Why Does Spin Down Have Lower Energy Than Spin Up in a Magnetic Field?

[docnet]

TL;DR

is spin up or spin down higher in energy?

I am confused about why spin down has a lower energy than spin up. What is the correct interpretation of the equations?

If we have a spin $\frac{1}{2}$ particle in a magnetic field $B_0$ that is applied in the positive z direction

The spin states of the particle are
$$\ket{up} = \begin{bmatrix} 1\\0\end{bmatrix}$$
$$\ket{down} = \begin{bmatrix} 0\\1\end{bmatrix}$$

and the hamiltonian for the spin in the magnetic field is $$\hat{H}=\frac{\hbar\omega}{2}\sigma_z$$

to find the energies of the spin states, we operate on $\ket{up}$ and $\ket{down}$ with $\hat{H}$

$$\hat{H}\ket{up} =\frac{\hbar\omega}{2} \begin{bmatrix} 1 & 0\\0 & -1\end{bmatrix}\begin{bmatrix} 1\\0\end{bmatrix}=\frac{\hbar\omega}{2}\begin{bmatrix} 1\\0\end{bmatrix}=\frac{\hbar\omega}{2}\ket{up}$$
$$\hat{H}\ket{down} =\frac{\hbar\omega}{2} \begin{bmatrix} 1 & 0\\0 & -1\end{bmatrix}\begin{bmatrix} 0\\1\end{bmatrix}=-\frac{\hbar\omega}{2}\begin{bmatrix} 1\\0\end{bmatrix}=-\frac{\hbar\omega}{2}\ket{down}$$

so the energy of $\ket{up}$ is $\frac{\hbar\omega}{2}$ and the energy of $\ket{down}$ is $-\frac{\hbar\omega}{2}$. The energy of $\ket{down}$ is lower than up..and it is a negative value. are states allowed to have negative energies? why does down have a lower energy than up?

 

[PeterDonis]

the hamiltonian for the spin in the magnetic field is $$\hat{H}=\frac{\hbar\omega}{2}\sigma_z$$

You left out a factor of $B_0$ here. That factor is necessary; see further comments below. (There is also a factor of the magnetic moment of the particle, but we can assume that to be $1$ for this discussion, although its sign is still important; see below.)

it is a negative value.

Only because you left out the rest of the Hamiltonian. The energies you are evaluating are really the difference in energy between the energy without any magnetic field present (in which case spin up and spin down both have the same energy) and the energy with the field present (in which case the energy of spin down is reduced and the energy of spin down is increased from the "no field present" value).

If you included the entire Hamiltonian, which, to be really complete, has to include the rest energy of the particle (i.e., its mass in energy units), the energies would all be positive.

Summary:: is spin up or spin down higher in energy?

why does down have a lower energy than up?

Because the field is pointing in the positive $Z$ direction, and the state you are calling "up" has its spin pointing in the same direction, and the way you have written the Hamiltonian (with a positive sign) makes that add energy to the state. If you reversed the direction of the field, you would also reverse the sign of the energy differences. To see this, put $B_0$ back into your Hamiltonian, as noted above; then flipping the sign of $B_0$ (which corresponds to reversing the direction of the field) will also flip the signs of the energies.

In other words, what you are really seeing here is that, given your assumptions, it takes more energy to have the spin pointing in the same direction as the magnetic field, than to have it pointing in the opposite direction from the magnetic field. Note that your assumptions implicitly include the charge of the particle (which appears in the sign of its magnetic moment, which, as above, also needs to be included as a factor in the Hamiltonian); if you flip the sign of the particle's charge, leaving everything else the same, you also flip the sign of the energies.

 

[vanhees71]

More precisely the corresponding part of the Hamiltonian is $\hat{H}=-\hat{\vec{\mu}} \cdot \vec{B}$. The relation between spin and magnetic moment (in non-relativistic QM) is $\hat{\vec{\mu}}=g q \hat{\vec{s}}/(2m)$, where $g$ is the gyromagnetic (or Lande) factor. For an elementary spin-1/2 particle like an electron it's close to two (the tree-level value from the Dirac equation/relativistic QED or in the non-relativistic approximation, i.e., the Pauli equation).