Phy5645/Transformations and Symmetry Problem: Difference between revisions
No edit summary |
No edit summary |
||
| Line 57: | Line 57: | ||
We therefore find that the states <math>|\psi\rangle</math> are also eigenstates of <math>\hat{H}</math> with eigenvalue <math>2\cos{k}\!</math>. | We therefore find that the states <math>|\psi\rangle</math> are also eigenstates of <math>\hat{H}</math> with eigenvalue <math>2\cos{k}\!</math>. | ||
'''(c)''' | '''(c)''' Similarly to the previous case, we may conclude from the fact that the effect of <math>\hat{F}</math> is | ||
<math>F|j\rangle = | | <math>\hat{F}|j\rangle = |N+1-j\rangle</math> | ||
that the operator may be written as | |||
<math>F | <math>\hat{F} = \sum_{n=1}^{N} |N+1-n\rangle\langle n|.</math> | ||
<math> | The Hermitian adjoint <math>\hat{F}^{\dagger}</math> is | ||
<math>\ | <math>\hat{F}^{\dagger} = \sum_{n=1}^{N} |n\rangle\langle N+1-n|,</math> | ||
and | |||
<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> | |||
<math> | <math>\hat{F}</math> is therefore unitary. 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>= \sum_{n=1}^{N} |N+1-n\rangle \langle n-1| + \sum_{n=1}^{N} |N+1-n\rangle \langle n+1|</math> | ||
<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+2-n\rangle \langle n| + \sum_{n=1}^{N} |N-n\rangle \langle n|</math> | ||
| Line 83: | Line 85: | ||
<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>= \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)=== | ===(d)=== | ||
Revision as of 14:51, 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) 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\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 \rangle = \sum_{n=1}^{N} e^{ikn}|n\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\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\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\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}\!} .
(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 Failed to parse (SVG (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}} 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|,}
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{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}.}
Failed to parse (SVG (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 therefore unitary. 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|}
Failed to parse (SVG (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|}
The expressions 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 \hat{F}\hat{H}} 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{H}\hat{F}} are identical, and thus 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 Failed to parse (SVG (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|n\rangle = |N+1-n\rangle }
by intuition let us construct states 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 |\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.