Answer to Question #282727 in Quantum Mechanics for BIGIRMANA Elie

Question

7. Show that if the linear operators Aˆ and Bˆ do not commute, the operators

(AˆBˆ + BˆAˆ) and i[A, ˆ Bˆ] are Hermitian.

Answer 1

$Answer$

According to given question

$\hat{A}^{+}=\hat{A}$

$\hat{B}^{+}=\hat{B}$

$[\hat{A}, \hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A} \mathrlap{\,/}{=}0$

Now checking hermitian Or not

$(\hat{A}\hat{B}+\hat{B}\hat{A}) ^{+}\\=(\hat{A}\hat{B}) ^{+}+(\hat{B}\hat{A}) ^{+}\\= \hat{B}^{+}\hat{A}^{+}+\hat{A}^{+}\hat{B}^{+}\\=\hat{B}\hat{A}+\hat{A}\hat{B}$

So this is hermitian.

Now another part

$(i[\hat{A}, \hat{B}]) ^{+}\\=(-i)[\hat{A}, \hat{B}] ^{+}\\=(-i)(\hat{A}\hat{B}-\hat{B}\hat{A})^{+}\\=(-i)(\hat{B}^{+}\hat{A}^{+}-\hat{A}^{+}\hat{B}^{+}) \\=(-i)(\hat{B}\hat{A}-\hat{A}\hat{B})$

Taking negative sign common

$=i( \hat{A}\hat{B}-\hat{B}\hat{A}) =i[\hat{A},\hat{B}]$

So this is also hermitian.