Commutation relation between different components

Commutation Relation between energy and linear momentum.

Q. Show that, \([\hat P_x, \hat H] = -i\hbar \frac{\partial \hat V}{\partial x}=-i\hbar \frac{\partial \hat H}{\partial x}\)\

Proof:

L.H.S \([\hat P_x, \hat x] \psi(x)= \biggl[\hat P_x, \frac{\hat P_x^2}{2m}+ \hat V(x)\biggr] \psi(x)\)

Where, \(\hat H\)= Hamiltonian operator\( =\frac{\hat P_x^2}{2m}+ \hat V(x)\)

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

$$=\frac{1}{2m} [\hat P_x, \hat P_x^2] \psi(x)+ [ \hat P_x, \hat V(x)]\psi(x)$$

$$=\frac{1}{2m}[\hat P_x^3- \hat P_x^3)\psi(x)+ (\hat P_x\hat V(x)- \hat V(x)\hat P_x)\psi(x)$$

$$=0+ (-i\hbar)\biggl[\frac{\partial}{\partial x}[\hat V(x)\psi(x)]-\hat V(x)\frac{\partial}{\partial x}\psi(x)]$$

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

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

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

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

Again,

$$\frac{\partial H}{\partial x}= \frac{\partial }{\partial x} \biggl[\frac{P_x^2}{2m}+ V(x)\biggr]$$

$$=0+\frac{\partial V}{\partial x} $$

$$\Rightarrow \frac{\partial H}{\partial x}= \frac{\partial v}{\partial x}$$

$$\therefore [\hat P_x, \hat H] = - \hbar \frac{\partial V}{\partial x}= -i\hbar \frac{\partial H}{\partial x}$$

$$Proved$$

The non-commutation of \([\hat P_x, \hat H]\) indicates that we can not measured inner momentum and Total energy of a particle accurately and simultaneously. [ For bounded particle ].

For free particle:

$$\Rightarrow \frac{\partial V}{\partial x}=0= \frac{\partial H}{\partial x}$$

$$\therefore\; [\hat P_x, \hat H] =0$$

For free particle, we can measure linear momentum and T.E accurately and simultaneously.

In General, \([\hat P, \hat H]= - \hbar \nabla V[ In \;3- dimension ]\)

Commutation Relation Between position operator and Hamiltonian operator.

Show that : \([\hat x, \hat H] = i\hbar \frac{\hat P_x}{m}= i\hbar \frac{\partial \hat H}{\partial \hat P_x}\)

\(\Rightarrow\) Solution:

Proof:

$$L.H.S \; [\hat x, \hat H]$$

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

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

$$=\frac{1}{2m}[\hat x, \hat P_x^2]+ [\hat x\hat V(x)- \hat V(x)\hat x]$$

$$=\frac{1}{2m}2i\hbar \hat P_x^{2-1}$$

$$=i\hbar \frac{\hat P_x}{m}$$

Again, \(\frac{\partial H}{\partial P_x}= \frac{\partial }{\partial P_x}\biggl[\frac{\hat P_x^2}{2m}+ V(x)\biggr]= \frac{\hat P_x}{m}\)

P.E is not a function of linear momentum.

$$\therefore\; \frac{\partial H}{\partial P_x}= \frac{\partial\hat P_x}{m}$$

$$\therefore\; [\hat x, \hat H] = i\hbar \frac{\hat P_x}{m}= i\hbar \frac{\partial \hat H}{2\hat P_x}$$ $$Proved.$$

In General, \([\hat r, \hat H] = i\hbar \frac{\partial \hat H}{\partial \hat P}\)

In general, position and T.E of particle can not be measured accurately and simultaneously.

Angular momentum Operator:

Cartesian co-ordinate system: Angular momentum of a particle is defined as moment of its linear momentum about given axis of rotation.

Mathematically,

$$\vec L= \vec r\times \vec p \dotsm(1)$$

The corresponding angular momentum operator is given by,

$$\hat L= \hat r\times \hat p\dotsm(2)$$

Where \(\hat r= \hat i \hat x+ \hat j \hat y + \hat k \hat z\) = Position operator

\(\hat i , \hat j \; and\; \hat k\) are three unit vectors.

\(\hat x, \hat y\) &\(hat z \)= Components of position operators.

\(\hat P = \hat i \hat P_x+ \hat j \hat P_y + \hat k \hat P_z \) = Linear momentum operator.

\(\hat P_x, \hat P_y\) and \( \hat P_z\) are component of linear momentum operators.

\(\hat L= \hat i \hat L_x+ \hat J \hat L_y+ \hat K \hat L_z\)= angular momentum operator.

$$\hat i\hat L_x+ \hat J \hat K_y+ \hat k \hat L_z= (\hat i \hat x+ \hat j \hat y+ \hat k \hat z)\times (\hat i \hat P_x+ \hat j \hat P_y+ \hat k \hat P_z)$$

$$=\hat k \hat x \hat P_y- \hat j \hat x \hat P_z- \hat k \hat y \hat P_x+ \hat i \hat y \hat P_z+ \hat j \hat z \hat P_x- i\hat z \hat P_y$$

$$=\hat i (\hat y \hat P_z- \hat z \hat P_y)+ \hat J (\hat z \hat P_x- \hat x \hat P_z)+ \hat K( \hat x \hat P_y- \hat y \hat P_x)\dotsm(3)$$

Comparing the coefficient of \(\hat i , \hat jj \) and \( \hat k\) on both sides,we get,

$$L_x= (\hat y \hat P_z- \hat z \hat P_y)= (-\hbar)[\hat y\frac{\partial }{\partial z}- \hat z \frac{\partial }{\partial y}]\dotsm(4(a))$$

$$\hat L_y= ( \hat z \hat P_x- \hat x \hat P_z)= (-i\hbar ) \biggl[ \hat z \frac{\partial}{\partial x}- \hat x \frac{\partial }{\partial z}\biggr]\dotsm(4(b))$$

$$\hat L_z= ( \hat x \hat P_y - \hat y \hat P_x) = (-i\hbar)\biggl[x\frac{\partial}{\partial y}- y\frac{\partial}{\partial x}\biggr]\dotsm(4(c))$$

Equation (4) (a) (b) (c) gives the cartesian component of angular momentum operator, IN terms of position and linear momentum operators.

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.