Phy5645/Transformations and Symmetry Problem: Difference between revisions

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===Symmetries(Problem taken from a quantum assignment in the Department of Physics, UF)===
'''(a)''' Given the action of <math>\hat{T}</math> on a state,
====Problem====
Consider an <math>N</math> state system with the states labeled as <math>|1\rangle , |1\rangle , ..., |N\rangle</math>. Let the hamiltonian for this system be


<math>\sum_{n=1}^{N} (|n\rangle \langle n+1| + |n+1\rangle \langle n|</math>  
<math>\hat{T}|n\rangle = |n+1\rangle,</math>


Notice that the Hamiltonian, in this form, is manifestly hermitian. Use periodic boundary condition, i.e, <math>|N+1\rangle = |1\rangle</math>. You can think of these states as being placed around a circle.
we find that the matrix elements of <math>\hat{T}</math> are


(a) Define the translation operator, <math>T</math> as taking <math>|1\rangle \to |2\rangle, |2\rangle \to |3\rangle ,...,|N\rangle \to |1\rangle</math> .Write T in
<math>\langle i|T|j\rangle = \delta_{1,j+1}.</math>
a form like <math>H</math> in the first equation and show that <math>T</math> is both unitary and commutes with <math>H</math> . It is thus a symmetry of the hamiltonian.


(b) Find the eigenstates of T by using wavefunctions of the form
We may therefore write <math>\hat{T}</math> as


<math>|\psi \rangle = \sum_{n=1}^{N} e^{ikn}|n\rangle</math>
<math>\hat{T} = \sum_{n=1}^{N} |n+1\rangle\langle n|,</math>


What are the eigenvalues of these eigenstates? Do all these eigenstates have to be eigenstates of <math>H</math> as well? If not, do any of these eigenstates have to be eigenstates of <math>H</math>? Explain your answer.
where <math>|N+1\rangle = |1\rangle.</math> The Hermitian adjoint is


(c) Next Consider <math>F</math> which takes <math>|n\rangle \to |N+1-n\rangle.</math> Write F in a form like <math>H</math> in the first equation and show that <math>F</math> both is unitary and commutes with <math>H</math>. It is thus a symmetry of the hamiltonian.
<math>\hat{T}^{\dagger} = \sum_{n=1}^{N} |n\rangle\langle n+1|,</math>


(d) Find a complete set of eigenstates of <math>F</math> and their eigenvalues. Do all these eigenstates have to be eigenstates of <math>H</math> as well? If not, do any of these eigenstates have to be eigenstates of <math>H</math>? Explain your answer.
so


<math>\hat{T}\hat{T}^\dagger = \sum_{m=1}^{N}\sum_{n=1}^{N}(|m+1\rangle\langle m|)(|n\rangle\langle n+1|) = \sum_{m=1}^{N}\sum_{n=1}^{N}\delta_{m,n}|m+1\rangle\langle n+1|= \sum_{n=1}^{N}|n\rangle\langle n| = \hat{I}.</math>


====Solution====
We have thus shown that <math>\hat{T}</math> is unitary.


<math>T|n\rangle = |n+1\rangle</math>
Let us now find the commutator of <math>\hat{T}</math> with the Hamiltonian.


<math>\langle i|T|j\rangle = \delta_{1,j+1}.</math> So
<math>[\hat{T},\hat{H}] = \sum_{m=1}^{N}\sum_{n=1}^{N}[(|m+1\rangle\langle m|)(|n\rangle \langle n+1| + |n+1\rangle \langle n|) - (|n\rangle \langle n+1| + |n+1\rangle \langle n|)(|m+1\rangle\langle m|)]</math>


<math>T = \sum_{n=1}^{N} |n+1\rangle\langle n|</math> with <math>|N+1\rangle = |1\rangle</math>
<math>= \sum_{n=1}^{N}[|n+1\rangle\langle n+1| + |n+2\rangle\langle n| - |n\rangle\langle n| - |n+1\rangle\langle n-1|]</math>
 
<math>= \sum_{n=2}^{N+1}|n\rangle\langle n| + \sum_{n=2}^{N+1}|n+1\rangle\langle n-1| - \sum_{n=1}^{N}|n\rangle\langle n| - \sum_{n=1}^{N}|n+1\rangle\langle n-1|</math>
 
<math>= \sum_{n=1}^{N}[|n+2\rangle\langle n| - |n+1\rangle\langle n-1| = 0 </math>
 
Therefore, <math>\hat{T}</math> commutes with the Hamiltonian.
 
