# Hamiltonian of an electron in a magnetic field

Suppose I have an electron in a magnetic field given by:

$$\vec{B}=B\hat{z}$$

The potential energy of this system is given by:

$$U=-\vec{\mu} \cdot \vec{B}=\frac{g\mu_B}{\hbar}\vec{S} \cdot \vec{B}$$

Here, $$\vec{\mu}$$ is the magnetic moment of the electron, $$g$$ is the Lande $$g$$-factor, $$\mu_B$$ is the Bohr Magneton, and $$\vec{S}$$ is the spin of the electron.

This shows that from a statistical mechanics perspective, electrons with spin oriented in the direction of the magnetic field have higher energy than the ones with spin antiparallel to $$B$$. Moreover, magnetic moment and spin point in the opposite directions.

Anyway, when solving, we simply note that:

$$\vec{S}\cdot\vec{B}=\hat{S_3} B_z$$

Since the electrons are either in $$|\uparrow\rangle$$ or in the $$|\downarrow\rangle$$ state, the expectation value of this is nothing but the eigenvalues corresponding to these states, i.e. $$\frac{\hbar}{2}$$ and $$-\frac{\hbar}{2}$$ respectively.

This is how we obtain the energy for parallel and antiparallel configurations of the spin and the magnetic field.

From what I understand, till now, we were basically finding the expectation value of $$\hat{U}$$ for parallel and antiparallel spins.

In general, we should have,

$$\hat{U}=\frac{g\mu_B}{\hbar}\vec{S} \cdot \vec{B}=\frac{g\mu_B}{\hbar}\frac{\hbar}{2}\vec{\sigma} \cdot \vec{B}\approx\mu_B\space\hat{\sigma_3}{B_z}$$

To obtain the energy of the parallel configuration, we would take $$\langle\uparrow |\hat{U}|\uparrow\rangle$$.

Similarly, we can obtain an expression for the antiparallel configuration.

My question is, whether:

$$\hat{U}=+\mu_B\space \hat{\sigma_3} B_z$$

or is it:

$$\hat{U}=-\mu_B\space \hat{\sigma_3} B_z \ \ \ ?$$

Since magnetic moment and angular momentum should be in the opposite direction for negatively charged particles, I believe it should be the former. Wikipedia agrees with this viewpoint.

However, in many texts on quantum statistical mechanics, like Pathria for example, the latter is said to be true.

Can someone point out which one of the two expressions is correct?

• I don't understand what you mean by "replacing" $\vec S$ by $\frac{\hbar}{2}$. You have $\vec B = B\hat{z}$ and so $\vec S \cdot \vec B = BS_z\hat{z}$ (no approximation!). The possible eigenvalues of $S_z$ are $\pm\frac{\hbar}{2}$. What exactly is the question? Jun 11, 2022 at 15:17
• @ACuriousMind That was my first question, and I think I've found the problem. My issue was, as you said, $\vec{S}.\vec{B}=B_zS_z$. However, $S_z$ is an operator, and we are replacing it directly with its eigenvalue. I think I was just confused by the notation, as I see now, that we are essentially calculating the expectation value when the electron has a parallel spin, and the expectation value when the spin is antiparallel. Jun 11, 2022 at 15:23
• @ACuriousMind I was essentially confused about how we were replacing an operator by it's eigenvalue. I noticed later that we are evaluating $U$ and not $\hat{U}$. Essentially it is the expectation value. I'm editing out my first question, but can someone solve my second question ? Jun 11, 2022 at 15:25

For spin 1/2 fermions

$$$$\boldsymbol S = \frac{\hbar}{2}\boldsymbol\sigma$$$$

where $$\boldsymbol\sigma$$ is a vector of the three Pauli matrices $$(\sigma_x, \sigma_{y}, \sigma_z).$$ Furthermore $$\boldsymbol \mu = -\frac{g_{S}\mu_{B}}{\hbar} \boldsymbol S$$, where the spin g-factor of an electron is approximately 2. For $$\boldsymbol B = (0, 0, B_{z})$$ your Hamiltonian is given by

$$$$H = \mu_{B}B_{z}\sigma_{z}$$$$

Note that the sign has changed because the charge of the electron is negative. This means electrons and protons actually behave in opposite manners in magnetic fields, i.e. they have opposite magnetic moments.

The two Zeeman Hamiltonians you have written are equivalent up to a unitary transformation $$UHU^{-1} = -H$$ where $$U=\sigma_{x}$$. This essentially amounts to a relabelling of the spin components ( $$\uparrow \rightarrow \downarrow$$ and vice-versa).

• The article on Wikipedia disagrees. According to that, $\vec{\mu}=-\frac{g\mu_B}{\hbar}\vec{S}$. The same is true for the article on hyperphysics website. I have no idea, where this difference is coming from. Jun 11, 2022 at 15:40
• For example, in Wikipedia, $g\approx 2$. Moreover, $\mu_B > 0$. Is there a different sign in the answer you provided? Jun 11, 2022 at 15:41
• Apologies, I've edited the answer to account for the negative signs that I erroneously omitted. I had a brief look at the Parthia text you mentioned and it does appear that there should be a sign difference in their definition. Of course, it doesn't make any difference as they end up diagonalising the Hamiltonian and calculating the density matrix for the canonical ensemble. Jun 12, 2022 at 8:55
• It does make some sense now. If one calculates the expectation value, then yes, it doesn't make a difference. However, if someone wants to calculate the average number of particles, with spin aligned parallel to the field, they would get the wrong answer by Pathria's convention, unless they relabel the spin, as you've mentioned. That does seem like an unnecessary complication. Jun 12, 2022 at 14:11