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GR9277 #50
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\prob{50}
The state of a quantum mechanical system is described by a wave function $\psi$. Consider two physical observables that have discrete eigenvalues: observable A with eigenvalues $\{\alpha\}$, and observable B with eigenvalues $\{\beta\}$. Under what circumstances can all wave functions be expanded in a set of basis states, each of which is a simultaneous eigenfunction of both A and B?

1. Only if the values $\{\alpha\}$ and $\{\beta\}$ are nondegenerate
2. Only if A and B commute
3. Only if A commutes with the Hamiltonian of the system
4. Only if B commutes with the Hamiltonian of the sytem
5. Under all circumstances

Quantum Mechanics$\Rightarrow$}Simultaneous Eigenstates

QM in verse...

Two operators, both alike in state functions,
In fair bases, where we lay our scene,
From ancient grudge break new mutiny...
Two operators unlike in eigenvalues

Yet star-crossed lovers commute.
So anyway, the problem gives $A | a \rangle = \alpha | a \rangle$ and $B | a \rangle = \beta | a \rangle$. That is, both A and B share the same eigenstate $| a \rangle$.

Consider $\langle a^{'} | [A,B]| a^{''} \rangle =\langle a^{'}| AB-BA| a^{''} \rangle = \langle a^{'} |(A\beta^{'}-B\alpha^{''}) | a^{''} \rangle = \langle a^{'} |(\beta^{'}A-\alpha^{''}B) | a^{''} \rangle = \langle a^{'} |(\alpha^{''}\beta^{''}-\beta^{''}\alpha^{''}) | a^{''} \rangle = (\alpha^{''}\beta^{''}-\beta^{''}\alpha^{''}) \langle a^{'} | a^{''} \rangle$.

But, the scalar term is 0. This implies that in order for both operators to have the same eigenstate, $[A,B]=0$.

(Also, one knows that $\langle a^{'} | a^{''} \rangle = \delta_{',''}$ by definition of orthonormal eigenfunctions.)

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bass
2014-07-03 02:30:39
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liamo
2010-09-29 22:08:05
I believe the answer is: under all circumstances.

The above answer does not make sense to me.

The wavefunction ( Psi ) is a wavefunction (Psi) of both A and B we are told. A basis is a place the wavefunctions reside. Therefore the wavefunctions are composed of the basis states.
 neon372010-11-12 00:19:38 I think you misunderstood the question. The question says some state is described by $\psi$ (described by a function). So, there are two observable A and B (described by matrices). Under what circumstance can $\psi$ be eigenfunction of both A and B at the same time. Bah! Its hard to word it any other way, hope that was helpful. But in essence to understand this, you can think of the following way. We have generalized uncertainty principle, ${\sigma_A}^2{\sigma_B}^2\geq (\frac{1}{2i}[\hat{A}, \hat{B}])$ If A and B dont commute then, there would some uncertainty while measuring the two observables. So you cant get exact values for both observations, ie the two eigenvalues. If they do commute then there is no uncertainty and you can get two exact values for both observations.
 mpdude82012-04-19 21:11:06 This is a weak argument, but I always shy away from answers like "under all circumstances" on the GRE regardless of the question. Especially when dealing with Quantum or particle physics questions, I haven't really ever seen an answer where a generalizing word like "all" or "none" is correct. Usually there's some exception or some situation where a rule or concept does or doesn't hold. This question is no different -- A and B must commute.
caffeinated
2008-04-09 11:31:38
How poetic.
bucky0
2007-11-01 14:51:56
To clarify, the answer is (B)

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