'''(b)''' Let us act on the state <math>|\psi(k)\rangle</math> with <math>\hat{T}:</math>
 
<math>\hat{T}|\psi(k)\rangle = \sum_{n=1}^{N} e^{ikn}|n+1\rangle=\sum_{n=2}^{N+1}e^{ik(n-1)}|n\rangle=e^{-ik}\left [\sum_{n=2}^{N} e^{ikn}|n\rangle + e^{ik(N+1)}|1\rangle\right ]</math>
 
The expression in the brackets is almost <math>|\psi\rangle,</math> except that the state <math>|1\rangle</math> in general has the "wrong" coefficient.  We may obtain the correct coefficient, and thus an eigenstate of <math>\hat{T},</math> if
 
<math>e^{ikN} = 1.\!</math>
 
This is satisfied if
 
<math>k=\frac{2\pi n}{N},</math>
 
where <math>0\leq n<N</math> is an integer.  This range of values gives all of the unique eigenstates of <math>\hat{T},</math> each with eigenvalue <math>e^{-ik}.\!</math>
 
We may show that these are also eigenstates of the Hamiltonian, as follows:
 
<math>\hat{H}|\psi(k)\rangle=\sum_{n=1}^{N} (|n\rangle \langle n+1| + |n+1\rangle \langle n|) \sum_{m=1}^{N} e^{ikm}|m\rangle</math>
 
<math>=\sum_{n=1}^{N} e^{ik(n+1)}|n\rangle + \sum_{n=1}^{N} e^{ikn}|n+1\rangle </math>


<math>T^{\dagger} = \sum_{n=1}^{N} |n\rangle\langle n+1|</math>
<math>= 2\cos{k}|\psi(k)\rangle</math>
So


<math>TT^\dagger = \sum_{m=1}^{N}\sum_{n=1}^{N}(|m+1\rangle\langle m|)(|n\rangle\langle n+1|) = \sum_{m=1}^{N}\sum_{n=1}^{N}\delta_{m,n}|m+1\rangle\langle n+1|</math>
We therefore find that the states <math>|\psi(k)\rangle</math> are also eigenstates of <math>\hat{H}</math> with eigenvalue <math>2\cos{k}.\!</math>  Note that all of these states, except for <math>k=0\!</math> and <math>k=\pi,\!</math> are two-fold degenerate - if we replace <math>k\!</math> with <math>2\pi-k,\!</math> then we obtain a state with the same eigenvalue.


<math>TT^\dagger = \sum_{n=1}^{N}|n\rangle\langle n| = \left [ I \right ]_{NXN}</math>  
'''(c)''' Similarly to the previous case, we may conclude from the fact that the effect of <math>\hat{F}</math> is


So <math>T</math> is unitary.
<math>\hat{F}|j\rangle = |N+1-j\rangle</math>


<math>\left [ T,H \right ] = \sum_{m=1}^{N}\sum_{n=1}^{N}[(|m+1\rangle\langle m|)(|n\rangle \langle n+1| + |n+1\rangle \langle n|) - (|n\rangle \langle n+1| + |n+1\rangle \langle n|)(|m+1\rangle\langle m|)]</math>
that the operator may be written as


<math>= \sum_{n=1}^{N}[|n+1\rangle\langle n+1| + |n+2\rangle\langle n| - |n\rangle\langle n| - |n+1\rangle\langle n-1|]</math>
<math>\hat{F} = \sum_{n=1}^{N} |N+1-n\rangle\langle n|.</math>
 
The Hermitian adjoint is
 
<math>\hat{F}^{\dagger} = \sum_{n=1}^{N} |n\rangle\langle N+1-n|=\hat{F}.</math>
 
We may see that the last equality is true by replacing <math>n\!</math> with <math>N+1-n,\!</math> and thus that <math>\hat{F}</math> is Hermitian.  We may see that it is also unitary as follows:
 
<math>\hat{F}\hat{F}^{\dagger} = \sum_{i=1}^{N} \sum_{j=1}^{N}|N+1-i\rangle \langle i|j \rangle \langle N+1-j|=\hat{I}.</math>
 
These two facts imply that <math>\hat{F}^2=\hat{I}.</math>
 
Let us now determine if it commutes with the Hamiltonian.
 
<math>\hat{F}\hat{H} = \sum_{m=1}^{N} |N+1-m\rangle\langle m| \sum_{n=1}^{N} (|n\rangle \langle n+1| + |n+1\rangle \langle n|)</math>
 
<math>= \sum_{n=1}^{N} |N+1-n\rangle \langle n-1| + \sum_{n=1}^{N} |N+1-n\rangle \langle n+1|</math>
 
<math>\hat{H}\hat{F} = \sum_{n=1}^{N} (|n\rangle \langle n+1| + |n+1\rangle \langle n|) \sum_{m=1}^{N} |N+1-m\rangle\langle m|</math>
 
<math>= \sum_{n=1}^{N} |N+2-n\rangle \langle n| + \sum_{n=1}^{N} |N-n\rangle \langle n|</math>
 
<math>= \sum_{n=1}^{N} |N+1-n\rangle \langle n-1| + \sum_{n=1}^{N} |N+1-n\rangle \langle n+1|</math>
 
The expressions for <math>\hat{F}\hat{H}</math> and <math>\hat{H}\hat{F}</math> are identical, and thus they commute.
 
