A solved problem for time reversal symmetry: Difference between revisions
(New page: Source: Problem 4-10,'Modern Quantum Mechanics'; J.J. Sakurai '''problem:''' suppose a spinless particle is bound to a fixed center by a potential V(x) so asymmetrical that no energy leve...) |
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\psi _{n}(x)=\langle\ x|\ n\rangle=\sum_{lm}\langle\ x|\ l,m\rangle\langle\ l,m|\ n\rangle=\sum_{l}\sum_{m}F_{lm}(r)Y_{l}^{m}(\theta ,\varphi ) | \psi _{n}(x)=\langle\ x|\ n\rangle=\sum_{lm}\langle\ x|\ l,m\rangle\langle\ l,m|\ n\rangle=\sum_{l}\sum_{m}F_{lm}(r)Y_{l}^{m}(\theta ,\varphi ) | ||
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<math>\psi _{n}(x)=e^{-i\delta }\langle\ x|\ m\rangle=e^{-i\delta }\psi _{n}^{*}(x)=e^{-i\delta }\sum_{m}F_{lm}^{*}(r)Y_{l}^{m}(\theta ,\varphi )=e^{-i\delta }\sum_{m}F_{lm}^{*}(r)(-1)^{m}Y_{l}^{-m}(\theta ,\varphi )=e^{-i\delta }\sum_{m}F_{l,-m}^{*}(r)(-1)^{-m}Y_{l}^{m}(\theta ,\varphi )</math> | <math>\psi _{n}(x)=e^{-i\delta }\langle\ x|\ m\rangle=e^{-i\delta }\psi _{n}^{*}(x)=e^{-i\delta }\sum_{m}F_{lm}^{*}(r)Y_{l}^{m}(\theta ,\varphi )=e^{-i\delta }\sum_{m}F_{lm}^{*}(r)(-1)^{m}Y_{l}^{-m}(\theta ,\varphi )=e^{-i\delta }\sum_{m}F_{l,-m}^{*}(r)(-1)^{-m}Y_{l}^{m}(\theta ,\varphi )</math> | ||
Latest revision as of 20:11, 25 April 2010
Source: Problem 4-10,'Modern Quantum Mechanics'; J.J. Sakurai
problem: suppose a spinless particle is bound to a fixed center by a potential V(x) so asymmetrical that no energy level is degenerate. Using time-reversal invariance prove <L>=o
for any energy eigenstate. (This is known as quenching of orbital angular momentum.) If the wave function of such a non degenerate eigenstate is expanded 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 \sum_{l}\sum_{m}F_{lm}(r)Y_{l}^{m}(\theta ,\varphi )}
what kind of phase restriction do we obtain on 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_{lm}(r)} ?
Solution: Under reversal time 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 P\to -P} 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 r\to -r} , [H,K]=0 Assume 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 HK|\ n \rangle=EK|\ n \rangle} 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 \rangle} is the eigenket of Hamiltonian. 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 \rangle} is also a eigenket of H with the same eigenvalue.
If we do not have any degeneracy, 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 \rangle=e^{i\delta } |\ n \rangle=|\ m \rangle}
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 \delta} is a real number. 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\ n |L|\ n\rangle=\langle\ m |KLK^{-1}|\ m\rangle=-\langle\ m |L|\ m\rangle=-e^{-i\delta }\langle\ n |L|\ n\rangle e^{i\delta }=-\langle\ n |L|\ 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 \Rightarrow \langle\ n |L|\ n\rangle=0 \Rightarrow <L>=o} 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 _{n}(x)=\langle\ x|\ n\rangle=\sum_{lm}\langle\ x|\ l,m\rangle\langle\ l,m|\ n\rangle=\sum_{l}\sum_{m}F_{lm}(r)Y_{l}^{m}(\theta ,\varphi ) }
Also
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 _{n}(x)=e^{-i\delta }\langle\ x|\ m\rangle=e^{-i\delta }\psi _{n}^{*}(x)=e^{-i\delta }\sum_{m}F_{lm}^{*}(r)Y_{l}^{m}(\theta ,\varphi )=e^{-i\delta }\sum_{m}F_{lm}^{*}(r)(-1)^{m}Y_{l}^{-m}(\theta ,\varphi )=e^{-i\delta }\sum_{m}F_{l,-m}^{*}(r)(-1)^{-m}Y_{l}^{m}(\theta ,\varphi )}
with comparison two different form of function, we get
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_{lm}(r)=F_{l,-m}^{*}(r)(-1)^{m}e^{-i\delta }}