# Finding this eigenstate

1. Jan 21, 2014

### unscientific

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

Given the a|n> = α|n-1>, show that α = √n :

2. Relevant equations

3. The attempt at a solution

$$<n|a^{+}\hat {a}|n> = \alpha <n|a^{+}|n-1> = | \hat a|n>|^2$$

$$\alpha = \frac{<n|a^{+}\hat {a}|n>}{<n|a^{+}|n-1>}$$

Taking the complex conjugate of both sides:

$$\alpha* = \frac{<n|a^{+}\hat {a}|n>}{\alpha<n-1|n-1>}$$
$$|\alpha^2| = \frac{<n|a^{+}\hat {a}|n>}{<n-1|n-1>}$$

Where $\hat {a} a^{+} = \frac{1}{2m\omega} \left( (m\omega x)^2 + p^2 + im\omega [\hat{x},\hat{p}] \right)$

Last edited: Jan 21, 2014
2. Jan 21, 2014

### BruceW

nice work. I think you're getting there. although remember that $\alpha$ isn't necessarily real, so when you took the complex conjugate of both sides, you should also take complex conjugate of $\alpha$. For the final equation you wrote down, I think that is the right idea. But it's definitely missing some $\hbar$ and the square root should not still be there. it's just a minor mistake.

3. Jan 21, 2014

### unscientific

The problem is, I have no idea how to proceed now..

4. Jan 21, 2014

### BruceW

once you've fixed the minor mistake, you have the operator $aa^\dagger$. what other operator does this operator look like? hint: the physical model of this system is simple harmonic motion.

5. Jan 21, 2014

### unscientific

Okay, The square root has gone away, due to it appearing twice from a and a+ and i've fixed the square term in the bracket.

6. Jan 21, 2014

### unscientific

USing $[\hat x,\hat p] = i\hbar$

$$\hat a a^{+} = \frac{\hat H}{\hbar \omega} - \frac{1}{2}$$

$$|\alpha^2| = \frac{<n| \frac{\hat H}{\hbar \omega} - \frac{1}{2}|n>}{<n-1|n-1>}$$

How do I proceed from here?

7. Jan 21, 2014

### BruceW

hmm. wait hold on, since they ask you to consider $|\hat{a}|n\rangle |^2$ I think you should use the operator that you were originally working with, $\hat{a}^\dagger \hat{a}$

edit: sorry about that, the rest of your working looks good. you've got the right idea.

8. Jan 21, 2014

### unscientific

How do I evaluate the inner product:

$$|\alpha^2| = \frac{<n| \frac{\hat H}{\hbar \omega} - \frac{1}{2}|n>}{<n-1|n-1>} = \frac{<n|a^{+}\hat {a}|n>}{<n-1|n-1>}$$

It should give $n$ on the RHS.

9. Jan 21, 2014

### BruceW

you're almost there now. In the question, they tell you $E_n = (n+1/2)\hbar \omega$ they mean this is the energy of the $|n\rangle$ eigenstate. so this should be enough to evaluate that inner product.

10. Jan 21, 2014

### unscientific

How do I evaluate the numerator?

$$<n| \frac{\hat H}{\hbar \omega} - \frac{1}{2}|n>$$
$$= \frac{1}{\hbar \omega} <n|\hat H|n> - \frac{1}{2}<n|n>$$
$$= \frac{1}{\hbar \omega} <n|\hat H|n> - \frac{1}{2}<n|n>$$
$$= \frac{E_n}{\hbar \omega}<n|n> - \frac{1}{2}<n|n>$$
$$= \left (\frac{E_n}{\hbar \omega} - \frac{1}{2} \right) <n|n>$$

Thus,
$$|\alpha|^2 = \left (\frac{E_n}{\hbar \omega} - \frac{1}{2} \right)\frac{ <n|n>}{<n-1|n-1>}$$
$$|\alpha|^2 = n \frac{ <n|n>}{<n-1|n-1>}$$

Last edited: Jan 21, 2014
11. Jan 21, 2014

### BruceW

yep. that is good so far for the numerator. But the denominator is not (n-1)2. each of the $|n\rangle$ are states, right? so what (conventionally), is the inner product of the state with itself?

edit: and now, there is a similar question for what is left over in the numerator.

12. Jan 21, 2014

### unscientific

Can I assume they are linearly independent, and have norm = 1?

13. Jan 21, 2014

### BruceW

yeah! I guess it is not obvious really. But I think in these kinds of problems, you can assume that if they give you a set of states, those states will be orthonormal. (unless they say otherwise).