I had recently been studying the Dirac equation, and as an example of how the equation is used, I was given a problem about the Landau levels of graphene (but I personally have no knowledge about graphene).

I was given the Hamiltonian

$$H = \begin{pmatrix} 0 & v_F(p_x - i(p_y + \frac{e}{c}B x)) \\ v_F(p_x - i(p_y + \frac{e}{c}B x)) & 0 \end{pmatrix}$$

using the Landau gauge $\textbf{A} = (0,Bx,0)$ ($B$ has only a z-component) and $v_F$ is the Fermi velocity. And I would like to solve for the energy levels and the 2-component wave functions.

I personally solved this problem by explicitly multiplying out the matrix and found that this Hamiltonian gives two decoupled harmonic oscillator equations. But the official solution solved it by introducing a pair of ladder operator right from the beginning

$$a = \sqrt{\frac{c}{2e\hbar B}}(p_x - i(p_y + \frac{e}{c}B x)) $$ $$a^\dagger = \sqrt{\frac{c}{2e\hbar B}}(p_x + i(p_y + \frac{e}{c}B x)) $$

and rewriting the Hamiltonian as

$$ H = \sqrt{\hbar \omega}\begin{pmatrix} 0 & a \\ a^\dagger & 0 \end{pmatrix} , \ \omega = \frac{2eBv_F^2}{c}$$

Then the energy levels are found to be $E = \pm \sqrt{n \hbar \omega}$ by exploiting the ladder operators.

While I fully understood how to use the algebra of $a$ and $a^{\dagger}$ to obtain the wave function and the energy levels, but I just have no clue how one motivates them in the first place. More precisely, I would like to ask

  1. How can I be sure that the $a$ and $a^{\dagger}$ I introduced here will actually behave like in the harmonic oscillator problem, i.e they are indeed the same raising and lowering operators? (except from the fact that the Hamiltonian is actually a pair of harmonic oscillator when multiplied out)

  2. How do I decide the constant in front of my $a$ and $a^{\dagger}$? The choice seems really arbitrary to me, and the official solution did not explain how it come up with this constant factor.

I had also been given a second problem for a massive Hamiltonian ("effective mass" $m^*$)

$$H = -\frac{1}{2m^*}\begin{pmatrix} 0 & (p_x - i(p_y + \frac{e}{c}B x))^2 \\ (p_x - i(p_y + \frac{e}{c}B x))^2 & 0 \end{pmatrix}$$

and I would like to know do the same method applies directly to this Hamiltonian as well (using the same $a$ and $a^{\dagger}$ as above?) ? In this case when the Hamiltonian is multiplied out, it will no longer gives a harmonic oscillator, but something of power 4, and I don't see how $a$ and $a^{\dagger}$ can still be useful.

I had nonetheless tried it out, and got the energy levels to be $E = n\hbar \omega$, where $\omega = \frac{eB}{m^*c}$ and the wave function $\psi = (|n>, -(n-1)|n-2>)$, where $|n>$ are the n-th harmonic oscillator wave function. It will also be very helpful if I can know whether my results are correct.

This is the first time I encountered this kind of method, on top of the already difficult enough topic of Dirac equation, and I am really confused. Thanks for any help.

  • $\begingroup$ Hi I know this is an old post but I am having the same problem. The solutions for my class say that $E = \hbar\omega\sqrt{n}$ but I keep getting the same answer you have. Would you be able to elaborate more on how you find the eigenvalues of the Hamiltonian using the ladder operators? Thank you! $\endgroup$ Commented Apr 1, 2023 at 20:46

1 Answer 1


I know that this answer is fairly late but if you are still wondering, all physics lies in the commutation relation, the ladder operators that Dirac introduced satisfy the relation $$[\hat{a},\hat{a}^\dagger] = 1$$ so I believe that as soon as your operator also satisfy this everything falls into place (this is where the arbitrary constant comes from). Well you might ask why this proves that your introduced ladder operators is essentially the same as the harmonic oscillator, if I just lay out the proof for the ladder operations you can see that as long as the commutation relation is fulfilled between the creation and annihilation operator it is essentially the same ladder operators.

First use the commutation relation to prove that: $$[\hat{N}, \hat{a}^\dagger] = \hat{a}^\dagger, \quad [\hat{N}, \hat{a}] = -\hat{a}$$ where $\hat{N} = \hat{a}^\dagger\hat{a}$ being the number operator, and this would indeed prove that $$\hat{N}\hat{a}^\dagger|n> = (n+1)\hat{a}^\dagger|n>, \quad \hat{N}\hat{a}|n> = (n-1)\hat{a}|n>$$ which implies that $\hat{a}^\dagger|n>$ and $\hat{a}|n>$ is eigenstates of the number operator such that $a|n> = c_1|n-1>$ and $a^\dagger|n>=c_2|n+1>$ where $c_1$ and $c_2$ are calculated via the normalization.

For a more elaborated proof of the ladder operators look at e.g: Modern quantum mechanics by Sakurai.

  • $\begingroup$ Thanks. Honestly I left that QM class still not knowing the exact answer to my question...not sure why didn't I at least try to ask the professor. But your explanation makes sense -- it is all in the commutation relation. $\endgroup$
    – AstroK
    Commented Mar 28, 2018 at 18:48

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