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We have the usual Hamiltonian for the 1D Harmonic Oscillator: $\hat{H_{0}}=\frac{\hat{P^2}}{2m} + \frac{1}{2}m \omega \hat{X^2}$

Now a new term has been added to the Hamiltonian, $\hat{H} = \hat{H_0} + \mu B\hat{S_z}$

The system has a spin degree of freedom with $s=\frac{1}{2}$

What are the new eigenstates of the Hamiltonian and what are the energy eigenvalues? We have been denoting the stationary states as $|n,s_z\rangle$ and the non-spin eigenvalues as usual $E = h \omega (n+1/2)$

Can anyone give me some help with this question? I'm guessing that the eigenstates don't change and then it's straightforward to find the eigenvalues but I'm not sure if this is correct (or why it would be correct).

Brandon Enright
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KatieC25
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  • Hi, if by eigenstates you mean the states $|n,s_z\rangle$ (you can type \rangle to gen a nicer ket), then you are correct. Just try acting with your hamiltonian on these states (and recall the definition of an eigenstate). – Peter Kravchuk May 02 '13 at 19:30
  • Thanks, sorted the ket! So why is it exactly that the eigenstates don't change with the added term? That's the bit I'm tripping up on I think. – KatieC25 May 02 '13 at 19:42
  • Ok, I'll write up an answer. – Peter Kravchuk May 02 '13 at 19:49

2 Answers2

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Here is the precise treatment for determining the eigenvectors of the full Hamiltonian. You will probably find the two physics.SE posts I link to at the end useful for understanding this stuff (which basically boils down to understanding tensor products):

Let $\mathcal H_0$ denote the harmonic oscillator Hilbert space and $\mathcal H_{1/2}$ denote the spin Hilbert space then the total Hilbert space of the system is their tensor product $\mathcal H = \mathcal H_0\otimes \mathcal H_{1/2}$. The notation you are using here is really a shorthand for defining the total Hamiltonian as an operator on $\mathcal H$ $$ \hat H = \hat H_0\otimes I_{1/2} + \mu BI_0\otimes \hat S_z $$ where $I_0$ is the identity operator in the harmonic oscillator hilbert space, and $I_{1/2}$ is the identity operator in the spin Hilbert space. If we define $$ \hat H_0|n\rangle = E_n|n\rangle, \qquad \hat S_z|s_z\rangle = \hbar s_z|s_z\rangle $$ then the states $|n\rangle$ form a basis for $\mathcal H_0$ consisting of eigenvectors of $H_0$ and the states $|s_z\rangle$ form a basis for $\mathcal H_{1/2}$ consisting of eigenvectors of $\hat S_z$. If we define $$ |n, s_z\rangle = |n\rangle\otimes |s_z\rangle $$ Then it is a standard result on tensor products of Hilbert spaces that the states $|n, s_z\rangle$ form a basis for the total Hilbert space $\mathcal H = \mathcal H_0\otimes \mathcal H_{1/2}$ of the system. Moreover, one can show (which you've probably already essentially done by the comments above) using these definitions that $$ \hat H|n,s_z\rangle = (E_n+\hbar \mu Bs_z)|n,s_z\rangle $$ So that the states are a basis for the total Hilbert space $\mathcal H$ consisting of eigenvectors of $\mathcal H$. We have therefore identified that the "stationary states" you wrote down originally, if correctly mathematically interpreted, can be proven to be the eigenstates of the full Hamiltonian.

Other useful posts:

  1. How to tackle 'dot' product for spin matrices

  2. Should it be obvious that independent quantum states are composed by taking the tensor product?

along with anything else you find about tensor products.

Community
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joshphysics
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    Wow, how many things can happen while you are typing an answer.) – Peter Kravchuk May 02 '13 at 19:58
  • Yeah. If I had seen your comment earlier, I would have told you I was almost done with the response. In any case, your presentation might be more appealing to people who cringe upon hearing the phrase "tensor product." – joshphysics May 02 '13 at 20:01
  • @joshphysics, You've mixed the identity operators a bit. Yes, tensor product is a more mathematical way. However, I think that it can hinder the physics a bit. From my point of view, you first pick a system, then determine a full set of commuting observables, and then associate a Hilbert space to it. And then it turns out that it is a tensor product. – Peter Kravchuk May 02 '13 at 20:13
  • @PeterKravchuk Thanks; I had forgotten that I had switched notation. I've always found that tensor products have enhanced my physical understanding. In particular, I think they contribute to intuition for why, the dimension of the full Hilbert space increases (as a product of dimensions in a sense) in the way that it does. You certainly make good points, and I hope people read both responses. – joshphysics May 02 '13 at 20:19
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First of all, states of the spinless oscillator are $|n\rangle$ such that: $$ H_0|n\rangle=E_n|n\rangle=\hbar\omega(n+1/2)|n\rangle $$ Then you introduce the spin $S_z$, so you now have a different space of states -- you have more degrees of freedom, not only should you specify the energy, but also the spin. You can choose a basis in this space $|n,s_z\rangle$ such that $$ H_0|n,s_z\rangle=E_n|n,s_z\rangle\\ S_z|n,s_z\rangle=\hbar s_z|n,s_z\rangle $$ You are now concerned with the eigenstates of the operator $H=H_0+\mu BS_z$. Lets just try to act with it upon our basis: $$ H|n,s_z\rangle=(H_0+\mu B S_z)|n,s_z\rangle=H_0|n,s_z\rangle+\mu BS_z|n,s_z\rangle=\\ =E_n|n,s_z\rangle+\hbar\mu Bs_z|n,s_z\rangle=(E_n+\hbar\mu Bs_z)|n,s_z\rangle $$ So it turns out that $|n,s_z\rangle$ is an eigenstate of $H$ with eigenvalue $E_n+\hbar\mu Bs_z$. As soon as these states form a basis, there are no other independent eigenstates.

Peter Kravchuk
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