Derivation of normal Zeeman-Effect

[PhysicsRock]

TL;DR

The Zeeman-Effect can be derived from coupling the Schrödinger-Hamiltonian to an external electromagnetic field $\vec{A}$. However, when doing so, I always seem to miss a factor of $2$.

I was / am trying to derive the energy shift resulting from the normal Zeeman-Effect by coupling the Hamiltonian to the external field $\vec{A}$, that carries the information about the field $\vec{B}$ via $\vec{B} = \nabla \times \vec{A}$. Let $q = -e$ be the charge of the electron and $M_e$ its mass, where I chose an upper case M to avoid confusion with the magnetic quantum number. Since we are not going to consider an external electric field, we can directly set $\Phi = 0$, where $\Phi$ is the electric scalar potential. The kinetic Hamiltonian then becomes

$$
\begin{align*}
\hat{H}_\text{kin} &= \frac{1}{2M_e} \left( \hat{p} - q \vec{A} \right)^2 \\
&= \frac{1}{2M_e} \left( -i\hbar\nabla - q \vec{A} \right)^2 = \frac{1}{2M_e} \left( i \hbar \nabla + q \vec{A} \right)^2 \\
&= \frac{1}{2M_e} \left( -\hbar^2 \Delta + q^2 A^2 + i q \hbar \nabla \cdot \vec{A} + i q \hbar (\vec{A} \cdot \nabla) \right).
\end{align*}
$$

We can now do some simplifications. First, for a homogenous magnetic field along the $z$-axis, which is what I am considering, the vector potential can be written as (taken from our lecture notes and additionally from various sources I found online)

$$
\vec{A} = -\frac{1}{2} \left( \vec{r} \times \vec{B} \right).
$$

It is then apparent that $\nabla \cdot \vec{A} = 0$, and if we assume $B \equiv \vec{B}_z$ to be sufficiently small, we can also ignore the term $\propto \vec{A}^2 \propto B^2$. What we are left with is

$$
\begin{align*}
\hat{H}_\text{kin} &= -\frac{\hbar^2}{2M_e} \Delta + \frac{iq \hbar}{2M_e} (\vec{A} \cdot \nabla).
\end{align*}
$$

Sorry for the long introduction, here comes the problem: Even without explicitly computing the term $\vec{A} \cdot \nabla$ we can immediately tell that, since $\vec{A}$ also carries a factor of $1/2$, we would get the hamiltonian to be of the form

$$
\hat{H}_\text{kin} = -\frac{\hbar^2}{2M_e} \Delta + \frac{\text{something}}{4 \cdot \text{something}}.
$$

That four in the denominator is what's confusing me. The second $1/2$ is seemingly being ignored completely in our lecture notes and I've only managed to find one derivation online that used this method, where it just randomly disappeared from one line to another. Obviously, it would have to be $1/2$, not $1/4$, but I can't figure out where an additional factor of $2$ is supposed to come from, or where I introduced one $1/2$ too much. I really hope someone can explain to me what the issue is. I have speculated that it might be some correction factor I ignored, however, since we are not considering spin (normal Zeeman-Effect) it most likely isn't the gyromagnetic ratio, although it would deliver the missing $2$.

 

[anuttarasammyak]

Remenber that $\nabla$ is differentialtion operator, the last term in RHS should be doubled. i.e.
<br /> \begin{align*}<br /> <br /> \hat{H}_\text{kin} &amp;= \frac{1}{2M_e} \left( \hat{p} - q \vec{A} \right)^2 \\<br /> <br /> &amp;= \frac{1}{2M_e} \left( -i\hbar\nabla - q \vec{A} \right)^2 = \frac{1}{2M_e} \left( i \hbar \nabla + q \vec{A} \right)^2 \\<br /> <br /> &amp;= \frac{1}{2M_e} \left( -\hbar^2 \Delta + q^2 A^2 + i q \hbar (\nabla \cdot \vec{A}) + 2i q \hbar (\vec{A} \cdot \nabla) \right).<br /> <br /> \end{align*}
where $(\nabla \cdot \vec{A})$ means that it is closed, not operating differentiaition on wavefunction.

 

[PhysicsRock]

Remenber that $\nabla$ is differentialtion operator, the last term in RHS should be doubled.

How exactly do the fact that $\nabla$ is a differential operator and the doubling of the last term go together?

 

[pines-demon]

$$\frac1{2m}\left(\mathbf p - q\mathbf A\right)^2=\frac1{2m}\left(\mathbf p^2 - q\mathbf A\cdot \mathbf P - q\mathbf p \cdot \mathbf A+q^2\mathbf A^2\right)$$
$$\frac1{2m}\left(\mathbf p^2 - 2q\mathbf A\cdot \mathbf p - q[ p_i, A_i]+q^2\mathbf A^2\right)$$
$$\frac1{2m}\left(\mathbf p^2 - 2q\mathbf A\cdot \mathbf p +i\hbar q (\nabla \cdot \mathbf A)+q^2\mathbf A^2\right)$$
where I used that
$$[p_i, f(x_i)]=-i\hbar \frac{\partial f(x_i)}{\partial x_i}$$

Edit: if you try to verify this last relation you will see why the 2 factor true, try for example $[p_z,z^3]=-3i\hbar z^2$. The naive calculation will you a wrong extra term.

 

[pines-demon]

Here is another way to see the factor 2 appear more evidently

$$(a x - b p)^2=a^2x^2 + b^2p^2 - a b xp - ab px$$
where $a,b$ are constants. Now compare the two following results:
$$(a x - b p)^2\neq a^2x^2 + b^2p^2 - a b xp + ab i\hbar$$
$$(a x - b p)^2=a^2x^2 + b^2p^2 - 2 a b xp + ab i\hbar$$
the former is similar to your result, while the second one is right. Can you guess why?

 

[PhysicsRock]

Here is another way to see the factor 2 appear more evidently

$$(a x - b p)^2=a^2x^2 + b^2p^2 - a b xp - ab px$$
where $a,b$ are constants. Now compare the two following results:
$$(a x - b p)^2\neq a^2x^2 + b^2p^2 - a b xp + ab i\hbar$$
$$(a x - b p)^2=a^2x^2 + b^2p^2 - 2 a b xp + ab i\hbar$$
the former is similar to your result, while the second one is right. Can you guess why?

Yes, of course. That makes it pretty clear. Would've been nice to see this mentioned somewhere, instead of just doing it and not even addressing the trick. Thank you for your help!