Boundary condition

Boundary Condition:

(1) For finite value of P,E or finite jump in ( discontinuity in ) P.E, the wave function \(\psi(r)\) and it's gradient \(\frac{d\psi}{dr}\) or \(\biggl(\frac{\partial \psi}{\partial x}, \frac{\partial \psi}{\partial y}, \frac{\partial \psi}{\partial z}\biggr)\) remains continuous.

\(\Rightarrow\) Proof:

Let \(\psi(x)\) be the wave function for a particle obtained from the solution of time independent Schrodinger equation.

$$\frac{-\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+ v(x)\psi(x)=E\psi(x)\dotsm(1)$$

Where, m= mass of particle

E= T.E of particle

v(x)= P.E of the particle.

Let V(x) has finite jump of the form,

$$V(x)= v_0\; For\; x\geq 0$$

$$= 0\; For \; x<0\dotsm(2)$$

Discountinuity is at x=0 in P.E by \(V_0\).

The motion of particle in the form of wavepacket is described by the independent Schrodinger equation. i.e. Equation (1) .

From equation (1)

$$\frac{d^2\psi(x)}{dx^2}= \frac{2m}{\hbar^2}[V(x)-E]\psi(x)\dotsm(3)$$

Integrating equation (3)) between \(0-\epsilon\) to \(0+\epsilon\) we get,

$$or,\;\; \int_{0-\epsilon}^{0+\epsilon}\frac{d^2\psi}{dx^2}\cdot dx= \frac{2m}{\hbar^2}\int_{0-\epsilon}^{0+\epsilon}(V(x)-E)\psi(x)dx$$

$$or,\;\; \int_{0-\epsilon}^{0+\epsilon}\frac{d}{dx}\biggl(\frac{d\psi}{dx}\biggr)dx=\frac{2m}{\hbar^2}\int_{0-\epsilon}^{0+\epsilon} (V(x)-E)\psi(x)dx$$

$$or,\;\ \biggl[\frac{d\psi(x)}{dx}\biggr]_{0-\epsilon}^{0+\epsilon}= \frac{2 m}{\hbar^2}\int_{0-\epsilon}^{0+\epsilon} (V(x)-E)\psi(x)dx$$

At boundary, i.e.

$$\lim_{\epsilon\to 0}\frac{d\psi(x)}{dx}\biggl|_{x=0+\epsilon}- \lim_{\epsilon\to 0} \frac{d\psi(x)}{dx}\biggr|_{x=0-\epsilon}$$

$$=\lim_{\epsilon \to 0}\int_{0-\epsilon}^{0+\epsilon} \frac{2m}{\hbar^2}[V(x)-E]\psi(x)dx$$

R.H.S of equation (4) in the limit \(\epsilon \to 0\dotsm(4)\) is zero. So,

$$\lim_{\epsilon\to 0}\frac{d\psi(x)}{dx}\biggr|_{x=0+\epsilon}- lim_{\epsilon\to 0}\frac{d\psi(x)}{dx}\biggr|_{x=0-\epsilon}=0$$

$$or,\;\; \lim_{\epsilon\to 0} \frac{d\psi(x)}{dx}\biggl|_{x=0+\epsilon}= \lim_{\epsilon\to 0}\frac{d\psi(x)}{dx}\biggr|_{x=0-\epsilon}$$

$$\Rightarrow \frac{d\psi_{II}(x)}{dx}\biggl|_{x=0}=\frac{d\psi_I(x)}{dx}\biggr|_{x=0}\dotsm(5)$$

Hence gradient of wave function across the boundary is continuous at change in potential is finite.

Where, \(\psi_I(x)\) and \(\psi_II(x)\) represents the wave function of particle in region I and II as shown in figure.

The continuity of gradient of wave function implies that the wave function itself must be continuous across the boundary.

$$\psi_I(x)\biggl|_{x=0}= \psi_{II}(x)\biggr|_{x=0}\dotsm(6)$$

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.