# Pauli equation: hermite adjoint when deriving probability density

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.

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 $$\mathbf{B}^T$$.
$$\left( \hat \sigma \cdot \mathbf{B} \right )= \left( \hat \sigma \cdot \mathbf{B} \right )^\dagger;$$ it is the sum of three Pauli matrices, with real coefficients, the components of the real magnetic field.