# I Expectation for the Harmonic Oscillator ( using dirac)

1. Jun 12, 2018

### Somali_Physicist

I've been trying to form a proof using , using majorly dirac notation.There has been claims that its much better to use in QM.

The question i wanted to generally show that the expected value is Zero for all odd energy levels.I believe i have solved the question but im a bit Iffy about a step i took:

<x>n = <Ψn|x|Ψn> = L
for a given Ψn = (A+)n(n!)-2
Energy eigen functions have definite parity, assume for all odd n's if one is zero the rest should follow.
Take n = 1
=> L = <(A+)(n!)-2|x|(A+)(n!)-2>
= (n!)-1 <(A+)|x|(A+)>
B = <(A+)|x|(A+)>
Define A+ = Lx + iC : B,C are Real
=> <Lx+iC|x|Lx+iC>
(Bit iffy after these steps)
= <Lx|x|Lx|> + <iC|x|iC>
= <L|x3|L>+<C|x|C>
as ∫x2n+1dx for limits [-∞,∞] and n =0,1,2,3....
=> 0
we find B=0
therefore
<x>n = 0
For odd ns.

Last edited: Jun 12, 2018
2. Jun 12, 2018

### kuruman

What about even values of $n$? You are going about it backwards. Instead of replacing $\psi_n$, write operator $x$ in terms of $a^{\dagger}$ and $a$ and use your knowledge of what $a^{\dagger}|\psi_n>$ and $a|\psi_n>$ are equal to.

3. Jun 14, 2018

### Somali_Physicist

Surely you wouldnt get an actual value for even values.That would be extremely counter intuitive, that said I will try your advice.

4. Jun 14, 2018

### Staff: Mentor

You wrote a couple of things that don't make sense.
This should be
$$|\psi_n\rangle = \frac{(A^\dagger)^n}{\sqrt{n!}} | 0 \rangle$$

The notation here doesn't work. You can't have an operator in a bra or a ket. You should end up with something like
$$\langle 0 | A x A^\dagger |0 \rangle$$
and so on.

5. Jun 14, 2018

### kuruman

Why is it so counter intuitive? In your proof, which needs fixing as @DrClaude suggested, you have the integral ∫x2n+1dx where n is odd. What would happen to this integral if n were even? Say n = 2k?

6. Jun 14, 2018

### Demystifier

What do you mean by "energy", i.e. what is the Hamiltonian?