Solution of Schrodinger equation

Solution of Equation:

$$i\hbar\frac{\partial \phi(t)}{\partial t}= E\phi(t)$$

$$\Rightarrow i\hbar\frac{\partial phi(t)}{\partial t}=E\phi(t)\dotsm(1)$$

$$or,\;\; \frac{\partial\phi(t)}{\phi(t)}=\frac{E}{i\hbar}\partial t$$

Integrating above equation,

$$\int\frac{\partial \phi(t)}{\phi(t)}=\frac{E}{i\hbar}\int \partial t$$

$$\therefore\;\; log_e[\phi(t)]=\frac{E}{i\hbar}t+K_1\dotsm(2)$$

using boundary condition;

At t=0, \(\phi(t)=\phi(0)\)

$$log_e[\phi(0)]=K_1$$

From (2)

$$\therefore\;\; log_e[\phi(t)]=\frac{Et}{i\hbar}+log_e\phi(0)$$

$$or,\;\; log_e\biggl[\frac{\phi(t)}{\phi(0)}\biggr]=\frac{Et}{i\hbar}$$

$$or,\;\; log_e\biggl(\frac{\phi(t)}{\phi(0)}\biggr)=\frac{i}{i^2}\frac{Et}{\hbar}=\frac{-iEt}{\hbar}$$

$$or,\;\; \frac{\phi(t)}{\phi(0)}=e^{\frac{-iEt}{\hbar}}$$

$$\therefore\;\; \phi(t)= \phi(0) e^{\frac{-iEt}{\hbar}}\dotsm(3)$$

Equation (3) gives the time-evolution of wave function ( packet ).

Solution of Equation: Schrodinger equation in one dimension.

The one dimensional time independent Schrodinger equation is,

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

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

$$or,\;\;\frac{d^2\psi(x)}{ds^2}=\frac{-2m}{\hbar^2}[E-V(x)]\psi(x)$$

$$or,\;\; frac{d^2\psi(x)}{dx^2}+\frac{2m}{\hbar^2}[E-V(x)]\psi(x)=0\dotsm(4)$$

Solution of equation (4) depends upon the value of V(x) and specified boundary condition.

Case (1): For V(x)=0 [Free particle]

$$\frac{d^2\psi(x)}{dx^2}=\frac{2m}{\hbar^2}E\psi(x)=0$$

$$\frac{d^2\psi(x)}{dx^2}+k^2\psi(x)=0\dotsm(5)$$

$$k^2=\frac{2mE}{\hbar^2}$$

$$\Rightarrow k= \sqrt{\frac{2mE}{\hbar^2}}$$

k is identified as wave number.

Solution of equation (5) is

$$\psi(x)= Ae^{ikx}+Be^{-ikx}\dotsm(6)$$

$$OR$$

$$[\psi(x)= Asinkx+Bcoskx]$$

Equation (6) is the required oscillatory solution.

Here, A and B represents amplitude.

Here \(Ae^{ikx}\) represents, particle travelling along +ve X-axis. i.e. towards right.

and \(Be^{-ikx}\) represents, particle travelling along -ve x-axis i.e. towards left.

Case II: For \(V(x)= V_0<E\) [Where \(V_0\)= Constant ]

Equation (4) becomes,

$$\frac{d^2\psi(x)}{dx^2}+\frac{2m}{\hbar^2}[E-V_0]\psi(x)=0$$

$$or,\;\; \frac{d^2\psi(x)}{dx^2}+ k^2 \psi(x)=0\dotsm(7)$$

Where, \(k= \sqrt{\frac{2m(E-V_0}{\hbar^2}}\) (real and +ve)

Solution of equation (7) is

$$[\psi(x)=Ce^{ikx}+De^{-ikx}]\dotsm(8)$$

(Oscillatory Solution )

Case(3): For \(V(x)= V_0>E\) [ Where \(V_0\)= Constant ]

$$\frac{d^2\psi(x)}{dx^2}+\frac{2m}{\hbar^2}(E-V_0)\psi(x)=0$$

$$or,\;\; \frac{d^2\psi(x)}{dx^2}-\frac{2m}{\hbar^2}(v_0-E)\psi(x)=0$$

$$or,\;\; \frac{d^2\psi(x)}{dx^2}-k^2\psi(x)=0\dotsm(9)$$

Where, \(k=\sqrt{\frac{2m(V_0-E}{\hbar^2}}\)

Solution of equation (9) is

$$[\psi(x)= Ce^{kx} + De^{-kx}]\dotsm(10)$$

Equation (10) represents non-oscillatory solution i.e exponentially decreasing or growing solution.

Finally solution (1) for free particle

$$\psi(x,t)=\phi(t)\psi(x)$$

$$=\phi(0)e^{\frac{-iEt}{\hbar}}\cdot[Ae^{ikx}+Be^{-ikx}]$$

For +ve x-axis. \(\psi(x,t)= \phi(0)e^{\frac{-iEt}{\hbar}}Ae^{ikx}\)

$$=\phi(0) Ae^{i(kx-\frac{Et}{\hbar})}$$

$$\psi(x,t)=Ne^{i(kx-\omega t)}$$

For negative X-axis,

$$\psi(x,t)= Ne^{i(-kx-\omega t)}$$

Where, \(N=\phi(0)B\)

For a bounded particle:

$$\psi(x,t)=\phi(0)e^{\frac{-iEt}{\hbar}}[Ae^{ikx}+Be^{-ikx}]\longrightarrow (E>V_0)$$

$$\psi(x,t)=\phi(0)e^{\frac{-iEt}{\hbar}}[ce^{kx}+De^{-kx}]\longrightarrow (E<V_0)$$

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.