'''(d)''' We saw from the previous part that <math>\hat{F}^2=\hat{I}.</math>.  The eigenvalues of <math>\hat{F}</math> are thus <math>\pm 1.</math>
 
Since <math> F|n\rangle = |N+1-n\rangle, </math> we may construct eigenstates by intuition; let us try
 
<math>|\psi_\pm\rangle = |n\rangle \pm |N+1-n\rangle,</math>
 
where <math>n=1,2,3,\ldots,\tfrac{1}{2}N.</math>  We now act on these states with <math>\hat{F}:</math>
 
<math> F|\psi_\pm\rangle = |N+1-n\rangle \pm |N-N+n\rangle </math>
 
<math> = \pm( |n\rangle \pm |N+1-n\rangle ) </math>
 
<math> = \pm|\psi_\pm\rangle </math>
 
Therefore, the states <math> |\psi_\pm\rangle </math> are in fact eigenstates of <math>\hat{F}</math>. It is noted that we have <math>\tfrac{1}{2}N</math> symmetric eigenstates given by
 
<math>|1\rangle + |N\rangle, |2\rangle + |N-1\rangle, |3\rangle +|N-2\rangle, \ldots</math>
 
and <math>\tfrac{1}{2}N</math> anti-symmetric eigenstates given by
 
<math>|1\rangle - |N\rangle, |2\rangle - |N-1\rangle, |3\rangle -|N-2\rangle, \ldots</math>
 
<math>\hat{H}|\psi_\pm\rangle=\sum_{m=1}^{N} (|m+1\rangle \langle m| + |m\rangle + \langle m+1|)(|n\rangle \pm |N+1-n\rangle) </math>
 
<math> =\sum_{m=1}^{N}(|m+1\rangle \langle m|n\rangle + |m\rangle \langle m+1|n\rangle \pm |m+1\rangle \langle m|N+1-n\rangle \pm |m\rangle \langle m+1|N+1-n\rangle) </math>
 
<math> = |n+1\rangle + |n-1\rangle \pm |N+2-n\rangle \pm |N-n\rangle </math>


<math>= \sum_{n=2}^{N+1}|n\rangle\langle n| + \sum_{n=2}^{N+1}|n+1\rangle\langle n-1| - \sum_{n=1}^{N}|n\rangle\langle n| - \sum_{n=1}^{N}|n+1\rangle\langle n-1|</math>
Thus we find that a given eigenstate of <math>\hat{F}</math> may not necessarily be an eigenstate of <math>\hat{H}.</math>  This is because the eigenstates of <math>\hat{F}</math> are degenerate.  There are, however, linear superpositions of eigenstates of <math>\hat{F}</math> that are also eigenstates of <math>\hat{H}.</math>  For example, <math>|\psi(0)\rangle</math> is an eigenstate of <math>\hat{F}</math> with eigenvalue <math>+1.\!</math>  In addition, one may easily verify that <math>\hat{F}|\psi(k)\rangle=e^{ik}|\psi(2\pi-k)\rangle.</math>  Since both the "input" and "output" states are degenerate, we see that <math>|\psi(k)\rangle\pm e^{ik}|\psi(2\pi-k)\rangle</math> is an eigenstate of both <math>\hat{H}</math> and of <math>\hat{F},</math> with an eigenvalue of <math>\pm 1</math> with respect to the latter.


<math>= \sum_{n=1}^{N}[|n+2\rangle\langle n| - |n+1\rangle\langle n-1|]</math>
Back to [[Transformations of Operators and Symmetry#Problem|Transformations of Operators and Symmetry]]
as <math>|N+1\rangle = |1\rangle</math>

Latest revision as of 13:26, 18 January 2014

(a) Given the action 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{T}} on 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 \hat{T}|n\rangle = |n+1\rangle,}

we find that the matrix elements 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{T}} are

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|T|j\rangle = \delta_{1,j+1}.}

We may therefore 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}} 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} = \sum_{n=1}^{N} |n+1\rangle\langle n|,}

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 |N+1\rangle = |1\rangle.} The Hermitian adjoint is

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}^{\dagger} = \sum_{n=1}^{N} |n\rangle\langle n+1|,}

so

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{T}^\dagger = \sum_{m=1}^{N}\sum_{n=1}^{N}(|m+1\rangle\langle m|)(|n\rangle\langle n+1|) = \sum_{m=1}^{N}\sum_{n=1}^{N}\delta_{m,n}|m+1\rangle\langle n+1|= \sum_{n=1}^{N}|n\rangle\langle n| = \hat{I}.}

We have thus shown 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{T}} is unitary.

