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.


1 Answer 1


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.

  • $\begingroup$ "you do the dot-product between $\vec{S}$ and $\vec{B}$" aren't $\vec{S}$ and $\vec{B}$ vectors here? $\endgroup$
    – Kashmiri
    Feb 4, 2022 at 8:49
  • 1
    $\begingroup$ @Kashmiri you can only perform a dot product between two vectors, resulting in a scalar. $\endgroup$
    – Mauricio
    Feb 4, 2022 at 9:17
  • $\begingroup$ @Mauricio, yes that's what I'm saying. They are vectors and not operators $\endgroup$
    – Kashmiri
    Feb 4, 2022 at 9:20
  • 4
    $\begingroup$ @Kashmiri You can have vector operators, $\mathbf O= (O_x,O_y,O_z)$ where each component is an operator. $\endgroup$
    – Mauricio
    Feb 4, 2022 at 9:33
  • $\begingroup$ @Mauricio, you suggest and $\mathbf S $ is a vector having components as operators ? What about $\mu$ $\endgroup$
    – Kashmiri
    Feb 4, 2022 at 11:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.