(a) Show that the sum of two hermitian operators is hermitian.

(b) Suppose$\hat{\mathit{Q}}$is hermitian, and$\mathit{\alpha }$is a complex number. Under what condition (on$\mathit{\alpha }$) islocalid="1655970881952" $\mathit{\alpha }\hat{\mathit{Q}}$hermitian?

(c) When is the product of two hermitian operators hermitian?

(d) Show that the position operator $\left(\hat{x}=x\right)$and the hamiltonian operator

localid="1655971048829" $\hat{\mathit{H}}\mathbf{=}\mathbf{-}\frac{{\sout{\mathit{h}}}^{\mathbf{2}}}{\mathbf{2}\mathit{m}}\frac{{\mathit{d}}^{\mathbf{2}}}{\mathit{d}{\mathit{x}}^{\mathbf{2}}}\mathbf{+}\mathit{V}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$are hermitian.

Hermitian operator:

$\hat{H}=-\frac{\sout{h^2}}{2m}\frac{d^2}{d{x}^2}+V\left(x\right)$

a)

Let $\hat{A}$and$\hat{B}$be two hermitian operators and$\hat{C}=\hat{A}+\hat{B}$

We must show that $\hat{C}$is hermitian, i.e.

The equation's left-hand side can be represented as

localid="1655972339408"

Since $\hat{A}$and $\hat{B}$are hermitian we further have

From this it follows

(b)

We now consider a hermitian operator $\hat{Q}$and complex number $\alpha$.

$\alpha Q$is hermitian only if

The left-hand side can be expressed as

Similarly, the right -hand equals

Only if the left and right sides are equal can we observe that the two sides are equal.

$\alpha =\alpha *$

i.e., only if $\alpha$is real.

(c)

Let$\hat{A}$ and$\hat{B}$ be two hermitian operators and $\hat{C}=\hat{A}\hat{B}$.

We must determine under which conditions is $\hat{C}$is hermitian:

The left-hand side can be written as

Since and are hermitian we further have

Therefore, in order for

we musthave

$\hat{B}\hat{A}=\hat{A}\hat{B}$

Finally, we have to show that the position operator $\hat{x}$and operator

$\hat{H}=-\frac{\sout{h^2}}{2m}\frac{d^2}{d{x}^2}+V\left(x\right)$

are hermitian.

Consider first the position operator. We have to show that

The equation's left-hand side can be represented as

Since $x=x*$(real number) we further find

We have

The first terms can be rewritten as

We can perform integration by parts using

$dv=\frac{d^{2}g}{d{x}^2};u=f*$

So,

$\int f*\frac{d^{2}g}{d{x}^2}dx={\overline{)f*\frac{dg}{dx}}}_{-\infty }^{+\infty }-\int \frac{dg}{dx}\frac{df*}{dx}dx$

Functions $f$and $g$are square-integrable which requires that they go to zero as $x\to \pm \infty$. The first terms above then vanish. The second integral can be calculated by performing part-by-part integration once more:

role="math" localid="1655974397201"

This guarantees that the Hamiltonian as a whole is hermitian.