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)=== | ===Symmetries(Problem taken from a quantum assignment in the Department of Physics, UF)=== | ||
====Problem==== | ====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 | Consider an <math>N</math> state system with the states labeled as <math>|1\rangle , |1\rangle , ..., |N\rangle</math>. Consider <math>N</math> to be even. Let the hamiltonian for this system be | ||
<math>\sum_{n=1}^{N} (|n\rangle \langle n+1| + |n+1\rangle \langle n|</math> | <math>\sum_{n=1}^{N} (|n\rangle \langle n+1| + |n+1\rangle \langle n|</math> | ||
| Line 22: | Line 22: | ||
====Solution==== | ====Solution==== | ||
====(a)==== | |||
<math>T|n\rangle = |n+1\rangle</math> | <math>T|n\rangle = |n+1\rangle</math> | ||
| Line 46: | Line 48: | ||
<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|]</math> | ||
as <math>|N+1\rangle = |1\rangle</math> | as <math>|N+1\rangle = |1\rangle</math> | ||
<math>= 0 </math> | |||
So <math>T</math> commutes with the Hamiltonian. | |||
====(b)==== | |||
<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>= \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=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> | |||
Now the state <math>|N+1\rangle \to |1\rangle</math> | |||
but the coefficient remains <math>e^{13ik}</math> | |||
We need that to be <math>e^{ik(1)}|n\rangle</math> | |||
So <math>e^{12ik} = e^{2\pi in}</math> | |||
So <math>k = n\pi/6 , n = 1,2,....,11</math> | |||
Hence <math>\psi\rangle</math> is an eigenstate of <math>T</math> with eigenvalue <math>e^{-ik}</math> | |||
Since <math>H,T</math> commute, any eigenstate of <math>T</math> is an eigenstate of <math>H</math> | |||
explicit proof: | |||
<math>\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} \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</math> | |||
<math>= 2\cos(k)|\psi\rangle</math> | |||
So they are eigenstates of <math>H</math> also. | |||
===(c)=== | |||
<math>F|j\rangle = |n+1-j\rangle</math> | |||
So <math>F = \sum_{n=1}^{N} |N+1-n\rangle\langle n|</math> | |||
<math>F^\dagger = \sum_{n=1}^{N} |n\rangle\langle N+1-n|</math> | |||
<math>FF^\dagger = \sum_{n=1}^{N} \sum_{m=1}^{N} |N+1-m\rangle\langle m||n\rangle\langle N+1-n|</math> | |||
<math>\sum_{n=1}^{N} \sum_{m=1}^{N}|N+1-m\rangle \langle i|j \rangle \langle N+1-j|</math> | |||
<math>= 1</math> | |||
So <math>F</math> is unitary. | |||
<math>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|</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>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|</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> | |||
Hence they commute. | |||
===(d)=== | |||
It has already been proved that F is both unitary and hermitian. Thus the eigenvalues of F are given by <math> \pm 1 <\math> | |||
Since <math> F|n\rangle = |N+1-n\rangle <\math> by intuition let us construct states given by | |||
<math> |\psi\pm\rangle = |n\rangle \pm |N+1-n\rangle n = 1,2,3,….N/2. <\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> | |||
Thus <math> |\psi\pm\rangle <\math> thus constructed are eigenstates of F. It is noted that we have <math> N/2 <\math> symmetric eigenstates given by | |||
<math> |1\rangle + |N\rangle ; |2\rangle + |N-1\rangle ; |3\rangle +|N-2\rangle … <\math> and <math> N/2 <\math> anti-symmetric eigenstates given by | |||
<math> |1\rangle - |N\rangle ; |2\rangle - |N-1\rangle ; |3\rangle -|N-2\rangle ; … <\math> | |||
<math> H (|n\rangle \pm |N+1-n\rangle) <\math> | |||
<math> =\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> | |||
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. | |||
Revision as of 12:13, 10 December 2009
Symmetries(Problem taken from a quantum assignment in the Department of Physics, UF)
Problem
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 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 , |1\rangle , ..., |N\rangle} . Consider Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N} to be even. Let the hamiltonian for this system 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 \sum_{n=1}^{N} (|n\rangle \langle n+1| + |n+1\rangle \langle n|}
Notice that the Hamiltonian, in this form, is manifestly hermitian. Use periodic boundary condition, 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} . You can think of these states as being placed around 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 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 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 H} in the first equation and show 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 T} is both unitary and commutes 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 H} . It is thus a symmetry of the hamiltonian.
(b) Find the eigenstates of T by using wavefunctions of 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 |\psi \rangle = \sum_{n=1}^{N} e^{ikn}|n\rangle}
What are the eigenvalues of these eigenstates? Do all these eigenstates have to be 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} as well? If not, do any of these eigenstates have to be 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} ? Explain your answer.
(c) Next Consider Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F} which takes Failed to parse (SVG (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\rangle \to |N+1-n\rangle.} Write F 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 H} in the first equation and show 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 F} both is unitary and commutes 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 H} . It is thus a symmetry of the hamiltonian.
(d) Find a complete set of 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 F} and their eigenvalues. Do all these eigenstates have to be 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} as well? If not, do any of these eigenstates have to be 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} ? Explain your answer.
Solution
(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 T|n\rangle = |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 \langle i|T|j\rangle = \delta_{1,j+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 T = \sum_{n=1}^{N} |n+1\rangle\langle 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\rangle = |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 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 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|}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle TT^\dagger = \sum_{n=1}^{N}|n\rangle\langle n| = \left [ I \right ]_{NXN}}
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 T} 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 \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|)]}
Failed to parse (SVG (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|]} 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 |N+1\rangle = |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 = 0 } 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 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 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 commute, any eigenstate of is an eigenstate of
explicit proof:
So they are eigenstates of also.
(c)
So
So is unitary.
Hence they commute.
(d)
It has already been proved that F is both unitary and hermitian. Thus the eigenvalues of F are given by <math> \pm 1 <\math>
Since <math> F|n\rangle = |N+1-n\rangle <\math> by intuition let us construct states given by
<math> |\psi\pm\rangle = |n\rangle \pm |N+1-n\rangle n = 1,2,3,….N/2. <\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>
Thus <math> |\psi\pm\rangle <\math> thus constructed are eigenstates of F. It is noted that we have <math> N/2 <\math> symmetric eigenstates given by
<math> |1\rangle + |N\rangle ; |2\rangle + |N-1\rangle ; |3\rangle +|N-2\rangle … <\math> and <math> N/2 <\math> anti-symmetric eigenstates given by
<math> |1\rangle - |N\rangle ; |2\rangle - |N-1\rangle ; |3\rangle -|N-2\rangle ; … <\math>
<math> H (|n\rangle \pm |N+1-n\rangle) <\math>
<math> =\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>
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.