Proof associated with Hermitian Operator

Show that:

\((a) \hat P_x = -i\hbar \frac{\partial }{\partial x}\) is Hermitian operator.

\((b) \biggl(\frac{-d}{dx}\biggr) \) is not Hermitain.

\((c) E_{op} = i\hbar \frac{\partial}{\partial t}\) is Hermitian.

Solution:

Let \(\psi(x) \) be valid wave function associated with a system.

Now, If \(\hat P_x\) is hermitian operator. Then,

$$\int_{-\infty}^\infty \psi^*(x)[\hat P_x \psi(x)]dx= < \hat P_x>$$

For any normalized wave function \(\psi\),

$$\int \psi^*\psi d\tau=1$$

Here, \(\hat P_x= -i\hbar \frac{\partial}{\partial x}\) and it's complex conjugate is

$$\hat P_x= i\hbar \frac{\partial}{\partial x}$$

differentating equation (1) with respect to 'x' on both side,

$$\frac{\partial}{\partial x}\int \psi^*\psi d\tau=0$$

$$or,\int \frac{\partial \psi^*}{\partial x}\psi d\tau= - \int \psi^*\frac{\partial\psi}{\partial x}d\tau$$

$$or, \int \psi^*\biggl(-i\hbar \frac{\partial \psi}{\partial x}d\tau= \int \biggl(-i\hbar \frac{\partial}{\partial x} biggr\psi )^* \psi d\tau$$

$$or, \int\psi^*(\hat P \psi)d\tau= \int (\hat p \psi)^* \psi d\tau$$

$$Or$$

\int \psi^*(x)[\hat P_x \psi(x)]dx= \int [\hat P_x \psi(x)]^* \psi(x) dx $$

Which show that \(\hat P_x= -i\hbar \frac{\partial}{\partial x}\) is hermitian operator.

Again, For any normalized wave function \(\psi\).

$$\psi^*(x)\psi(x)dx=1$$

Differentiating with respect to t, we get

$$\int \psi^* \frac{\partial \psi}{\partial t}dx+ \int \frac{\partial \psi^*}{\partial t}\psi dx=0$$

$$or,\; \; \int \psi^*\frac{\partial \psi}{\partial t} dx= -\int \frac{\partial \psi^*}{\partial t}\psi dx$$

$$or,\;\; \int \psi^*\biggl(i\hbar \frac{\partial \psi}{\partial t}\biggr)dx= \int \biggl(i\hbar \frac{\partial \psi}{\partial t}\biggr)^* \psi dx$$

$$or,\;\; \int \psi^*(\hat H \psi)dx= \int (\hat H \psi)^* \psi dx$$

Hence, \(\hat H = i\hbar \frac{\partial }{\partial t} \) is Hermitian

$$OR$$

$$\int \psi^*(r)[\hat H \psi(r)]d\tau= \int [\hat H \psi(r)]^* \psi(r)d\tau$$

$$\int \psi^*(r)[ \hat E_{op} \psi(r)]d\tau= \int [ \hat E_{op}\psi(r)]^* \psi(r)d\tau$$

Show that

\(\hat D= \frac{-d}{dx}\) is not Hermitian.

We have,

$$\int \psi^*\psi dx=1$$

Differentiating with respect to x

$$-\frac{d}{dx}\biggl(\int \psi^*\psi dx\biggr)=0$$

$$OR$$

$$\int \psi^*(x)[\hat D \psi(x)]dx= \int \psi^*(x)\biggl[\frac{-d}{dx} \psi(x)\biggr]dx$$

$$=(-1)\int_{-\infty}^\infty \psi^*(x)\frac{d}{dx}\psi(x)dx$$

$$=(-1)\biggl([\psi^*(x)\psi(x)]_{-\infty}^\infty+ \int_{-\infty}^\infty \frac{d\psi^*}{dx}\psi(x)dx\biggr)$$

$$=0+ (-1) \int_{-\infty}^\infty \frac{d}{dx}\psi^*(x)\psi(x)dx$$

$$=-\int_{-\infty}^\infty \biggl[\frac{-d}{dx}\psi(x)\biggr]^* \psi(x)dx$$

$$\therefore\;\; \int_{-\infty}^\infty \psi^*(x)[\hat D \psi(x)]dx\ne \int_{-\infty}^\infty [\hat D \psi(x)]^* \psi(x)dx$$

Hence, \(\hat D= \frac{-d}{dx}\) is not Hermitian.

$$[\hat A, \hat B^n ] = n[ A, B]B^{n-1}$$

Solution:

$$[\hat A, \hat B^n]= [\hat A, \hat B\cdot \hat B^{n-1}]$$

$$=\hat B[ \hat A, \hat B^{n-1}]+ [ \hat A, \hat B] \hat B^{n-1}\dotsm(1)$$

$$= [ \hat A, \hat B]\hat B^{n-1}+ \hat B[ \hat A \hat B \cdot \hat B^{n-2}]$$

$$=[\hat A, \hat B]\hat B^{n-1}+ \hat B\biggl( \hat B[\hat A, \hat B^{n-1}]+ [ \hat A, \hat B] \hat B^{n-2}\biggr)$$

$$=[\hat A, \hat B]\hat B^{n-1}+ \hat B^2[ \hat A, \hat B^{n-1}]+ [ \hat A, \hat B] \hat B^{n-1}$$

$$= 2[\hat A, \hat B] \hat B^{n-1}+ \hat B^2 [ \hat A, \hat B^{n-2}]$$

Continuing this process for r-times,

$$[\hat A, \hat B^n]= r[\hat A, \hat B]\hat B^{n-1}+ \hat B^r [\hat A, \hat B^{n-r}]$$

Put r=n-1

$$= (n-1) [ \hat A, \hat B] \hat B^{n-1}+ \hat B^{n-1}[\hat A, \hat B]$$

$$=n[\hat A, \hat B] \hat B^{n-1}- [ \hat A, \hat B] \hat B^{n-1}+ [ \hat A, \hat B]\hat B^{n-1}$$

$$= n[\hat A, \hat B] \hat B^{n-1}$$

$$\therefore\;\; [ \hat A, \hat B^n]= n[ \hat A, \hat B] \hat B^{n-1}$$

Non- Commutating Operators:

Let \(\hat A\) and \(\hat B\) be two operators such that, \(\hat A= \frac{d}{dx}\) and \(\hat B= \hat x\) then,

$$\hat A\hat B\psi= \frac{d}{dx}(\hat x \psi)= \psi+ x\frac{d\psi}{dx}$$

In this case the eigen value of two operator can not be measure simultaneously if one quantity is measure accurate then there will be greater error in another quantity.

and

$$\hat B \hat A \psi= x\frac{d\psi}{dx}$$

$$[\hat A\hat B- \hat B\hat A]\psi= \psi$$

$$\therefore\; [ \hat A \hat B- \hat B\hat A]=1\dotsm(1)$$

Any operator satisfying the equation (1) is called non-commutating.

Reference:

1. Mathews, P.M and K Venkatesan. A Text Book of Quantum Mechanics. New Delhi: Tata McGraw Hill Publishing Co. Ltd, 1997.
2. Merzbacher, E. Quantum Mechanics . New York: John Wiley, 1969.
3. Prakash, S and S Salauja. Quantum Mechanics. Kedar Nath Ram Nath Publishing Co, 2002.
4. Singh, S.P, M.K and K Singh. Quantum Mechanics. Chand & Company Ltd., 2002.