# Operators on a Harmonic oscillator ground state

1. May 17, 2013

### tomwilliam2

1. The problem statement, all variables and given/known data

Calculate the expectation value for a harmonic oscillator in the ground state when operated on by the operator:
$$AAAA\dagger A\dagger - AA\dagger A A\dagger + A\dagger A A A\dagger)$$

2. Relevant equations

$$AA\dagger - A\dagger A = 1$$
I also know that an unequal number of raising and lowering operators gives a zero expectation value due to orthogonality requirements.

3. The attempt at a solution
I guess that the first term in brackets gives a zero expectation value as it leads to a function which is orthogonal to $$\psi_0$$
If I say n=1 then:
$$A\dagger\psi_0 = \psi_1$$
And
$$A \psi_1 = \psi_0$$

I've tried taking the third term and saying:
$$A\dagger A(1+ A\dagger A) = A\dagger A + A\dagger A A\dagger A$$

Then doing the same thing with the second term to get
$$1+ 2A\dagger A +A\dagger A A\dagger A$$
Then I subtract this term from the third and I get
$$-A A\dagger$$

But this, operating on $$\psi_0$$ seems to give me an expression which results in an infinity when integrated over all space.

Can someone tell me where I went wrong?

Last edited: May 17, 2013
2. May 17, 2013

### vela

Staff Emeritus
You shouldn't need to do an integral, but I am curious what integral you're trying to evaluate.

3. May 17, 2013

### tomwilliam2

It's a sandwich integral to get the expectation value:

$$\frac{\hbar^4}{4a^4} \int_{-\infty}^{\infty} \psi_0^* (...) \psi_0 dx$$
Where the dots in brackets are the operator terms I mentioned above.
Can you see anything wrong with my reasoning above?

Last edited: May 17, 2013
4. May 17, 2013

### vela

Staff Emeritus
Your reasoning looked fine. I'm not sure now why you think that integral diverges.

5. May 17, 2013

### tomwilliam2

Actually, it doesn't give me infinity, it gives me the wrong answer, using:

$$\psi_0 = \left ( \frac{1}{\sqrt{\pi}a}\right )^{1/2} e^{-x^2/2a^2}$$

6. May 17, 2013

### vela

Staff Emeritus
You need to show more details if you're having trouble evaluating the integral explicitly. You shouldn't need to, however, if you understand what $A$ and $A^\dagger$ do.

7. May 17, 2013

### tomwilliam2

Well, I realise that the $$A$$ and $$A\dagger$$ operators convert energy eigenfunctions from the state $$n$$ to $$n\pm0.5$$ where $$E_n = (n + 0.5) \hbar \omega_0$$, but I am specifically told to use the integral above to calculate the expectation value (it's actually the expectation value of $$p_x^4$$).

So I get the integrand as \psi_0 (-\psi) and it turns out as -1 by my reckoning. I should get 3 in order to get the required result.

P.s. how do you type Latex on the same line as text?

8. May 17, 2013

### vela

Staff Emeritus
Not exactly. They take the nth state to the (n±1)th state.

Your mistake occurred somewhere earlier. For one thing, you shouldn't have had any terms with an odd number of annihilation/creation operators.

9. May 18, 2013

### tomwilliam2

Ok, thanks for your help. If I set out what I'm doing in clear stages, maybe you could tell me in which stage my mistake lies:

a) I note that of the three terms in the operator above, the first has an unequal number of raising/lowering operators, so must give an expectation value of zero...so I remove it from the integrand.

b) I take the second term $-AA\dagger A A\dagger$ and I use the commutation relation to rewrite it as $-(1+A\dagger A)(1+A\dagger A)=-(1 + 2A\dagger A + A\dagger A A\dagger A)$

c) I note that two of these terms end in the lowering operator, which, when acting on the ground state will give zero. So the middle term becomes -1

d) I take the third term $A\dagger A A A\dagger$ and use the commutation relation to rewrite as $A\dagger A(1 + A\dagger A)$.

e) I note that both terms end in the lowering operator, so when acting on the ground state will be zero. I remove the third term entirely.

f) my integrand is left as I mentioned in my previous post.

So could you possibly point out which of these stages involves faulty reasoning? Most appreciated, thanks for the help once again.

10. May 18, 2013

### vela

Staff Emeritus