Harmonic oscillator expectation values

[land]

I'm given that there is a harmonic oscillator in a state that is a superposition of the ground and first excited stationary states given by \psi = \frac{1}{\sqrt{2}}\abs{\psi_0(x,t) + \psi_1(x,t)}, where \psi_0 = \psi_0(x)e^{\frac{-iE_0t}{\hbar}} and \psi_1 = \psi_1(x)e^{\frac{-iE_1t}{\hbar}}. I need to calculate expectation values for position, momentum, and total energy.

Here's what I've done: I'm assuming this is a simple harmonic oscillator, and for the x operator I have \frac{i}{\sqrt{2m\omega}}(a_- - a_+). I think that a- operating on Psi0 would be zero, and on Psi1 would be Psi0, and a+ operating on Psi0 would be Psi1, and on Psi1 would be Psi2. So... I have for <x>:

<x> = \frac{i}{2\sqrt{2m\omega}}<\psi_0 + \psi_1 | \psi_0 - \psi_1 - \psi_2>.. but.. uh.. how do you do this? I feel like I must have done something wrong. A push in the right direction would be much appreciated.

Thanks so much!

[land]

There's no way that expectation value could be zero, right? because <\psi_n | \psi_{n+1}> = 0, I think.. so all that stuff inside the <> would be zero. No?

[George Jones]

I'm given that there is a harmonic oscillator in a state that is a superposition of the ground and first excited stationary states given by \psi = \frac{1}{\sqrt{2}}\abs{\psi_0(x,t) + \psi_1(x,t)}

Is that supposed to be

\Psi = \frac{1}{\sqrt{2}} \left( \abs{\Psi_0(x,t) + \Psi_1(x,t)} \right)?

Brackets are important, as are distinctions between upper case and lower case. For example,

\psi_0 = \psi_0(x)e^{\frac{-iE_0t}{\hbar}}

doesn't make sense without such a distinction. If fact, in this case, it doesn't make sense to write

\psi = \psi_{0} + \psi_{1},

since the sum of two energy eigenstates is not an energy eigenstate.

and on Psi1 would be Psi2.

No - don't forget the numerical factor.

Now calculate

\left< \Psi \left| x \right | \Psi \right>

by, as you did, replacing x with the appropriate, but don't to forget to include all the time dependences and other numerical factors.

Also, use the orthogonality condition that you gave in your last post.

I don't mean to sound so critical, but, in this calculation, it's important to get the details right.

[land]

I did mean to put brackets around it. I'm not sure why I didn't. I have trouble getting LaTeX to display things correctly sometimes.

Edit: nevermind, I'm an idiot and figured out what you were talking about. See next post for confusion.

[land]

OK. I tried to go back through and put in the terms you were talking about, and I've gotten something.. but I'm still confused. I've gotten down to \frac{i}{2\sqrt{2m\omega}}<\psi_0 + \psi_1 | \sqrt{\hbar\omega}\psi_0 - \sqrt{\hbar\omega}\psi_1 - \sqrt{2\hbar\omega}\psi_2>.


<\psi_0 | \psi_0> I know how to do, but <\psi_0 | \psi_1> I don't. So you said to use the orthoganility condition I brought up in my last post.. so is this just zero since it's Psi0 and Psi1? If not, here's what I'm getting for that, plugging in the x and t dependences: -\sqrt{\hbar\omega}<\psi_0(x)e^{\frac{-iE_0t}{\hbar}} | \psi_1(x)e^{\frac{-iE_1t}{\hbar}}>. I have no idea how to do that. So I hope it's zero.

So if it is, then the <Psi0 | Psi0> part of that would just be \sqrt{\hbar\omega}. That seems way too easy though.