math> we say that it is a symmetry of the Hamiltonian if it leaves the Hamiltonian invariant; i.e., if <math>\hat{H}'=\hat{U}\hat{H}\hat{U}^\dagger=\hat{H}.</math> We will now show that, if a transformation is a symmetry of the Hamiltonian, then it commutes with the Hamiltonian.  To see this, let us take the relation,


Since <math>| n></math>
<math>\hat{H}|\psi\rangle=|\phi\rangle,</math>


is an arbitrary ket vector, we can conclude that
and act on both sides with <math>\hat{U}:</math>


<math>\left [ H,U \right ] = 0 </math>
<math>\hat{U}\hat{H}|\psi\rangle=\hat{U}|\phi\rangle</math>


==Problem on symmetry[http://wiki.physics.fsu.edu/wiki/index.php/Phy5645/symmetryprob1]==
Now, if <math>\hat{U}</math> is a symmetry of the Hamiltonian, then it must also be true that


==Commutators & symmetry ==
<math>\hat{H}\hat{U}|\psi\rangle=\hat{U}|\phi\rangle.</math>


We can define an operator called the parity operator, <math>\hat{P}</math> which does the following:
Subtracting these two equations, we see that, because <math>|\psi\rangle</math> is arbitrary, the Hamiltonian commutes with the transformation operator; i.e., <math>[\hat{H},\hat{U}]=0.</math>
 
This is a very important result; we know that, if two operators commute, then it is possible to simultaneously diagonalize them.  This implies that every symmetry of the Hamiltonian has a "good quantum number" associated with it that we may use to describe the eigenstates of the Hamiltonian.
 
To help illustrate this fact, let us consider the parity, or inversion, operator, <math>\hat{P}:</math>


<math>\hat{P}f(x)=f(-x).</math>
<math>\hat{P}f(x)=f(-x).</math>


The parity operator commutes with the Hamiltonian <math>\hat{H}</math> if the potential is symmetric, <math>\hat{V}(r)=\hat{V}(-r)</math>.  Since the two commute, the eigenfunctions of the Hamiltonian can be chosen to be eigenfunctions of the parity operator.  This means that if the potential is symmetric, the solutions can be chosen to have definite parity (even and odd functions).
The parity operator commutes with the Hamiltonian <math>\hat{H}</math> if the potential is symmetric; i.e., <math>\hat{V}(x)=\hat{V}(-x)</math>.  Since the two commute, the eigenfunctions of the Hamiltonian can be chosen to be eigenfunctions of the parity operator.  This means that, if the potential is symmetric, then the eigenstates of the Hamiltonian can be chosen to have definite parity (even or odd).
 
==Problem==
 
(From a Quantum Mechanics assignment in the Department of Physics, UF)
 
Consider an <math>N</math> state system, with <math>N</math> even and the states labeled as <math>|1\rangle, |2\rangle, \ldots, |N\rangle,</math> described by the Hamiltonian,
 
<math>\hat{H}=\sum_{n=1}^{N} (|n\rangle \langle n+1| + |n+1\rangle \langle n|).</math>
 
Notice that the Hamiltonian, in this form, is manifestly Hermitian.  Assume periodic boundary conditions; i.e, <math>|N+1\rangle = |1\rangle.</math>  One may therefore think of this Hamiltonian as describing a particle on a circle.
 
'''(a)''' Define the translation operator, <math>\hat{T},</math>  as taking <math>|1\rangle \to |2\rangle, |2\rangle \to |3\rangle ,...,|N\rangle \to |1\rangle.</math>  Write <math>\hat{T}</math> in a form like <math>\hat{H}</math> in the first equation and show that <math>\hat{T}</math> is both unitary and commutes with <math>\hat{H},</math> thus showing that <math>\hat{T}</math> is a symmetry of the Hamiltonian.
 
'''(b)''' Find the eigenstates of <math>\hat{T}</math> by using wavefunctions of the form,
 
<math>|\psi(k)\rangle = \sum_{n=1}^{N} e^{ikn}|n\rangle.</math>
 
What are the eigenvalues associated with these eigenstates?  Do all these eigenstates have to be eigenstates of <math>\hat{H}</math> as well?  If not, do any of these eigenstates have to be eigenstates of <math>\hat{H}?</math>  Explain your answer.
 
'''(c)''' Next, consider the operator <math>\hat{F}</math> which takes <math>|n\rangle \to |N+1-n\rangle.</math>  Write <math>\hat{F}</math> in a form like <math>\hat{H}</math> in the first equation and show that <math>\hat{F}</math> is both unitary and commutes with <math>\hat{H},</math> thus showing that <math>\hat{F}</math> is also a symmetry of the Hamiltonian.
 
'''(d)''' Find a complete set of eigenstates of <math>\hat{F}</math> and their associated eigenvalues. Do all these eigenstates have to be eigenstates of <math>\hat{H}</math> as well?  If not, do any of these eigenstates have to be eigenstates of <math>\hat{H}?</math> Explain your answer.
 
