Equation of motion in Quantum Mechanics

Equation of motion in Quantum Mechanics:

In Quantum Mechanics we can not pin-point the position and linear momentum of a particle. There is always uncertainity in the measurement of dynamical variables associated with system. So we take average or expectation value of measurement of the quantity. Equation of motion is Quantum Mechanics gives the rate of change of expectation value or average value of a physical quantity.

Let \(\psi(r,t) \) be the normalized state of a system then average value of quantity A is

$$<A>= \int_\tau \psi^*(\hat A \psi)d\tau\dotsm(1)$$

Where,

<A>= Average value of A.

\(<\hat A>\)= Operator associated with A

\(d\tau\)= Volume element.

Differentiating equation (1) with respect to time, we get

$$\frac{d<A>}{dt}=\frac{d}{dt}\biggl[\int_\tau \psi^*(\hat A\psi)d\tau\biggr]$$

$$or, \int_\tau\frac{d}{dt}[\psi^*(\hat A\psi)]d\tau= \int_\tau \frac{\partial \psi^*}{\partial t}(\hat A\psi)d\tau+ \int_\tau \psi^*\biggl(\frac{\partial \hat A}{\partial t}\biggr)\psi d\tau+ \int_\tau \psi^*\hat A\frac{\partial\psi}{\partial t}d\tau$$

$$=\int_\tau \biggl[\frac{\partial \psi^*}{\partial t}(\hat A\psi)+ \psi^* \hat A\frac{\partial \psi}{\partial t}\biggr]d\tau+ \int_\tau \psi^*\biggl(\frac{\partial \hat A}{\partial t}\biggr)\psi d\tau$$

$$=\int_\tau \biggl[\frac{\partial \psi^*}{\partial t}(\hat A\psi)+\psi^*\hat A\frac{\partial \psi}{\partial t}\biggr]d\tau+ \biggl< \frac{\partial \hat A}{\partial t}\biggr>\dotsm(2)\;[From\; (1)]$$

According to time dependent Schrodinger equation.

$$\hat H \psi= i\hbar \frac{\partial \psi}{\partial t}$$

$$\Rightarrow \frac{\partial \psi}{\partial t}= \frac{1}{i\hbar } (\hat H \psi)\dotsm(3)$$

Taking complex conjugate of (3)

$$\frac{\partial \psi^*}{\partial t}= \frac{1}{(-i\hbar)}(\hat H \psi)^*\dotsm(4)$$

Substituting from equation (3) and (4) in equation (2), we get

$$\frac{d<A>}{dt}= \int_\tau\biggl[\psi^*\hat A\frac{1}{i\hbar}(\hat H \psi)- \frac{1}{i\hbar}(\hat H \psi)^*(\hat A\psi)\biggr]d\tau+ \biggl<\frac{\partial \hat A}{\partial t}\biggr>$$

$$=\frac{1}{i\hbar}\int \biggl[\psi^*\hat A\hat H \psi- (\hat H \hat psi)^* (\hat A\psi)]d\tau+ \biggl<\frac{\partial \hat A}{\partial t}\biggr>$$

Since Hamiltonian operator (\(\hat H\)) is Hemitian operator . So

$$\frac{d<A>}{dt}= \frac{1}{i\hbar}\int\biggl[\psi^* \hat A\hat H \psi- \psi^* \hat H \hat A \psi\biggr]d\tau+ \biggl<\frac{\partial \hat A}{\partial t}\biggr>$$

$$or,\; \frac{d<A>}{dt}= \frac{1}{i\hbar}\int \psi^*[\hat A\hat H- \hat H \hat A]\psi d\tau+ \biggl<\frac{d\hat A}{\partial t}\biggr>$$

$$or, \; \frac{d<A>}{dt}=\frac{1}{\hbar}\int \psi^*[\hat A, \hat H]\psi d\tau + \biggl<\frac{\partial \hat A}{\partial t}\biggr>$$

$$\therefore\; \frac{d<A>}{dt}= \frac{1}{i\hbar}\biggl<[\hat A, \hat H]\biggr>+ \biggl<\frac{\partial \hat A}{\partial t}\biggr>\dotsm(5)$$

Equation (5) is called equation of motion in Quantum Mechanics. It gives the rate of change of <A>.

