Eigenvalues of a "unusual" Hamiltonian of a harmonic oscillator

[laser1]

Homework Statement

Description

Relevant Equations

N/A

for d), I am a bit confused. I have two trains of thoughts here

any thoughts on which answer is correct, and why the other one is incorrect? Both seem like valid solutions to me. Or is the question ambiguous?

thanks

 

[JimWhoKnew]

Please use LaTeX. I can't quote images, and also...

Given that $~\hat{b}~$ is a self-adjoint operator and $~\hat{b}^\dagger~$ is its adjoint, I'd expect their commutator to vanish. What am I missing?

I'd say some more if your question will become quotable. For now, note that the Hilbert space must be closed: if $~|\phi\rangle~$ is a state, so is $~b|\phi\rangle~$.

 

[laser1]

Let us consider the following Hamiltonian
$$
\hat{H}=\hbar\omega\left(\hat{b}^\dagger \hat{b}+\frac{1}{2}\right).
$$
The self-adjoint operator $\hat b$ and its adjoint $\hat b^\dagger$ fulfil the (unusual) commutation relation
$$
[\hat b,\hat b^\dagger]=\hat b\hat b^\dagger-\hat b^\dagger\hat b = 4.
$$
Define an (unusual) number operator by $\hat N=\hat b^\dagger \hat b$. The normalised eigenstates of $\hat N$ are $|\phi_n\rangle$ such that
$$
\hat N|\phi_n\rangle = n|\phi_n\rangle.
$$

(a) Evaluate the commutator $[\hat b,\hat N]$.

(b) Use the previous result to show that $\hat b|\phi_n\rangle$ is a (not-normalised) eigenstate of the number operator for $n\ge 4$. What is the corresponding eigenvalue?

(c) Let $|\Phi\rangle=g\,\hat b|\phi_n\rangle$ with $g\in\mathbb R$. Calculate the value of $g\ge 0$ such that $\|\Phi\|=1$.

(d) Use the previous results to argue what are the eigenvalues of this unusual number operator $\hat N$.

---

My working so far:

(a) I evaluated the commutator and found
$$
[\hat b,\hat N]=4\hat b.
$$

(b) I deduced that
$$
\hat N\hat b=\hat b\hat N-4\hat b,
$$
so that $\hat b|\phi_n\rangle$ is an eigenstate with eigenvalue $n-4$.

(c) Using the norm condition, I got $g=1/\sqrt{n}$.

---

For (d), I am confused. I have two trains of thought:

(1) The smallest eigenvalue is $0$ because repeatedly applying $\hat b$ will keep lowering the eigenvalue until it would become negative, which contradicts the property that the eigenvalue must satisfy $n\ge 0$ (since $\langle\psi|\hat N|\psi\rangle=\|\hat b|\psi\rangle\|^2\ge 0$). So then the only possible eigenvalues are $0,4,8,\dots$.

(2) Since part (b) is only stated for $n\ge 4$, I can start from e.g. $7$ and then stop when it hits $3$, because the relation doesn't apply again. So then the solutions also include $3,7,11,\dots$. The same logic can be applied to $1,5,9,\dots$ and $2,6,10,\dots$, which would suggest the eigenvalues are $0,1,2,3,\dots$.

Any thoughts on which answer is correct, and why the other one is incorrect? Both seem valid to me, or is the question ambiguous?

Given that b^ is a self-adjoint operator and b^† is its adjoint, I'd expect their commutator to vanish. What am I missing?

you are correct, I asked my professor, he said it is an error, $\hat{b}$ is not self-adjoint.

 

[JimWhoKnew]

(2) Since part (b) is only stated for $n\ge 4$, I can start from e.g. $7$ and then stop when it hits $3$, because the relation doesn't apply again.

Why not?
In (b) you were asked about $~n\ge 4~$, but the commutation relations are valid throughout, so your answer to (b) should apply to any participating $n~$ (edit: $~\hat{b}|\phi_0\rangle~$ is an exception). Suppose $~|\phi_7\rangle~$ exists. Then $~\hat{N}\hat{b}|\phi_7\rangle=3\hat{b}|\phi_7\rangle~$. But then $~\hat{b}\hat{b}|\phi_7\rangle~$ is also a state. Is it an eigenstate of $~\hat{N}~$?

BTW: since it had taken you such a short time to re-post in LaTeX, why didn't you just do so in #1?

 

[laser1]

Why not?
In (b) you were asked about $~n\ge 4~$, but the commutation relations are valid throughout, so your answer to (b) should apply to any participating $n~$ (edit: $~\hat{b}|\phi_0\rangle~$ is an exception). Suppose $~|\phi_7\rangle~$ exists. Then $~\hat{N}\hat{b}|\phi_7\rangle=3\hat{b}|\phi_7\rangle~$. But then $~\hat{b}\hat{b}|\phi_7\rangle~$ is also a state. Is it an eigenstate of $~\hat{N}~$?

BTW: since it had taken you such a short time to re-post in LaTeX, why didn't you just do so in #1?

hmm yes good point, the relations are valid throughout. So $|\phi_7\rangle$ exists, so $|\phi_3\rangle$ exists, which means $|\phi_{-1}\rangle$ exists, which can't be possible, so that whole ladder can't exist?

 

[JimWhoKnew]

This "unusual" HO follows from the "usual" by substituting $~\hat{a}=\hat{b}/2~$ and identifying $~|\phi_n^{(a)}\rangle~$ as $~|\phi_{4n}^{(b)}\rangle~$. Compare the energy spectra.

 

[laser1]

This "unusual" HO follows from the "usual" by substituting $~\hat{a}=\hat{b}/2~$ and identifying $~|\phi_n^{(a)}\rangle~$ as $~|\phi_{4n}^{(b)}\rangle~$. Compare the energy spectra.

so it is only 0, 4, 8, ...?

 

[JimWhoKnew]

so it is only 0, 4, 8, ...?

Yes, by the argument you wrote in #5.