[[Phy5645/Transformations and Symmetry Problem|Solution]]

Latest revision as of 13:27, 18 January 2014

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

Transformations of Operators

In the previous section, we discussed operators as transformations of vectors. In many cases, however, we will be interested in how operators, observables in particular, will transform under the action of another operator. Given an 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}} and a transformation 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{T},} we define the transformed 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}'} as follows. Given 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 \hat{A}|\psi\rangle=|\phi\rangle,}

between two 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} , 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}'} is the operator giving the relation between 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=\hat{T}|\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=\hat{U}|\phi\rangle;} 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 \hat{A}'|\psi'\rangle=|\phi'\rangle.}

To find 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 first act on both sides of the original relation 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{T}:}

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{T}\hat{A}|\psi\rangle=\hat{T}|\phi\rangle}

We now introduce the identity between 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 |\psi\rangle} in the form, 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{T}^{-1}\hat{T}:}

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{T}\hat{A}\hat{T}^{-1}\hat{T}|\psi\rangle=\hat{T}|\phi\rangle}

Using the above definitions 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} 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,} we may write this 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{T}\hat{A}\hat{T}^{-1}|\psi'\rangle=|\phi'\rangle}

We see then that the transformed 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}'=\hat{T}\hat{A}\hat{T}^{-1}.} In matrix form, this would simply correspond to a similarity transformation 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}.}

Of particular importance is the case in which 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{T}} is unitary 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}} is an observable. This is because, in addition to preserving the normalization of the state vectors, as mentioned in the previous section, unitary transformations also preserve the Hermitian nature 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}:}

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}'^{\dagger}=(\hat{T}\hat{A}\hat{T}^\dagger)^\dagger=\hat{T}\hat{A}^\dagger\hat{T}^\dagger=\hat{T}\hat{A}\hat{T}^\dagger=\hat{A}'}

Symmetry and its Role in Quantum Mechanics

Having discussed the transformation of operators, we will now apply our results to discuss symmetries of the Hamiltonian, a very important topic. As alluded to in the previous section, identifying the symmetries of the Hamiltonian will allow us to greatly simplify the problem at hand. In addition, in both classical and quantum mechanics, symmetry transformations become important due to their relation to conserved quantities via Noether's Theorem. In quantum mechanics, the importance of symmetries is further enhanced by the fact that measurements of conserved quantities can be exact in spite of the probabilistic nature of quantum predictions.

Given a unitary transformation 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{U},} we say that it is a symmetry of the Hamiltonian if it leaves the Hamiltonian invariant; i.e., 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{H}'=\hat{U}\hat{H}\hat{U}^\dagger=\hat{H}.} We will now show that, if a transformation is a symmetry of the Hamiltonian, then it commutes with the Hamiltonian. To see this, let us take 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 \hat{H}|\psi\rangle=|\phi\rangle,}

and act on both sides 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{U}:}

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{U}\hat{H}|\psi\rangle=\hat{U}|\phi\rangle}

Now, 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{U}} is a symmetry of the Hamiltonian, then it must also be true 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{H}\hat{U}|\psi\rangle=\hat{U}|\phi\rangle.}

Subtracting these two equations, we see that, 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 |\psi\rangle} is arbitrary, the Hamiltonian commutes with the transformation operator; 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 [\hat{H},\hat{U}]=0.}

This is a very important result; we know that, if two operators commute, then it is possible to simultaneously diagonalize them. This implies that every symmetry of the Hamiltonian has a "good quantum number" associated with it that we may use to describe the eigenstates of the Hamiltonian.

To help illustrate this fact, let us consider the parity, or inversion, 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{P}:}

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{P}f(x)=f(-x).}

The parity operator commutes with the 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 \hat{H}} if the potential is symmetric; 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 \hat{V}(x)=\hat{V}(-x)} . Since the two commute, the eigenfunctions of the Hamiltonian can be chosen to be eigenfunctions of the parity operator. This means that, if the potential is symmetric, then the eigenstates of the Hamiltonian can be chosen to have definite parity (even or odd).

Problem

(From a Quantum Mechanics assignment in the Department of Physics, UF)

Consider an 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} state system, 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 N} even and the states labeled 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 |1\rangle, |2\rangle, \ldots, |N\rangle,} described by the 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 \hat{H}=\sum_{n=1}^{N} (|n\rangle \langle n+1| + |n+1\rangle \langle n|).}

Notice that the Hamiltonian, in this form, is manifestly Hermitian. Assume periodic boundary conditions; 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 |N+1\rangle = |1\rangle.} One may therefore think of this Hamiltonian as describing a particle on a circle.

(a) Define the translation 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{T},} as taking 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 \to |2\rangle, |2\rangle \to |3\rangle ,...,|N\rangle \to |1\rangle.} Write 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{T}} in a form like 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{H}} in the first equation and show that is both unitary and commutes with thus showing that is a symmetry of the Hamiltonian.

(b) Find the eigenstates of by using wavefunctions of the form,

What are the eigenvalues associated with these eigenstates? Do all these eigenstates have to be eigenstates of as well? If not, do any of these eigenstates have to be eigenstates of Explain your answer.

(c) Next, consider the operator which takes Write in a form like in the first equation and show that is both unitary and commutes with thus showing that is also a symmetry of the Hamiltonian.

(d) Find a complete set of eigenstates of and their associated eigenvalues. Do all these eigenstates have to be eigenstates of as well? If not, do any of these eigenstates have to be eigenstates of Explain your answer.

Solution

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