Show that expecting value of energy is independent of time

[emeriska]

Sorry I failed to make Latex work, don't know why...

1. Homework Statement
We consider
for a given potential. Psi is also normalized...

Homework Equations

Show that expectation value of energy is independent of time.

The Attempt at a Solution

Well, I'm use to expecting values, but not of the energy...

I started with <H>psi=<E>psi
then if I compute <H>=<psi|H|psi> = E|Psi> = E

Then I told myself H=p^2/2m + V(x)

So I guess if I compute dH/dt and I find it equals to 0 I can say E is independent of time?

I feel like this is wrong tough... It looks too simple

Thanks a lot for checking that out guys!

 

[BvU]

Hello, and welcome to PF :-)
They aren't called expecting values, but expectation values...

<H>=<psi|H|psi> = E|Psi> = E

the thing in the middle doesn't look right . What exactly do you compute ?

 

[emeriska]

Thanks about that!

Well, I'm really not quite sure...found that somewhere and I tought it made sens but the more I look at it the more this doesn't help at all...

 

[Matterwave]

Are you given that $V=V(x)\neq V(x,t)$ is true, so that your Hamiltonian is explicitly time-independent? Otherwise, I don't think the statement in this problem is true.

I'm guessing this is assumed because you have written:

$$\Psi(x,t)=\sum_n c_n(t=0)e^{-iE_n t/\hbar}\psi_n(x)$$

If so, all you have to prove is that $\left<H\right>$ is independent of time, since that IS the expectation value of the energy. In other words, prove:

$$\frac{d\left<H\right>}{dt}=0$$

 

[BvU]

If $H$ is independent of time, the wave functions $\Psi({\bf r},t)$ in $i\hbar {\partial \Psi\over\partial t} = H\Psi$ can be written as $f(t) \;\psi({\bf r})$ and the equation can be re-written as $i\hbar {\partial f \over\partial t} \psi({\bf r}) = f(t) \;H\psi (\bf r)$ or $$i\hbar {\partial f \over\partial t} /f(t) = H\psi (\bf r)/\psi({\bf r})$$ Now left hand side is a function of t only, right hand of r only. Can only be if both sides are constant. name the constant E and you get the time-independent Schroedinger eqn $H\psi ({\bf r}) = E\;\psi({\bf r})$.
For the complete wave function there still is the time part, for which $i\hbar {\partial f \over\partial t} = E \; f(t)$ with solution $f(t) = e^{{-i E\over \hbar} t }$. That's where the $\Psi(x,t)$ in your exercise come from.

The full wording of your exercise probably says the $\psi_n$ are normalized ? And they satisfy $H|\psi_n> = E_n|\psi_n>$ ?

Now your <H>=<psi|H|psi> can be written out in full (by you) as $$\int_{-\infty}^{\infty}\; \Psi^*(x,t) \; H \; \Psi^*(x,t) \; dx $$to give the expectation value of E. Fill in the given $\Psi$, see that the exp (and hence the time dependence) disappears

 

[emeriska]

I had to read that a couple times but thanks a lot! that really helped!

 

[emeriska]

Oh, and while I'm there,

I also need to prove that $<H> <= E_{1}$, where $E_{1}$ is the fundamental energy. How will I proceed to do that?

Do I need to compute $E_{1}$? Cause if I do, I'll end up with terms in $\psi_{1}$ that I'm not quite sure how to compare with $\psi_{n}$

 

[Matterwave]

I think you mean $\left<H\right> \geq E_1$ right? $E_1$ is the ground state energy, so you shouldn't be getting energies lower than it.

 

[emeriska]

Yes sorry your right! I know it sound really obvious, but I need to prove it...That's what bugging me :S

 

[Matterwave]

Well start with the definition: $\left<H\right>=\left<\Psi|H|\Psi\right>$ and then break down the $\left|\Psi\right>$ into the superposition of eigenvectors of the Hamiltonian to see how this will work out.