Example of Equation of Motion:

Ehrenfest Theorem:-

(1) The average rate of change of position of wave packet is equals to expectation value of linear momentum per unit mass. i.e.

$$\frac{d<r>}{dt}= \frac{<p>}{m}\dotsm(6)[\; In-3-dimesion]$$

In one-dimension, along x-axis:-

$$\biggl[\frac{d<x>}{dt}= \frac{<P_x>}{m}\biggr]$$

\(\Rightarrow\) proof:

Put A=x and \(\hat A= \hat x\) in equation of motion. i.e equation (5)

$$\frac{d<x>}{dt}=\frac{1}{i\hbar}\biggl<[\hat x, \hat H]\biggr>+ \biggl<\frac{\partial \hat x}{\partial t}\biggr>$$

$$=\frac{1}{i\hbar} \biggl< [\hat x, \frac{\hat P_x^2 }{2m}+ \hat V(x)]\biggr>+0$$

$$=\frac{1}{i\hbar}\biggl[\biggl<[\hat x, \frac{P_x^2}{2m}]+[\hat x, \hat V(x)]\biggr>\biggr]$$

$$=\frac{1}{i\hbar 2m}<[\hat x, \hat P_x^2]>+0$$

$$=\frac{1}{i\hbar 2m}<2i\hbar \hat P_x>$$

$$=\frac{2i\hbar}{2i\hbar m}<P_x>$$

$$\frac{d<x>}{dt}= \frac{<P_x>}{m}$$

Similarly,

$$\frac{d<y>}{dt}= \frac{<P_y>}{m}\;and\; \frac{d<z>}{dt}=\frac{<P_z>}{m}$$

Adding above equation we get,

$$\frac{d<r>}{dt}=\frac{<P>}{m}$$

$$Proved$$

(2) The rate of change of average momentum f a particle associated with wave packet is equal to average force acting on it. i.e.

$$\frac{d<P>}{dt}= <-\nabla V>\dotsm(7)$$

$$\therefore\frac{d<P>}{dt}= <F>$$

In one-dimension,

$$\biggl[\frac{d<P_x>}{dt}=\biggl<-\frac{\partial V(x)}{\partial x}\biggr>= <F_x>\biggr]\dotsm(8)$$

Put $$A=P_x$$

$$\hat A= \hat P_x$$

In equation (5) we get,

$$\frac{d<P_x>}{dt}= \frac{1}{i\hbar}<[\hat P_x, \hat H]>+ \biggl<\frac{\partial P_x}{\partial t}\biggr>\dotsm(9)$$

Since, \(\hat P_x\) is not an explicit function of time,

$$\biggl<\frac{\partial P_x}{\partial t}\biggr>=0$$

and

$$[\hat P_x, \hat H]= [\hat P_x, \frac{\hat P^2}{2m}+ \hat V(x)]$$

$$=[\hat Px, \frac{\hat P_x^2}{2m}]+[\hat P_x, \hat V(x)]$$

$$\Rightarrow [\hat P_x\hat V(x)]=?\dotsm(10)$$

Now,

$$[\hat P_x, \hat V(x)]\psi(x)= (\hat P_x\hat V(x)- \hat V(x)\hat P_x)\psi(x)$$

$$=\biggl(-i\hbar\frac{\partial}{\partial x}\biggr)\biggl(V(x)\psi(x)\biggr)-V(x)\biggl(-i\hbar\frac{\partial}{\partial x}\psi(x)\biggr)$$

$$=-i\hbar\biggl[\frac{\partial V(x)}{\partial x}\psi(x)+ V(x)\frac{\partial \psi}{\partial x}- V(x)\frac{\partial }{\partial x}\psi(x)\biggr]$$

$$[\hat P_x, \hat V(x)]= -i\hbar \frac{\partial V(x)}{\partial x}\dotsm(11)$$

Using (11) in (9) , we get

$$\frac{d<P_x>}{dt}= \frac{1}{i\hbar}\biggl<-i\hbar \frac{\partial v}{\partial x}\biggr>$$

$$=\biggl<-\frac{\partial v}{\partial x}\biggr>$$

$$=<F_x>$$

Similarly, \(\frac{d<P_y>}{dt}= < F_y>\) and \(\frac{d<P_z}{dt}= <F_z>\)

Adding , we get

$$\frac{d<P>}{dt}= < F> = <-\nabla V> $$

$$Proved$$

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.