# Hamiltonian operator and spin operator

I've just started Quantum mechanics by McIntyre and have understood the following about operators :

• Each observable has an operator
• Operators act on kets to produce another kets.
• Only eigenvalues of an operator are possible values of a measurement.

Now the author introduces the Hamiltonian operator $$H$$ and says

The eigenvalues of the Hamiltonian are the allowed energies of the quantum system, and the eigenstates of $$H$$ are the energy eigenstates of the system.

I understood this.

Then the author discusses about a Spin 1/2 particle in a constant magnetic field along $$z$$ direction.

The Hamiltonian operator represents the total energy of the system... So to begin, we consider the potential energy of a single magnetic dipole (e.g., in a silver atom) in a uniform magnetic field as the sole term in the Hamiltonian. Recalling that the magnetic dipole is given by $$\mu=g \frac{q}{2 m_{e}} \mathbf{S}$$ the Hamiltonian is \begin{aligned} H &=-\mu \cdot \mathbf{B} \\ &=-g \frac{q}{2 m_{e}} \mathbf{S} \cdot \mathbf{B} \\ &=\frac{e}{m_{e}} \mathbf{S} \cdot \mathbf{B} \end{aligned} $$\mathbf{B}=B_{0} \hat{\mathbf{z}}$$ allows the Hamiltonian to be simplified to \begin{aligned} H &=\frac{e B_{0}}{m_{e}} S_{z} \\ &=\omega_{0} S_{z} \end{aligned} \tag 1 where $$\omega_{0} \equiv \frac{e B_{0}}{m_{e}}$$ The Hamiltonian is proportional to the $$S_{z}$$ operator.

• The way equation (1) was derived took $$H$$ to be energy and $$S$$ to be a vector therefore it isn't a operator relationship. Why then does the author say it is an operator relationship by saying that " The Hamiltonian is proportional to the $$S_{z}$$ operator" ?

• I understand that if a particle having a magnetic moment $$\mu$$ is in a magnetic field $$B$$ then it has energy $$E$$ (a scalar) given as $$E=-{\mu}. B$$

Now in QM we have an operator relationship between the Hamiltonian (operator) and magnetic moment (operator) exactly in the same form as $$H=-{\mu}.B$$

Why is that so?

Based on what the author has written so far as I've mentioned in starting of this post I cannot understand this correspondence.

Equation (1) is an operator relationship since, to derive that, you do the dot-product between $$\vec{S}$$ and $$\vec{B}$$, so you don't get a vector. So you have an expression of the operator $$H$$ in terms of another operator: $$S_z$$. In $$\hat{z}$$ direction because $$\vec{B}=B_0\hat{z}$$. So you can have your eingenstates and eingenvalues.
• "you do the dot-product between $\vec{S}$ and $\vec{B}$" aren't $\vec{S}$ and $\vec{B}$ vectors here? Commented Feb 4, 2022 at 8:49
• @Kashmiri You can have vector operators, $\mathbf O= (O_x,O_y,O_z)$ where each component is an operator. Commented Feb 4, 2022 at 9:33
• @Mauricio, you suggest and $\mathbf S$ is a vector having components as operators ? What about $\mu$ Commented Feb 4, 2022 at 11:31