Expectation for the Harmonic Oscillator ( using dirac)

[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 I am a bit Iffy about a step i took:

<x>_n = <Ψ_n|x|Ψ_n> = L
for a given Ψ_n = (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.

 

[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.

 

[Somali_Physicist]

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.

Surely you wouldn't get an actual value for even values.That would be extremely counter intuitive, that said I will try your advice.

 

[DrClaude]

You wrote a couple of things that don't make sense.

Ψ_n = (A⁺)ⁿ(n!)^-2

This should be
$$
|\psi_n\rangle = \frac{(A^\dagger)^n}{\sqrt{n!}} | 0 \rangle
$$

=> L = <(A⁺)(n!)^-2|x|(A⁺)(n!)^-2>

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.

 

[kuruman]

Surely you wouldn't get an actual value for even values.That would be extremely counter intuitive, that said I will try your advice.

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?

 

[Demystifier]

The question i wanted to generally show that the expected value is Zero for all odd energy levels.

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