Spin Hamiltonian of a hydrogen atom in a magnetic field

[Nullity]

Homework Statement

5.1 of the first picture is the homework statement and the second is the solution. So my question is in the solution, how did they compute <1|w_0S_z|1>?

Relevant Equations

The book uses Dirac notation for matrix computation but this homework is not for a class and is strictly for recreation so feel free to solve it using the mathematical methods you feel comfortable with.

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[TSny]

Can you give us some indication of why you are having trouble evaluating $\langle 1 |\omega_0 \hat S_{1z }|1\rangle$? If you have made an attempt, please show us what you have done so we can tell where you are having difficulty. If you cannot get started with an attempt, then it is likely that you are having trouble with the notation. For example, are you clear on the meaning of the state $|1 \rangle$? Can you describe the spins of the electron and the proton in this state?

One possible source of confusion is with the notation of the operator $\hat S_{1z }$ . The subscript 1 denotes the electron. But the $1$ in the state $|1\rangle$ denotes a particular state of the electron-proton system. When $\hat S_{1z }$ acts on the state $|1 \rangle$, the operator $\hat S_{1z }$ acts only on the electron part of the state $|1\rangle$. It is understood that this operator does not act on the proton. Or, in other words, the operator $\hat S_{1z }$ acts as the identity operator $\mathbb { I}_2$ when acting on the proton part of the state. The subscript 2 here refers to the proton. The proper way to write the second term of the Hamiltonian would be as a direct product of operators: $\hat S_{1z } \otimes \mathbb {I}_2$. But for convenience, the $\otimes \mathbb {I}_2$ part is not written and is assumed to be understood.

Thus, consider a state $|\Psi \rangle = |\alpha_1 \rangle |\beta_2 \rangle$ in which the electron is in some spin state $|\alpha_1 \rangle$ while the proton is in spin state $|\beta_2 \rangle$. Then $$\langle \Psi|\hat S_{1z}|\Psi \rangle = \langle \beta_2|\langle \alpha_1|\hat S_{1z }\otimes \mathbb {I}_2|\alpha_1 \rangle |\beta_2 \rangle = \langle \alpha_1|\hat S_{1z }|\alpha_1\rangle \langle \beta_2|\mathbb {I}_2|\beta_2\rangle = \langle \alpha_1|\hat S_{1z }|\alpha_1\rangle \langle \beta_2|\beta_2\rangle$$When you get used to this, you will mentally go directly from the far left to the far right of this series of steps without having to write the intermediate steps.