# Change of operator in the Hamiltonian [closed]

We are told that the particle has mass m and charge e and is moving in 2 dimensions.

The position operator $\mathbf{X}=(X_{1},X_{2})$ and momentum operator $\mathbf{P}=(P_{1},P_{2})$

We are given that the Hamiltonian is:

$H=\frac{1}{2m}((P_{1}+\frac{1}{2}eBX_{2})^{2}+(P_{2}-\frac{1}{2}eBX_{1})^{2})$

Where $B$ is a constant and $eB\neq 0$.

We are required to show that the energy levels are of the form $(n+\frac{1}{2})\frac{\hbar\left | eB \right |}{m}$.

My thoughts so far: the energy levels look fairly similar to that of a harmonic oscillator, so I'm thinking I may need to define new operators to try and get to a Hamiltonian that looks more like that. I'm just not sure how I go about doing that?

I think that the correct spectrum is: $E_n = \frac{\hbar |eB|}{m}(n+1/2)$.

To solve the problem we start noticing that, defining: $$x' := P_{1}+\frac{1}{2}eBX_{2}\quad \mbox{and}\quad p':= P_{2}-\frac{1}{2}eBX_{1}$$ we have: $$[x',p'] = i\hbar eB I$$Assuming $eB >0$ (otherwise it is sufficient to swap the definitions of $x$ and $p$), we can redefine $$x := |eB|^{-1/2}x'\quad \:, \quad p := |eB|^{-1/2}p'$$ so that, they both hold: $$[x,p] = i \hbar I\qquad (1)$$ and $$H=\frac{|eB|}{2m}(x^2+p^2)\:.\qquad (2)$$ consequently we have the standard levels of a harmonic oscillator: $$E_n = \frac{\hbar |eB|}{m}(n+1/2)\:,\qquad (3)$$

NOTE. The answer could stop here. However something further can be said also taking into account the kind correction by nervxxx to my previous version, concerning the degeneracy of levels.

If the CCR representation (1) were irreducible we would have a harmonic oscillator with eigenspaces of dimension 1. However this representation is not irreducible because the Hilbert space is ${\cal H}= {\cal H_1} \otimes {\cal H_2} = L^2(R) \otimes L^2(R)$ as initially assumed and there are operators commuting with $x$ and $p$. One of them is:

$$\tilde{H} =\frac{|eB|}{2m}(\tilde{x}^2+\tilde{p}^2)\:,$$

where we have defined (using the same strategy exploited to define $x$ and $p$ but changing the internal signs) $$\tilde{x} := |eB|^{-1/2} \left(P_{1}-\frac{1}{2}eBX_{2}\right)\quad \mbox{and}\quad \tilde{p}:= |eB|^{-1/2} \left( P_{2}+\frac{1}{2}eBX_{1}\right)$$

We conclude that in each eigenspace of $\tilde{H}$ we have a harmonic oscillator described by the restriction of $H$ to this invariant subspace. Since the eigenvalues of $\tilde{H}$ are (countably) infinitely many we conclude that the degeneracy of each level of $H$ is infinite.

• The answer is not entirely correct, unfortunately. The spectrum is right, but the degeneracy is wrong. Like, way off. Note that this Hamiltonian is that of a particle constrained to move in 2D with a perpendicular B field, i.e. the Quantum Hall effect. Anyway, one can decompose the Hilbert space into the tensor product of the Hilbert spaces of the cyclotron degree of freedom and the guiding center degree of freedom. One can define raising and lowering operators $a_i, a_i^\dagger$ which obey $[a_i, a_j^\dagger]=\delta_{ij}$ where $i = 1$ corresponds to the cyclotron part and $i = 2$ corresponds – nervxxx Jan 11 '14 at 1:19
• to the guiding center part. Note that the Hamiltonian expressed in these operators is ONLY a function of $a_1, a_1^\dagger$. This means that any operator comprised of $a_2, a_2^\dagger$ commutes with the Hamiltonian. Corresponding we see that there is a massive (infinite, bounded by sample size) degeneracy in each level which you derived above, which btw, is called a Landau level. – nervxxx Jan 11 '14 at 1:21
• For a physical picture of the massive degeneracy: classically, the particle undergoes a cyclotron orbit of a certain radius. But where we place this orbit in our sample is arbitrary. Consequently you can put the particle pretty much anywhere in the sample - this is the origin of degeneracy in each Landau level. – nervxxx Jan 11 '14 at 1:24
• Many thanks! In the previous version of my answer I stupidly confused the direct sum with the tensor product in constructing the Hilbert space of the system. I agree with you, the degeneracy is infinite. However I think that here it is countably infinite (because I suspect that each eigenspace of $\tilde{H}$ is irreducible for $x$ and $p$ but I have no time to try to prove it) while in Landau levels is continuously infinite. However I am by no means familiar with (quantum) Hall effect so I could be completely wrong. – Valter Moretti Jan 11 '14 at 9:16
• Yes it is countably infinite :) – nervxxx Jan 12 '14 at 18:47