91Ó°ÊÓ

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 $\mathrm{Î\pm }$.

$\mathrm{Î\pm }Q$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.

$\mathrm{Î\pm }=\mathrm{Î\pm }*$

i.e., only if $\mathrm{Î\pm }$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,

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

Functions $f$and $g$are square-integrable which requires that they go to zero as $x→Â\pm ∞$. 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.