- #1

physicsxanime

- 13

- 3

- Homework Statement
- See screenshot next

- Relevant Equations
- hermitian operator ##A^\dagger = A##

antihermitian operator ##A^\dagger = -A##

A. I can show that A is either hermitian or antihermitian by

$$(B^\dagger B=1-A^2)^\dagger$$

$$B^\dagger B=1-A^\dagger A^\dagger$$

comparing, we know that

$$A^\dagger = \pm A$$

I don't know how I can make use of the communtation relation to get hermiticity of B. But I know that A and B must have same hermiticity because hermitian * antihermitian = 0 and the communtation relation will not make sense.

B. The fact that the question hints on only the other eigenstate is weird, does the above imply A and B are 2x2?

Anyway, consider the state ##B|a=0>##

$$AB|a=0> = BA|a=0> + B|a=0> = 0 + B|a=0>$$

Therefore, ##B|a=0>## is a eigenstate of eigenvalue 1.

C. This need result in A.

note that we can work in the basis of A

##|a=0> =

\begin{pmatrix}

1\\

0

\end{pmatrix}

##

##B|a=0> =

\begin{pmatrix}

0\\

1

\end{pmatrix}

##

$$

We see ##B

\begin{pmatrix}

0\\

1

\end{pmatrix}

= B^2|a=0> = \pm B^\dagger B|a=0> = \pm(1-A^2)|a=0> = \pm |a=0>

##

Very simple matrix to diagonalize