Equation of continuity in quantum mechanics

Equation of continuity in quantum mechanics:

The motion of a particle in Quantum Mechanics is descried by the propagation of wave packet. The equation of motion for wave packet is Schrodinger equation

$$i\hbar \frac{\partial (\vec r,t)}{\partial t}=\frac{-\hbar^2}{2m}\nabla^2 \psi (\vec r, t)+ V(r) \psi ( r,t)\dotsm(1)$$

Taking complex conjugate of equation (1)

$$-\hbar\frac{\partial^* (\vec r, t)}{\partial t}= \frac{-\hbar^2}{2m}\nabla^2 \psi*(\vec r, t)+ V(r) \psi* ( r,t) \dotsm(2)$$

Multiplying equation (1) by \(\psi*(r, t)\) and (2) by \( \psi(r,t)\)

$$i\hbar \frac{\partial (\vec r,t)}{\partial t}=\frac{-\hbar^2}{2m}\nabla^2 \psi (\vec r, t)+ V(r) \psi ( r,t)\times [\psi^*(r,t)]$$

$$-\hbar\frac{\partial^* (\vec r, t)}{\partial t}= \frac{-\hbar^2}{2m}\nabla^2 \psi*(\vec r, t)+ V(r) \psi* ( r,t) \times [\psi(r,t)]$$

and then subtracting (2) from (1) we get,

$$i\hbar\biggl[\psi^\frac{\partial \psi}{\partial t}+\psi\frac{\partial \psi^*}{\partial t}\biggr]=\frac{-\hbar^2}{2m}\biggl[\psi^* \nabla^2 \psi-\psi\nabla^2\psi^*\biggr]+0$$

$$or,\;\; \frac{\partial (\psi^*\psi}{\partial t}= \frac{-\hbar^2}{2m\hbar i}\biggl[\psi^* \nabla^2\psi- \psi \nabla^2\psi^*]$$

$$or,\;\;\frac{\partial}{\partial t}(\psi^*\psi)+\frac{\hbar}{2im}[\psi^* \nabla^2\psi-\psi \nabla^2\psi^*]=0$$

$$or,\;\; \frac{\partial }{\partial t}(\psi^*\psi)+\nabla\cdot\biggl[\frac{\hbar}{2im}[\psi^*\nabla\psi-\psi\nabla\psi^*]\biggr]=0\dotsm(3)$$

Equation(3) is in the form of equation of continuity electrodynamics.:

$$\frac{\partial rho}{\partial t}+\nabla\cdot \vec J=0\dotsm(4)$$

From (3) and (4)

$$\rho= \psi^*\psi= \psi^*(r,t)\psi(r,t)\dotsm(5)=|\psi(r,t)|^2$$

$$\vec J= \frac{\hbar}{2im}\biggl[\psi^*\nabla\psi-\psi\nabla \psi^*]\dotsm(6)$$

Equation (3) is known as equation of continuity in quantum mechanics

According to Max-Born interpretation

\(\rho= \psi^*\psi \) is interpreted as probability density of finding particle at point \(\vec r\) in time t.

unit of \(\rho=\psi^*\psi=|\psi|^2 \) is \(m^{-3}\) [ 3-dimension

\(\therefore\) Probability of finding particle in volume dV around point \(\vec r\) in time 't' is

$$dp=\rho(r,t) dV\dotsm(7)$$

Integrating equation (7) over all space, we get the total probability of finding particle over all space ( i.e. 1)

$$P_{total}=1=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \rho(r,t)dxdydz$$

Here, dv= dxdydz

$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \psi^* (r,t) \psi(r,t) dxdydz=1\dotsm(8)$$

Any wave function or wavepacket which satisfy equation (8) is said to be Normalised wave function.

$$ If\;\; \iiint \psi^*(r,t) \psi(r,t)dxdydz\ne 1$$

then \(\psi(r,t)\) is said to be unnormalised wave function.

