I have been working through a problem. It has asked me to determine the eigenstates and corresponding eigenvalues of the number operator in a quantum harmonic oscillator; $$\hat{n}=\hat{a}_+\hat{a}_-$$ I have been looking for some literature on it but I can't seem to find anything! I know what solution I am expecting. I believe since the Hamiltonian and the number operator commute, then we can say that they must share the same family of eigenstates, which in the case of the Quantum harmonic oscillator come in the form of the Hermite Gauss Polynomials. My problem is where to really start. I'm unsure as to how this 2 term equation can blow up into the long expression for the eigenstates of the Hamiltonian operator in the quantum harmonic oscillator. I have also found some videos that go through the calculation, they find that the corresponding eigenvalues can be any positive integer. Any help would be greatly appreciated!
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3Related: https://physics.stackexchange.com/q/23028/2451 and links therein. – Qmechanic Sep 25 '20 at 10:21
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2@Qmechanic I have read through the related post. I am in an introduction to Quantum mechanics course and are yet to be introduced to the notions of a 'Fork Space'. I found following the discussion rather difficult. I believe I want to solve the eigenvalue problem $$\hat{n}|\psi\rangle = \lambda|\psi\rangle$$ For eigenvalues $\lambda$ and eigenstates $|\psi\rangle$ based off the definition of the number operator. – JayP Sep 25 '20 at 10:46
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1The subject is covered extensively in many basic QM textbooks and notes (that can bee found in google.) I personally learned it from the Haken's "Quantum Field theory of solids" which may sound a bit advanced, but the first few chapters are the basic QM for the harmonic oscillator. – Roger V. Sep 25 '20 at 11:14
1 Answers
Given that this is a homework problem, I'll just outline the general procedure that you need to follow: your argument is correct, since the Hamiltonian and the Number operator commute, there must be at least one simultaneous eigenbasis for the two operators. (There is no guarantee in general in such a case for every eigenbasis to be the same, of course the form of the Hamiltonian should give you a clue as to whether or not the "standard" energy eigenbasis would work.)
Suppose now you want to do this as generally as possible. Keep in mind that the eigenstates of the Harmonic Oscillator form a complete set. Let's call them $|n\rangle$. Since they do, any arbitrary state $|\psi\rangle$ can be written as a linear combination of these states.
In particular, a candidate for an eigenvector of $\hat{a}_{+}\hat{a}_{-}$ can also be written as $$|\psi\rangle = \sum_{n=0}^\infty c_n |n\rangle,$$
and your job is now to calculate the possible $c_n$s and $\lambda$s that would allow the following relation to hold:
$$\hat{a}_{+}\hat{a}_{-} |\psi\rangle = \lambda |\psi\rangle.$$
If you simply plug in the form of $|\psi\rangle$ given above, and remember that the different $|n\rangle$ states are linearly independent, you should be able to find that only certain values of $\lambda$ and $c_n$ are allowed. That should give you the states $\psi_k$ and eigenvalues $\lambda_k$ of the $\hat{N}$ operator. (You can verify your answer by keeping in mind that it's called the number operator for a reason!)
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Could you use the fact that in the eigenvalue problem $$\hat{a}+\hat{a}-|\psi_n\rangle = \lambda|\psi_n\rangle $$ that by applying the operators on the LHS, we obtain $$n|\psi_n\rangle = \lambda|\psi_n\rangle $$ To then conclude that $\lambda=n$? And in this case it is indeed $n\in\mathbb{Z}^+$ since we know that raising and lowering operators can not annihilate below the ground state, and only raises and lowers the quantum number by integer values? – JayP Sep 25 '20 at 12:04
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What guarantee do you have that the eigenvalues are indeed $n$? In other words, how do you know that $\hat{a}{+}\hat{a}{-} |\psi_n\rangle = n|\psi_n\rangle$? Isn't that what you need to find? So assuming it seems to be a little tautological. – Philip Sep 25 '20 at 12:19
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Because wouldn't you have from $$\hat{a}+\hat{a}-|\psi_n\rangle=\lambda|\psi_n\rangle$$ $$\hat{a}+(\sqrt{n}|\psi{n-1}\rangle)=\lambda|\psi_n\rangle$$ And then, $${n}|\psi_{n}\rangle=\lambda|\psi_n\rangle$$ Meaning that $\lambda=n$ being the eigenvalue? – JayP Sep 25 '20 at 12:29
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You are assuming that the states $|\psi_n\rangle$ are the eigenstates of the $\hat{H}$ operator as well as the $\hat{N}$ operator. A priori there is no special reason for this. It is possible in general to have states that are not the eigenstates of one operator, but are the eigenstates of another even when the two operators commute. (This generally happens when there is some degeneracy in the problem. For example, consider a free particle. The state $|p\rangle + |-p\rangle$ is a state of definite energy but not a state of definite momentum.) – Philip Sep 25 '20 at 14:00