0
$\begingroup$

When trying to derive the probability density from the Pauli equation, I face a problem. Starting from the Pauli equation $$ i\hbar \frac{\partial \Psi}{\partial t}=\hat H_0 \Psi +\mu_B \ \hat \sigma \cdot \mathbf{B} \Psi, $$ I need to adjoint it: $$ -i\hbar \frac{\partial \Psi^+ }{\partial t}=\hat H_0^* \Psi^+ +\mu_B \ \left( \hat \sigma \cdot \mathbf{B} \Psi \right)^+. $$ So, I'm trying to calculate the multiplication in brackets, using the properties of conjugation $(AB)^+=B^+A^+$: $$ \left( \hat \sigma \cdot \mathbf{B} \Psi \right)^+ \equiv \bigg( \left( \hat \sigma \cdot \mathbf{B} \right) \Psi \bigg)^+= \Psi^+ \left( \hat \sigma \cdot \mathbf{B} \right)^+ = \Psi^+ \mathbf{B}^+ \hat \sigma^+= \Psi^+ \mathbf{B}^T \hat \sigma. $$ Here I've used the facts that the magnet field is real $(\mathbf{B}^+=\mathbf{B}^T)$ and pauli matrices are hermitian $(\hat \sigma^+ =\hat \sigma)$.

However, in the book (Greiner, Quantum Mechanics: an introduction) there is another answer: $$ \left( \hat \sigma \cdot \mathbf{B} \Psi \right)^+ = \Psi^+ \hat \sigma \cdot \mathbf{B}. $$ Where is the mistake? Thanks in advance.

P.S. I understand that $\hat \sigma$ is an operator, and so it must act some function, but... I still don't see my mistake.

$\endgroup$
0
$\begingroup$

Your textbook is right. In fact, your third equation mistake is not even a mistake, since you omitted a dot, so your answer is meaningless, as it stands.

What you need to understand is the Hermitian 2×2 matrix
$$ \left( \hat \sigma \cdot \mathbf{B} \right )= \left( \hat \sigma \cdot \mathbf{B} \right )^\dagger; $$ it is the sum of three Pauli matrices, with real coefficients, the components of the real magnetic field.

So a good Hermitian piece of the Hamiltonian. So its Hermitian conjugate is just itself. You may think of the vector consisting of the three Pauli matrices and that of the magnetic field, but the two are dotted, so you have a scalar, ignorant of transpositions. The only Hermitian conjugation involved is that of each and all Pauli matrices, so the book answer is trivially right. Try an explicit example.

$\endgroup$
1
$\begingroup$

There is no difference between your result $$\Psi^+\mathbf{B}^T\hat{\sigma}$$ and Greiner's result $$\Psi^+\hat{\sigma}\cdot\mathbf{B}.$$ Both evaluate to $$\Psi^+(\hat\sigma_x B_x+\hat\sigma_y B_y+\hat\sigma_z B_z).$$ Remember $B_j$ are just real numbers (1x1 matrices). Therefore it is pointless to distinguish between $\mathbf{B}$, $\mathbf{B}^+$ and $\mathbf{B}^T$.

$\endgroup$

Your Answer

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

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