1
$\begingroup$

I just wanted to check that I carried out this problem correctly. I got the correct answer, but I'm not sure if what I did to get it is completely correct. This is from the second part of problem 3.4 in the second edition of Modern Quantum Mechanics from Sakurai. We have a Hamiltonian given by \begin{equation} H=A\mathbf{S}^{(e^-)}\cdot\mathbf{S}^{(e^+)} \end{equation} acting on an electron-positron system in the state $\chi_{+}^{(e^-)}\chi_{-}^{(e^+)}$, which I took to be the tensor product of the individual spin states: \begin{equation} \chi_{+}^{(e^-)}\chi_{-}^{(e^+)}=\begin{pmatrix}1\\0\end{pmatrix}_{e^-}\otimes\begin{pmatrix}0\\1\end{pmatrix}_{e^+}. \end{equation} The two spin operators are \begin{equation} \mathbf{S}^{(e^-)}=\frac{\hbar}{2}\left(\sigma_x^{(e^-)}+\sigma_z^{(e^-)}+\sigma_z^{(e^-)}\right),\quad \mathbf{S}^{(e^+)}=\frac{\hbar}{2}\left(\sigma_x^{(e^+)}+\sigma_z^{(e^+)}+\sigma_z^{(e^+)}\right), \end{equation} so in taking their dot product, I assumed that the Hamiltonian in terms of $\sigma_i^{(e^-)}$ and $\sigma_i^{(e^=)}$ would be \begin{equation} H=\frac{A\hbar^2}{4}\left(\sigma_x^{(e^-)}\otimes\sigma_x^{(e^+)}+\sigma_y^{(e^-)}\otimes\sigma_y^{(e^+)}+\sigma_z^{(e^-)}\otimes\sigma_z^{(e^+)}\right). \end{equation} Is this the proper way of taking the dot product of the two spin operators? I wasn't quite sure how to handle the dot product since the two operators should live in two different Hilbert spaces (or they simply act on their respective spaces and don't actually live in them?). When I have the Hamiltonian act on the state $\chi_+^{(e^-)}\chi_-^{(e^+)}$, I end up with \begin{align} H\chi_+^{(e^-)}\chi_-^{(e^+)} &= \frac{A\hbar^2}{4}\left[\begin{pmatrix}0\\1\end{pmatrix}_{e^-}\otimes\begin{pmatrix}1\\0\end{pmatrix}_{e^+}+\begin{pmatrix}0\\i\end{pmatrix}_{e^-}\otimes\begin{pmatrix}-i\\0\end{pmatrix}_{e^+}+\begin{pmatrix}1\\0\end{pmatrix}_{e^-}\otimes\begin{pmatrix}0\\-1\end{pmatrix}_{e^+}\right]\\ &= \frac{A\hbar^2}{4}\left[2\begin{pmatrix}0\\1\end{pmatrix}_{e^-}\otimes\begin{pmatrix}1\\0\end{pmatrix}_{e^+}-\begin{pmatrix}1\\0\end{pmatrix}_{e^-}\otimes\begin{pmatrix}0\\1\end{pmatrix}_{e^+}\right]\\ &= \frac{A\hbar^2}{4}(2\chi_-^{(e^-)}\chi_+^{(e^+)}-\chi_+^{(e^-)}\chi_-^{(e^+)}). \end{align} And when I take the expectation of the Hamiltonian, I get \begin{align} \langle H\rangle &= \chi_+^{(e^-)}\chi_-^{(e^+)}\cdot H\chi_+^{(e^-)}\chi_-^{(e^+)}\\ &= \frac{A\hbar^2}{4}(2\chi_+^{(e^-)}\chi_-^{(e^+)}\cdot\chi_-^{(e^-)}\chi_+^{(e^+)}-\chi_+^{(e^-)}\chi_-^{(e^+)}\cdot\chi_+^{(e^-)}\chi_-^{(e^+)})\\ &= -\frac{A\hbar^2}{4}, \end{align} which is the correct answer. But again, I'm not sure about some of the operations that I carried out in writing $\chi_+^{(e^-)}\chi_-^{(e^+)}$ and the spin operators explicitly as tensor products, and I hope I didn't get the right answer entirely by accident.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.