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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.

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1 Answer 1

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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.

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  • $\begingroup$ "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
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    $\begingroup$ @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
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    $\begingroup$ @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

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