Phy5645/Transformations and Symmetry Problem: Difference between revisions

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====(a)====
'''(a)'''


<math>T|n\rangle = |n+1\rangle</math>
Given the action of <math>\hat{T}</math> on a state,


<math>\langle i|T|j\rangle = \delta_{1,j+1}.</math> So
<math>\hat{T}|n\rangle = |n+1\rangle,</math>


<math>T = \sum_{n=1}^{N} |n+1\rangle\langle n|</math> with <math>|N+1\rangle = |1\rangle</math>
we find that the matrix elements of <math>\hat{T}</math> are


<math>T^{\dagger} = \sum_{n=1}^{N} |n\rangle\langle n+1|</math>
<math>\langle i|T|j\rangle = \delta_{1,j+1}.</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 may therefore write <math>\hat{T}</math> as


<math>TT^\dagger = \sum_{n=1}^{N}|n\rangle\langle n| = \left [ I \right ]_{NXN}</math>  
<math>\hat{T} = \sum_{n=1}^{N} |n+1\rangle\langle n|,</math>


So <math>T</math> is unitary.
where <math>|N+1\rangle = |1\rangle.</math> The Hermitian adjoint is


<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>
<math>\hat{T}^{\dagger} = \sum_{n=1}^{N} |n\rangle\langle n+1|,</math>
 
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>
 
We have thus shown that <math>\hat{T}</math> is unitary.
 
Let us now find the commutator of <math>\hat{T}</math> with the Hamiltonian.
 
<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>= \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=1}^{N}[|n+1\rangle\langle n+1| + |n+2\rangle\langle n| - |n\rangle\langle n| - |n+1\rangle\langle n-1|]</math>
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<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=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|]</math>
<math>= \sum_{n=1}^{N}[|n+2\rangle\langle n| - |n+1\rangle\langle n-1| = 0 </math>
as <math>|N+1\rangle = |1\rangle</math>
 
Therefore, <math>\hat{T}</math> commutes with the Hamiltonian.


<math>= 0 </math>
'''(b)'''
So <math>T</math> commutes with the Hamiltonian.


====(b)====
<math>|\psi \rangle = \sum_{n=1}^{N} e^{ikn}|n\rangle</math>
<math>\psi \rangle = \sum_{n=1}^{N} e^{ikn}|n\rangle</math>


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


<math>= \sum_{n=1}^{N}\sum_{m=1}^{N}e^{ikn}|m+1\rangle \delta{mn}</math>
<math>= \sum_{n=1}^{N}\sum_{m=1}^{N}e^{ikn}|m+1\rangle \delta_{mn}</math>


<math>= \sum_{n=1}^{N} e^{ikn}|n+1\rangle</math>
<math>= \sum_{n=1}^{N} e^{ikn}|n+1\rangle</math>


<math>= sum_{n=2}^{N+1} e^{ik(n-1)}|n\rangle</math>
<math>= \sum_{n=2}^{N+1} e^{ik(n-1)}|n\rangle</math>


<math> e^{-ik}[\sum_{n=2}^{N} e^{ikn}|n\rangle + e^{ik(13)}|1\rangle</math>
<math> e^{-ik}[\sum_{n=2}^{N} e^{ikn}|n\rangle + e^{ik(13)}|1\rangle</math>

Revision as of 13:45, 23 July 2013

(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)

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

Failed to parse (SVG (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^{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 = \sum_{n=2}^{N+1} e^{ik(n-1)}|n\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 e^{-ik}[\sum_{n=2}^{N} e^{ikn}|n\rangle + e^{ik(13)}|1\rangle}

Now 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 |N+1\rangle \to |1\rangle} but the coefficient remains Failed to parse (SVG (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^{13ik}} We need that to be Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{ik(1)}|n\rangle}

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 e^{12ik} = e^{2\pi in}}

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 k = n\pi/6 , n = 1,2,....,11}

Hence Failed to parse (SVG (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 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 T} 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}}

Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H,T} commute, any 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 T} is 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 H}

explicit proof:

Failed to parse (SVG (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\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} \sum_{m=1}^{N} e^{ikm}|n\rangle \langle n+1|n\rangle + \sum_{n=1}^{N} \sum_{m=1}^{N} e^{ikm}|n+1\rangle \langle n|n\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\rangle}

So they are 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 H} also.

(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 F|j\rangle = |n+1-j\rangle}

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 F = \sum_{n=1}^{N} |N+1-n\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 F^\dagger = \sum_{n=1}^{N} |n\rangle\langle N+1-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 FF^\dagger = \sum_{n=1}^{N} \sum_{m=1}^{N} |N+1-m\rangle\langle m||n\rangle\langle N+1-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} \sum_{m=1}^{N}|N+1-m\rangle \langle i|j \rangle \langle N+1-j|}

Failed to parse (SVG (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}

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 F} is unitary.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle FH = \sum_{m=1}^{N} |N+1-m\rangle\langle n| \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 HF = \sum_{n=1}^{N} (|n\rangle \langle n+1| + |n+1\rangle \langle n| \sum_{m=1}^{N} |N+1-m\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+2-n\rangle \langle n| + \sum_{n=1}^{N} |N-n\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|}

Hence they commute.

(d)

It has already been proved that F is both unitary and hermitian. Thus the eigenvalues of F are given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pm 1}


Since by intuition let us construct states given by

Failed to parse (syntax error): {\displaystyle |\psi\pm\rangle = |n\rangle \pm |N+1-n\rangle n = 1,2,3,….N/2. }

Thus thus constructed are eigenstates of F. It is noted that we have symmetric eigenstates given by

Failed to parse (syntax error): {\displaystyle |1\rangle + |N\rangle ; |2\rangle + |N-1\rangle ; |3\rangle +|N-2\rangle … } and anti-symmetric eigenstates given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |1\rangle - |N\rangle ; |2\rangle - |N-1\rangle ; |3\rangle -|N-2\rangle ; … }


Thus we find that the eigenstate of F may not necessarily be eigenstates oh H. This is because F is degenerate. However there may exist linear superposition of eigenstates of F which is also an eigenstates of H and vice versa.

Back to Transformations of Operators and Symmetry

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