GREPhysics.NET: GR9677 #100

GREPhysics.NET

GR9677 #100

Problem

GREPhysics.NET Official Solution

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Quantum Mechanics $\Rightarrow$ }Raising Operator

$a^\dag = a^* = \sqrt{\frac{m\omega_0}{2\hbar}}\left(x+-ip/(m\omega_0)\right) \neq a$ , and thus choice III must be true. This eliminates all but choice (C) and (E). Since one knows that the raising operator acts to raise the energy level, $[a,H]\neq 0$ implies that they don't commute. This leaves just choice (C).

(Choice II is false because, from above, the condition for a Hermitian operator is violated $a^\dag \neq a$ .)

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Comments

physicsisgod
2008-11-07 14:44:21

I think we can eliminate (II) because the eigenvalues of $\hat{a}$ are definitely not observables. The index of the stationary state, n, is not a physical quantity. The trick to this problem, then, is eliminating (I), and I think evanb provides the best reason for that.

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NEC

phys2718
2008-10-14 10:28:48

Formally if you take the commutator of $a$ and the Hamiltonian, $[x + ip, x^{2} + p^{2}]$ , it is a relatively simple calculation to see that you don't get $0$ .

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NEC

sawtooth
2007-10-30 10:17:44

Question: Why are we so sure to say $\hat{x}^\dagger = \hat{x}$ and more importantly $\hat{p}^\dagger = \hat{p}$ ? We all know that momentum and space are observables and so they have hermitian operators, but if you check, for the momentum we have: rnrn $\hat{p}^\dagger = (-i \hbar \frac{d}{dx} )^\dagger = i\hbar \frac{d}{dx} = - \hat{p}$ rnrnso the momentum operator is not hermitian. Ofcourse, with careful study, it turns out this is not the entirely correct way to take the adjoint, and the momentum operator is self-adjoint (hermitian) only with specific boundary conditions. So, my question is, do we assume these conditions hold in this harmonic oscillator problem (since we know it's an observable) and therefore we quickly get to the point where $\hat{\alpha}^\dagger \neq \hat{\alpha}$ due to the minus before the momentum operator?rnrnFinally, I can add, if somenone doesn't recognise the ladder operator, I think they can reject I since we know that a conservable quantity commutes with the hamiltonian. Momentum is conserved, but there's no such law for space (in this context, at least i think) so the operator -related observable (which is the sum of the two) is not conserved and hence the operator doesn't commute with H. Correct me if I am wrong.rnrnAlso, correct me if I am wrong^2, but doesn;t the fact that $\alpha$ and the hamiltonian have eigenvalues different in this way, mean that they don't commute? Wild guess, didnt think of it thoroughly.

evanb
2008-07-03 14:25:31

p --absolutely must-- be self-adjoint, because it is observable. The only boundary conditions that we need to apply is that $\psi$ is square-integrable (goes to zero quick enough that the boundary term is ignored), which holds true for a harmonic potential. From this, you can deduce that a is not hermitian, eliminated II.

For the last piece, "doesn't the fact that a and the hamiltonian have different eigenvalues in this way mean that they don't commute?" That's a weirdly-phrased question, mainly because you don't explain what "this way" is supposed to mean. However, if you know that the eigenstates of the ladder operators are coherent states and the eigenstates of the hamiltonian are NOT coherent states (they are stationary states), that is enough to say that H and a do not commute, eliminating I and leaving (C) as the answer.

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NEC

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You are replying to:

Question: Why are we so sure to say $\hat{x}^\dagger = \hat{x}$ and more importantly $\hat{p}^\dagger = \hat{p}$ ? We all know that momentum and space are observables and so they have hermitian operators, but if you check, for the momentum we have: rnrn $\hat{p}^\dagger = (-i \hbar \frac{d}{dx} )^\dagger = i\hbar \frac{d}{dx} = - \hat{p}$ rnrnso the momentum operator is not hermitian. Ofcourse, with careful study, it turns out this is not the entirely correct way to take the adjoint, and the momentum operator is self-adjoint (hermitian) only with specific boundary conditions. So, my question is, do we assume these conditions hold in this harmonic oscillator problem (since we know it's an observable) and therefore we quickly get to the point where $\hat{\alpha}^\dagger \neq \hat{\alpha}$ due to the minus before the momentum operator?rnrnFinally, I can add, if somenone doesn't recognise the ladder operator, I think they can reject I since we know that a conservable quantity commutes with the hamiltonian. Momentum is conserved, but there's no such law for space (in this context, at least i think) so the operator -related observable (which is the sum of the two) is not conserved and hence the operator doesn't commute with H. Correct me if I am wrong.rnrnAlso, correct me if I am wrong^2, but doesn;t the fact that $\alpha$ and the hamiltonian have eigenvalues different in this way, mean that they don't commute? Wild guess, didnt think of it thoroughly.

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