Here, \(\vec = \frac{\hbar}{2m}[ \psi^* \nabla \psi- \psi \nabla \psi^*], is a probability current density, it represents the flow of probability per unit area per second. It's unit is \(m^{-2}s^{-1}\).

Case (1):

For any real wave function:

$$\psi^*= \psi$$

So, $$ \vec J= \frac{\hbar}{2im} [\psi^* \nabla\psi^*-\psi* \nabla\psi*]=0$$

The probability current density for real wave function always zero.

Case (2):

for \(\psi\) complex \(\vec J\) may or may not be zero, such that

$$(a)\;\;\; \vec J= Im \; of \; [ \psi^* \frac{\hbar}{m}\nabla\psi]$$

$$(b)\;\;\; \vec J = Re \; of \; [\psi^* -\frac{i\hbar}{m}\nabla\psi]$$

$$(c)\;\;\; \vec J= \frac{i\hbar}{2m}[ \psi \nabla\psi^* - \psi^* \nabla\psi]$$

$$(d)\;\;\;\vec J= \frac{\hbar}{2im}[\psi^*\nabla\psi-\psi\nabla\psi^*]$$

\(\Rightarrow\) Let \(\psi= A+iB\) be the form of wave function. Where A and B are both function of \(\vec r\) and t and real but \\psi\) is the complex.

We have,

$$\vec J= \frac{\hbar}{2im}[\psi^*\nabla\psi-\psi\nabla\psi^*]$$

$$=\frac{\hbar}{2im}[(A-iB)\nabla(A+iB)-(A+iB)\nabla(A-iB)]$$

$$\frac{\hbar}{2im}[A\nabla A+ iA\nabla B-iB\nabla A - i^2B\nabla B- A\nabla A+ iA\nabla B- iB\nabla A+ i^2 B \nabla B]$$

$$=\frac{\hbar}{2im}2i[A\nabla B- B\nabla A]$$

$$\therefore \vec J= \frac{\hbar}{m}[ A\nabla B- B\nabla A]\dotsm(1)$$

Again,

$$[\psi^*\nabla\psi]= (A-iB)\nabla (A+ iB)$$

$$=A\nabla A+ iA\nabla B- iB\nabla A- i^2 B\nabla B$$

$$= ( A\nabla A+ B\nabla B)+ i(A\nabla B - B\nabla a)$$

$$\therefore Im \; of \; [\psi^*\frac{\hbar}{m}\nabla \psi]= \frac{\hbar}{m}[A\nabla B- B\nabla A]\dotsm(2)$$

From equation (1) and (2), We get,

$$\therefore \;\; \vec J= Im\; of \; [\psi^*\frac{\hbar}{m}\nabla \psi]\dotsm(3)$$

Again,

$$\frac{[\psi^*\nabla\psi]}{i}= \frac{(A- iB)\nabla ( A+ iB)}{i}$$

$$=\frac{(A\nabla A+ B\nabla B)}{i}+ \frac{i(A\nabla B- B\nabla A)}{i}$$

$$\therefore Real \; of \; [\psi^* \frac{\hbar}{im}\nabla \psi]=\frac{\hbar }{m} [A\nabla B- B\nabla A]\dotsm(4)$$

From (1) and (4)

$$\vec J= Re\; of \; [\psi^* \frac{\hbar}{im}\nabla\psi]$$

$$\therefore \; \vec J= Re\; of \; [\psi^*-\frac{-i\hbar}{m}\nabla\psi]$$

$$Proved$$

$$or,\;\; \vec J= Re\; of \; [\psi^*\frac{(-\hbar \nabla \psi)}{m}]$$

$$=Re\;of\; [\psi^* \frac{(\hat p \psi)}{m}]\dotsm(5)$$

Where, \(\hat p= -i\hbar \nabla\) = Momentum operator.

$$\therefore \; \vec J= Re\; of \; [ \psi^*\frac{(\hat p \psi}{m}]$$

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.