Orthogonality of wavefunction and eigen value

Orthogonal wave function:

The wave functions \(\psi_1(x)\) and \(\psi_2(x)\) are called orthogonal if,

$$\int \psi_1^*(x)\psi_2(x)dx=\int \psi_1(x)\psi_2^*(x)dx=0$$

Example: Show that \(\psi_1(x)=1\) and \(\psi_2(x)=sinx\) are orthogonal in the range \([-\pi,\pi]\).

Here, $$\int_{-\pi}^\pi \psi_1(x)\psi_2^*(x)dx=\int_{-\pi}^\pi 1.(sinx)^*dx$$

$$=|1-cos|_{-\pi}^\pi$$

$$=[-cos\pi+cos(-\pi)]$$

$$=0$$

Hence \(\psi_1(x)\) and \(\psi_2(x)\) are orthogonal.

Orthonormal wave function:

Two wave function \(\psi_1(x)\) and \(\psi_2(x)\) are called orthonormal if they are orthogonal to each other and self normalized. i.e

$$\int \psi_n^*(r,t)\psi_m(r,t)dT=0 \; If \; m\ne n$$

$$\int \psi_n^*(r,t)\psi_m(r,t)dT=1 \; If \; m= n$$

Example: Orthonormalize \(\psi_1(x)=1 \) and \(\psi_2(x)= sinx\) [\(-\pi,\pi\)]

Here, $$\int_{-\pi}^\pi \psi_1^*(x)\psi_2(x)dx=\int_{-\pi}^\pi sinxdx=0$$

and, $$\int_{-\pi}^\pi \psi_1(x)\psi_2^*(x)dx=\int_{-\pi}^\pi sinxdx=0$$

And,

$$\psi{1N}=A\psi_1$$

Then, $$\int_{-\pi}^\pi \psi_{1N}^* \psi_{1N}dx=1$$

$$\int_{-\pi}^\pi (A1)^*(A1)dx=1$$

$$\Rightarrow |A|^2\int_{-\pi}^\pi dx=1$$

$$\Rightarrow |A|^2=\frac{1}{2\pi}$$

$$\Rightarrow |A|=\frac{1}{\sqrt{2\pi}}$$

$$\therefore\;\; \psi_{1N}=A\psi_1=\frac{1}{\sqrt{2\pi}}; -\pi\leq x\leq \pi$$

Similarly, \(\psi_{2N}=A\psi_2=\frac{1}{\sqrt{\pi}} sinx\); \(-\pi\leq x\leq \pi\)

Question: Find the expectation value of \(\hat P_x\). If \(\psi(x)=\sqrt{\frac{2}{L}}sin\frac{n\pi x}{L}\) ; \(o\leq x\leq L\).

\(\Rightarrow\) Solution: Here, $$\int_0^L \psi^*(x)\psi(x)dx=1$$

$$\Rightarrow \frac 2L \int_0^L sin^2\frac{n\pi}{L} xdx$$

$$\Rightarrow \frac 2L \int_0^L\frac{(1-cos\frac{2n\pi}{L} x)}{2}dx$$

$$\Rightarrow \frac 2L \frac{x}{2}|_0^L$$

$$\Rightarrow \frac 2L \biggl[\frac L2- \frac 02\biggr]$$

$$\Rightarrow 1$$

i.e. \(\psi\) is normalized.

Now,

$$<P_x>=\int_0^L \psi^*(x)[\hat P_x \psi(x)]dx$$

$$=\frac 2L \int_0^L sin\frac{n\pi}{L} x\biggl[-\hbar \frac{\partial}{\partial x} sin\frac{n\pi}{L} x\biggr] dx$$

$$=\frac 2L (-i\hbar)\int_0^L sin\frac{n\pi}{L} x\cdot cos \frac{n\pi}{L} x\cdot \frac{n\pi}{L} dx$$

$$=\frac{-2i\hbar}{L^2}\cdot n\pi\int_0^L sin\frac{n\pi}{L}x\cdot cos\frac{n\pi}{L}xdx$$

$$=\frac{-i\hbar}{L^2}n\pi \int_0^L sin2\frac{n\pi}{L} xdx$$

$$=\frac{-i\hbar}{L^2} n\pi \biggl|\frac{-cos\frac{2n\pi}{L} x}{\frac{2n\pi}{L}}\biggr|_0^L$$

$$=\frac{-i\hbar}{2L} (-cos2n\pi+ cos 0)$$

$$<P_x>0$$

The orthogonality of two wavefunction indicates that they do not have common eigen values.

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.