Average values of operators of potential and kinetic energy

[LagrangeEuler]

In case of quantum LHO in eigen state of the system $|n \rangle$
\langle \hat{T} \rangle=\langle \hat{U} \rangle=\frac{1}{2}(n+\frac{1}{2})\hbar \omega
What will happened in some superposition of states? Does Ehrenfest theorem can tell me something more general? Is it possible to say that
\langle \hat{T} \rangle=\langle \hat{U} \rangle
in any prepared state?

 

[gonadas91]

What do you mean here by a supperposition of states? I mean when you take the average of these quantities, you are averaging them with one state of the harmonic oscillator. However, if your state is a superposition of states, the calculation can be done in the same way. I would start by taking a state which is a superposition of two, for example:

| n\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle )

and see what do you obtain. Then, you can generalize the case of an infinite superposition of states (a vector on the Hilbert space corresponding to the eigenstates of the H0, so just a linear combination of them).

 

[LagrangeEuler]

I mean superposition of two eigen states. For example if
$|\psi \rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$
Is then
$\langle \hat{T} \rangle_{\psi}=\langle \hat{V} \rangle_{\psi}$? In this case it is. But is it always the case?

 

[PeroK]

I mean superposition of two eigen states. For example if
$|\psi \rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$
Is then
$\langle \hat{T} \rangle_{\psi}=\langle \hat{V} \rangle_{\psi}$? In this case it is. But is it always the case?

If $<T> = <U>$ in all eigenstates, then clearly $<T> = <U>$ in any state (linear combination of eigenstates). The probability of each state is irrelevant, as whatever eigenstate you find, you will have $<T> = <U>$.

$<T> = \sum |c_n|^2 <T>_n = \sum |c_n|^2 <U>_n = <U>$ (at $t = 0$)

Edit: Apologies: this is rubbish, of course. That calculation only works for <E>, not for <T> and <V>. As has been pointed out below.

 

[Nikola Kolev]

Newbie here, can someone explain what these fancy brackets are?

 

[blue_leaf77]

Does Ehrenfest theorem can tell me something more general?

Ehrenfest or virial?

then clearly <T>=<U> = in any state (linear combination of eigenstates).

Are you sure? Virial theorem was proven under the assumption that the expectation values of kinetic and potential energy were to be taken w.r.t the stationary states, that is the H eigenstate. In fact $<U-T> = \frac{\omega \hbar}{2}<a^{+2}+a^2>$ which in general does not vanish if there are one or more pair of eigenstates in the superposition state which differ in the quantum number by 2.

<T>=∑|cn|2<T>n

Why have you assumed that $<T>_{mn} = 0$ for $m\neq n$?

 

[PeroK]

In case of quantum LHO in eigen state of the system $|n \rangle$
\langle \hat{T} \rangle=\langle \hat{U} \rangle=\frac{1}{2}(n+\frac{1}{2})\hbar \omega
What will happened in some superposition of states? Does Ehrenfest theorem can tell me something more general? Is it possible to say that
\langle \hat{T} \rangle=\langle \hat{U} \rangle
in any prepared state?

Ehrenfest or virial?

Are you sure? Virial theorem was proven under the assumption that the expectation values of kinetic and potential energy were to be taken w.r.t the stationary states, that is the H eigenstate. In fact $<U-T> = \frac{\omega \hbar}{2}<a^{+2}+a^2>$ which in general does not vanish if there are one or more pair of eigenstates in the superposition state which differ in the quantum number by 2.

Why have you assumed that $<T>_{mn} = 0$ for $m\neq n$?

Yes, you're correct. I took "always" to mean "for any combination" of states. Not "for any time". I've edited my post to clarify.

 

[blue_leaf77]

For harmonic oscillator and any time, $\langle m|T|n\rangle = -\hbar \omega/4 \langle m|a^{+2}-a^+a-aa^++a^2|n\rangle$ is in general nonzero.