QM - Deriving the Ladder Operators' Eigenbasis

[Tangent87]

I'm am trying to derive the relations:

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

using just the facts that [a,a⁺]=1 and N|n>=|n> where $$N=a^{\dagger}a$$ (which implies $$\langle n|N|n\rangle=n\geq 0$$). This is what I've done so far:

$$[a,a^{\dagger}]=1 \Rightarrow aa^{\dagger}|n\rangle=(n+1)|n\rangle$$

$$\Rightarrow a^{\dagger}a(a^{\dagger}|n\rangle)=(n+1)(a^{\dagger}|n\rangle)$$

Therefore, $$a^{\dagger}|n\rangle=K|n+1\rangle$$ for some K (possibly complex).

Taking the conjugate and then multiplying together gives:

$$\langle n|aa^{\dagger}|n\rangle=\langle n|1+N|n\rangle=n+1=|K|^2$$

So my question is how do we conclude from this that $$K=\sqrt{n+1}$$?

 

[fzero]

Are you asking why $$K$$ has no phase? Because of the orthogonality of the $$|n\rangle$$, there are no observables that could depend on such a phase, so it can be absorbed into the definition of the states. It's not that tricky to show this if you express $$|n\rangle$$ in terms of $$(a^\dagger)^n | 0\rangle$$.

 

[Tangent87]

Are you asking why $$K$$ has no phase? Because of the orthogonality of the $$|n\rangle$$, there are no observables that could depend on such a phase, so it can be absorbed into the definition of the states. It's not that tricky to show this if you express $$|n\rangle$$ in terms of $$(a^\dagger)^n | 0\rangle$$.

Yes that's exactly what I'm asking, like why can't K be say, $$e^{i\pi/276}\sqrt{n+1}$$? As this satisfies the condition |K|²=n+1. I don't really like the idea of showing it by expressing $$|n\rangle$$ in terms of $$(a^\dagger)^n | 0\rangle$$ because doesn't that already assume $$K=\sqrt{n+1}$$?

 

[fzero]

You can assume that

$$a^\dagger |n\rangle = \sqrt{n+1} e^{i\alpha_n} | n+1\rangle ,$$

then

$$|n\rangle = \frac{1}{n!} \exp\left(-\sum_n \alpha_n\right) (a^\dagger)^n | 0\rangle.$$

Now let $$\hat{A}$$ be some operator that we can express in terms of $$a$$ and $$a^\dagger$$. We can show that $$\langle 0 | \hat{A} |0\rangle$$ is either zero or independent of the $$\alpha_n$$.

Note, I don't think the HW problem expects you to go through all this, but it's probably worthwhile doing anyway. I think they just expect you to decide that there's no real reason to introduce a phase, so you don't.

 

[Tangent87]

You can assume that

$$a^\dagger |n\rangle = \sqrt{n+1} e^{i\alpha_n} | n+1\rangle ,$$

Agreed.

then

$$|n\rangle = \frac{1}{n!} \exp\left(-\sum_n \alpha_n\right) (a^\dagger)^n | 0\rangle.$$

Do you mean $$|n\rangle = \frac{1}{\sqrt{n!}} \exp\left(-i\sum_n \alpha_n\right) (a^\dagger)^n | 0\rangle.$$?

Now let $$\hat{A}$$ be some operator that we can express in terms of $$a$$ and $$a^\dagger$$. We can show that $$\langle 0 | \hat{A} |0\rangle$$ is either zero or independent of the $$\alpha_n$$.

Do you mean $$\langle n | \hat{A} |n\rangle$$? I can see why that wouldn't depend on the $$\alpha_n$$ because the exponentials would cancel so now we've shown that no observables can depend on the phase but I would be grateful if you could explain a little further how this tells us we can "absorb" the phase into the states. What would happen if we just left the phase outside the states? As we've shown it doesn't affect any observable results?

Note, I don't think the HW problem expects you to go through all this, but it's probably worthwhile doing anyway. I think they just expect you to decide that there's no real reason to introduce a phase, so you don't.

Yes sorry about this, in an exam I would be happy just to write "...since we can absorb the phase into the states" but I want to be thorough now so that I can properly understand this when it comes to the exam.

Thanks, Tangent.

 

[fzero]

Agreed.



Do you mean $$|n\rangle = \frac{1}{\sqrt{n!}} \exp\left(-i\sum_n \alpha_n\right) (a^\dagger)^n | 0\rangle.$$?

Yes, I left out the i by mistake.

Do you mean $$\langle n | \hat{A} |n\rangle$$?

Well I suppose we could consider $$\langle n | \hat{A} |m\rangle$$ to be most general, but the calculation is the same.

I can see why that wouldn't depend on the $$\alpha_n$$ because the exponentials would cancel so now we've shown that no observables can depend on the phase but I would be grateful if you could explain a little further how this tells us we can "absorb" the phase into the states. What would happen if we just left the phase outside the states? As we've shown it doesn't affect any observable results?

Since no observables depend on the phase, there's no measurement that we can do to detect the phase. So we can just set the phases to zero without losing any information. Put another way, we know that the objects $$c|n\rangle$$, where $$c$$ is a nonzero complex number, represent the same quantum state as $$|n\rangle$$. This is what let's us normalize states. So the states $$e^{\pm i \alpha}|n\rangle$$ and $$|n\rangle$$ are the same. This last way of looking at it is probably the most direct.

 

[Tangent87]

Since no observables depend on the phase, there's no measurement that we can do to detect the phase. So we can just set the phases to zero without losing any information. Put another way, we know that the objects $$c|n\rangle$$, where $$c$$ is a nonzero complex number, represent the same quantum state as $$|n\rangle$$. This is what let's us normalize states. So the states $$e^{\pm i \alpha}|n\rangle$$ and $$|n\rangle$$ are the same. This last way of looking at it is probably the most direct.

Ahh I see, thanks very much.