Let us now find the commutator 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{T}} 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{T},\hat{H}] = \sum_{m=1}^{N}\sum_{n=1}^{N}[(|m+1\rangle\langle m|)(|n\rangle \langle n+1| + |n+1\rangle \langle n|) - (|n\rangle \langle n+1| + |n+1\rangle \langle n|)(|m+1\rangle\langle m|)]}

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_{n=1}^{N}[|n+1\rangle\langle n+1| + |n+2\rangle\langle n| - |n\rangle\langle n| - |n+1\rangle\langle n-1|]}

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_{n=2}^{N+1}|n\rangle\langle n| + \sum_{n=2}^{N+1}|n+1\rangle\langle n-1| - \sum_{n=1}^{N}|n\rangle\langle n| - \sum_{n=1}^{N}|n+1\rangle\langle n-1|}

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_{n=1}^{N}[|n+2\rangle\langle n| - |n+1\rangle\langle n-1| = 0 }

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{T}} commutes with the Hamiltonian.

(b) Let us act on the 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(k)\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 \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}|\psi(k)\rangle = \sum_{n=1}^{N} e^{ikn}|n+1\rangle=\sum_{n=2}^{N+1}e^{ik(n-1)}|n\rangle=e^{-ik}\left [\sum_{n=2}^{N} e^{ikn}|n\rangle + e^{ik(N+1)}|1\rangle\right ]}

The expression in the brackets is almost 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,} except that the 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 |1\rangle} in general has the "wrong" coefficient. We may obtain the correct coefficient, and thus 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{T},} 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 e^{ikN} = 1.\!}

This is satisfied 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 k=\frac{2\pi n}{N},}

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 0\leq n<N} is an integer. This range of values gives all of the unique 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{T},} each with eigenvalue 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^{-ik}.\!}

We may show that these are also eigenstates of the Hamiltonian, 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{H}|\psi(k)\rangle=\sum_{n=1}^{N} (|n\rangle \langle n+1| + |n+1\rangle \langle n|) \sum_{m=1}^{N} e^{ikm}|m\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 =\sum_{n=1}^{N} e^{ik(n+1)}|n\rangle + \sum_{n=1}^{N} e^{ikn}|n+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\cos{k}|\psi(k)\rangle}

We therefore find that the 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(k)\rangle} are also 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{H}} with eigenvalue 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\cos{k}.\!} Note that all of these states, except 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 k=0\!} 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 k=\pi,\!} are two-fold degenerate - if we replace 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 k\!} 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 2\pi-k,\!} then we obtain a state with the same eigenvalue.

(c) Similarly to the previous case, we may conclude from the fact that the effect 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{F}} is

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{F}|j\rangle = |N+1-j\rangle}

that the operator may be written 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{F} = \sum_{n=1}^{N} |N+1-n\rangle\langle n|.}

The Hermitian adjoint is

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{F}^{\dagger} = \sum_{n=1}^{N} |n\rangle\langle N+1-n|=\hat{F}.}

We may see that the last equality is true by replacing 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\!} 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+1-n,\!} and thus 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{F}} is Hermitian. We may see that it is also unitary 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{F}\hat{F}^{\dagger} = \sum_{i=1}^{N} \sum_{j=1}^{N}|N+1-i\rangle \langle i|j \rangle \langle N+1-j|=\hat{I}.}

These two facts imply 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{F}^2=\hat{I}.}

Let us now determine if it 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{F}\hat{H} = \sum_{m=1}^{N} |N+1-m\rangle\langle m| \sum_{n=1}^{N} (|n\rangle \langle n+1| + |n+1\rangle \langle 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 = \sum_{n=1}^{N} |N+1-n\rangle \langle n-1| + \sum_{n=1}^{N} |N+1-n\rangle \langle n+1|}

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{F} = \sum_{n=1}^{N} (|n\rangle \langle n+1| + |n+1\rangle \langle n|) \sum_{m=1}^{N} |N+1-m\rangle\langle m|}

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_{n=1}^{N} |N+2-n\rangle \langle n| + \sum_{n=1}^{N} |N-n\rangle \langle n|}

The expressions for and are identical, and thus they commute.

(d) We saw from the previous part that . The eigenvalues of are thus

Since we may construct eigenstates by intuition; let us try

where We now act on these states with

Therefore, the states are in fact eigenstates of . It is noted that we have symmetric eigenstates given by

and anti-symmetric eigenstates given by

Thus we find that a given eigenstate of may not necessarily be an eigenstate of This is because the eigenstates of are degenerate. There are, however, linear superpositions of eigenstates of that are also eigenstates of For example, is an eigenstate of with eigenvalue In addition, one may easily verify that Since both the "input" and "output" states are degenerate, we see that is an eigenstate of both and of with an eigenvalue of with respect to the latter.

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