# Tag Info

5

In general, we can say that $C=AB$ will have real, imaginary and complex eigenvalues (complex of the form $z=a+ib$ where and $\{a,b\in \mathbb{R}\mid a,b \ne 0\}$ as shown in the comments by Mark and Qmechanic's answer). For example, if $$A=\begin{bmatrix} 0 &1 \\ 1& 0 \end{bmatrix}\ \ \text{and}\ \ B=\begin{bmatrix} 1 & 0\\ 0& -1 \end{... 4 Suppose A and B are Hermitian operators: what is the nature of the eigenvalues of [A,B]+\{B,A\}=2AB? First, "A is Hermitian" from the practical point of view means that: A=A^\dagger, it is unitarily diagonalizable and has real eigenvalues. The same for B. Now, (AB)^\dagger = B^\dagger A^\dagger =BA: this tells us that, in general AB\... 4 TL;DR: Assuming that A,B are self-adjoint, the product AB does not need to be diagonalizable. And if AB is diagonalizable, the eigenvalues need not be real or imaginary. Example 1: AB is not diagonalizable:$$A~=~\begin{pmatrix} 0 & 1 \cr 1 & 0 \end{pmatrix} \quad\wedge\quad B~=~\begin{pmatrix} 1 & 0 \cr 0 & 0 \end{pmatrix}\quad\...

3

As @Cream points out in a comment, no, you do not have enough information to draw that conclusion. What you are asking for is if: $$[[\hat{A},\hat{B}], \hat{A}] = 0 \quad \quad \stackrel{\color{red}{?}}{\implies} \quad \quad [\hat{A}, \hat{B}] = 0.$$ But what you have "shown" instead is that: $$[\hat{A},\hat{B}] = 0 \quad \quad \implies \quad \quad ... 1 You are mixing properties that are defined for |\uparrow_z\rangle  with those defined for |\uparrow_x\rangle . The right relations are$$S_z|\uparrow_z\rangle = \frac\hbar2 |\uparrow_z\rangle;S_z|\downarrow_z\rangle = -\frac\hbar2 |\downarrow_z\rangle,$$and thus$$S_z|\uparrow_x\rangle = \alpha\frac\hbar2 |\uparrow_z\rangle-\beta \frac\hbar2 |\...

1

You will need: Commutation relation of $\mathbf r,\mathbf p$, given by $[r_i,p_j]=i\hbar\delta_{ij}$ where $\delta_{ij}$ is Kronecker's delta The following identity for the Pauli matrices $$\sigma_i\sigma_j=\delta_{ij}+i\epsilon_{ijk}\sigma_k$$ where $\epsilon_{ijk}$ is the Levi-Civita symbol. The rest is just willful work.

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