# Matrix elements in the quantum oscillator

1. Oct 5, 2014

### andre220

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

For a quantum oscillator find all non-zero matrix elements of the operators $\hat{x}^3$ and $\hat{x}^4$

2. Relevant equations

$\hat{x} = \sqrt{\frac{\hbar}{2m\omega}}\left(a+a^\dagger\right)$

$a^\dagger |n\rangle = \sqrt{n+1}|n+1\rangle$
$a |n\rangle = \sqrt{n}|n-1\rangle$

3. The attempt at a solution
Okay so first it is necessary to compute $\langle n |\hat{x}^3|n'\rangle$ and then from that outcome determine the non-zero elements.

$$x^3 = \left(\frac{\hbar}{2m\omega}\right)^{3/2}(a + a^\dagger)^3 =\left(\frac{\hbar}{2m\omega}\right)^{3/2}(a + a^\dagger)(a _ + a^\dagger)(a + a^\dagger) =$$

$$=\left(\frac{\hbar}{2m\omega}\right)^{3/2}(a a a + a^\dagger a a + a a^\dagger a + a^\dagger a^\dagger a + a a a^\dagger + a^\dagger a a^\dagger + a a^\dagger a^\dagger + a^\dagger a^\dagger a^\dagger)$$

So then we compute

$$\langle n|\hat{x}^3|n' \rangle =$$
$$\left(\frac{\hbar}{2m\omega}\right)^{3/2} \langle n|(a a a + a^\dagger a a + a a^\dagger a + a^\dagger a^\dagger a + a a a^\dagger + a^\dagger a a^\dagger + a a^\dagger a^\dagger + a^\dagger a^\dagger a^\dagger)|n'\rangle$$

And then from here I am having trouble evaluating this, however, I am pretty sure that the majority of the terms above go to zero.

Any help would be greatly appreciated.

2. Oct 6, 2014

### Fightfish

You just have to evaluate the action of the operators on the states sequentially. Lets take the term $a a^{\dagger} a$ for example:
$$\left\langle n\right| a a^{\dagger} a \left|n'\right \rangle = \left\langle n\right| a a^{\dagger} \sqrt{n'} \left|n' - 1\right \rangle = \left\langle n\right| a \sqrt{n'} \sqrt{n'} \left|n'\right \rangle = \left\langle n\right| \sqrt{n'} \sqrt{n'} \sqrt{n'} \left|n'-1\right \rangle = \left(n'\right)^{3/2} \delta_{n,n'-1}$$

3. Oct 6, 2014

### andre220

Thanks for your response. Why wouldn't the above be: $$\langle n|aa^\dagger a|n'\rangle = \langle n|aa^\dagger \sqrt{n'}|n'-1\rangle = \langle n|a\sqrt{n'+1}\sqrt{n'}|n'\rangle = \langle n|\sqrt{n'+1}n'|n'-1\rangle = n'\sqrt{n'+1}\delta_{n,n'-1}$$

Becuase you would apply the lowering operator and then the raising and then the lowering, which would give you that factor $\sqrt{n'+1}$ instead of 3 factors of $\sqrt{n'}$. Or am I misguided in my thinking?

4. Oct 6, 2014

### BvU

I seem to remember that $a^\dagger a = N$ and $a a^\dagger = N+1$ ...

5. Oct 6, 2014

### andre220

Okay so just to make sure that I am going about this right here is what I have for the first four terms:
$$\langle n|a a a|n'\rangle = \langle n|a a \sqrt{n'}|n'-1\rangle = \cdots = (n')^{3/2}\delta_{n,n'-3}$$
$$\langle n|a^\dagger a a|n'\rangle = \langle n|a^\dagger a \sqrt{n'}|n'-1\rangle = \langle n|(n')^{3/2}|n'-2\rangle = (n')^{3/2}\delta_{n,n'-2}$$
$$\langle n|a a^\dagger a|n'\rangle = \langle n|a n'|n'\rangle = (n')^{3/2}\delta_{n,n'-1}$$
$$\langle n|a^\dagger a^\dagger a|n'\rangle = \langle n |a^\dagger n'|n'\rangle = n'\sqrt{n'+1}\delta_{n,n'+1}$$

6. Oct 6, 2014

### BvU

As I said (you can check with your post #1) $a^\dagger a = N$, so $a^\dagger a \, a |n> = a^\dagger a \, \sqrt N\, |n-1> = \sqrt N a^\dagger a \, |n-1> = (N-1)\sqrt N |n-1>$, meaning I get something else for your second term...
And for the coefficient of the third ...
And for the fourth I get the same as you :-)

7. Oct 6, 2014

### Fightfish

Because you are applying the raising operator on the state $\left|n'-1\right\rangle$ instead of $\left|n'\right\rangle$, and so you obtain a factor $\sqrt{(n'-1) +1} = \sqrt{n'}$.
This also sort of explains why you got the wrong answer for the second term: you need to take note of the state that you are operating on.

8. Oct 6, 2014

### andre220

Okay after about five minutes of staring at it I think it makes sense now. What you are both saying is that $$\langle n|a^\dagger a a|n'\rangle =\underbrace{ \langle n |\sqrt{n'}a^\dagger a |n'-1\rangle}_\textrm{Now we are acting on n-1 state and thus the outcome of the operator will be not just n' but n'-1}$$

It clicked now, the notation was slightly throwing me off, but also I clearly wasn't getting it. Thank you for your help.

9. Oct 6, 2014

### BvU

Nice exercise. In my book (Merzbacher, QM, 1970 :-) ) there is a matrix for x that still looks decent. But for higher powers it will look a bit less nice; still it's a diagonal band matrix.