Commutation relation between position and angular momentum, linear momentum and angular momentum, components of angular momentum operators.

Commutation relation between position and Angular momentum.

$$[\hat x , \hat L_x]= [ \hat x , \hat y \hat P_z- \hat Z\hat P_y]$$

$$=[\hat x , \hat y \hat P_z]- [ \hat x, \hat z \hat P_y]$$

$$\hat y[\hat x, \hat P_z]+ [\hat x, \hat y]\hat P_z-- \hat z [ \hat x, \hat P_y] - [ \hat x, \hat z]\hat P_y$$

$$= 0+ 0-0-0$$

$$=0$$

Similarly, $$[\hat y, \hat L_y]= [ \hat z, \hat L_z]=0$$

Again,$$[\hat x, \hat L_y]= [ \hat x, \hat z \hat P_x- \hat x\hat P_z] = [\hat x, \hat z \hat P_x]- [\hat x , \hat x \hat P_z]$$

$$=\hat Z[ \hat x, \hat P_x]+ [ \hat x, \hat z ] \hat P_x - \hat x [ \hat x, \hat P_z]- [ \hat x , \hat x ]\hat P_z$$

$$= \hat z ( i \hbar ) + 0-0-0$$

$$=\hat Z(i\hbar)$$

$$\therefore\;\; [ \hat x , \hat L_y]= \hat z i\hbar$$

Similarly,

$$[\hat x, \hat L_z]= -\hat y i\hbar$$

$$[\hat y, \hat L_y]=0, [\hat y, \hat L_x]=-\hat z i\hbar, [\hat y, \hat L_z]= \hat x i\hbar$$

and

$$[\hat z, \hat L_x]= \hat y i\hbar, [\hat z, \hat L_y]= -\hat x i\hbar, [\hat z, \hat L_z]=0$$

Hence, same component of position and angular momentum can be measured accurately and simultaneous;y but different component can not be measured accurately and simultaneously.

Commutation relation between linear momentum and angular momentum:

\(\Rightarrow\)

\([\hat P_x, \hat L_x]=?\) \([\hat P_x, \hat L_y]=?\) \([\hat P_x, \hat L_z]=?\)

\([\hat P_y, \hat L_y]=?\) \([\hat P_y , \hat L_x]=?\) \([\hat P_y, \hat L_z]=?\)

\([\hat P_z, \hat L_z]=?\) \([\hat P_z, \hat L_x]=?\) \([\hat P_z, \hat L_y]=?\)

Solution: $$[\hat P_,x, \hat L_x]= [\hat P_x, \hat y P_z- \hat Z P_y]$$

$$=[\hat P_x, \hat y \hat P_z]-[\hat P_x, \hat z P_y]$$

$$=\hat y [ \hat P_x, \hat P_z]+ [\hat P_x, \hat y] \hat P_z- \hat Z [ \hat P_x, \hat P_y]- [\hat P_x, \hat z] \hat P_y$$

$$=0+0-0-0$$

$$=0$$

Similarly, \([\hat P_y, \hat L_y]=0=[\hat P_x, \hat L_z]\) and so on.

Hence same component of linear momentum and angular momentum can be measured accurately and simultaneously.

Again, $$[\hat P_x, \hat L_y]= [ \hat P_x, \hat Z\hat P_x- \hat x \hat P_z]$$

$$=[\hat P_x, \hat z\hat P_x]-[\hat P_x, \hat x \hat P_z]$$

$$= \hat Z[ \hat P_x, \hat P_x]+ [ \hat P_x, \hat Z] \hat P_x- \hat x [ \hat P_x, \hat P_x] - [ \hat P_x, \hat x ] \hat P_x$$

$$= 0+0-0- (i\hbar)\hat P_z$$

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

$$=i\hbar \hat P_z$$

$$\therefore\;\; [\hat P_x, \hat L_y]= i\hbar \hat P_z$$

Similarly,

\([\hat P_x, \hat L_z]= -i\hbar \hat P_y, [\hat P_y, \hat L_z] = i\hbar \hat P_x\)

\([\hat P_y, \hat L_x]= i\hbar \hat P_z, [\hat P_x, \hat L_y]= i\hbar \hat P_z\)

\([\hat P_z, \hat L_x] = i\hbar \hat P_y, [\hat P_z, \hat L_y]=-\hbar \hat P_x\)

Hence different component of linear and angular momentum can not be measured accurately and simultaneously.

Commutation relation between components of angular momentum operator:

\(\rightarrow\) \([\hat L_x, \hat L_x]=0\; \; [\hat L_y, \hat L_x]= (-i\hbar \hat L_z)[ \hat L_z, \hat L_x]= (i\hbar)\hat L_y\)

\([\hat L_x, \hat L_y]= (i\hbar \hat L_z) [ \hat L_y, \hat L_y]=0\;\; [ \hat L_z, \hat L_y]= ( -i\hbar)\hat L_x\)

\([\hat L_x, \hat L_z] = ( -i\hbar \hat L_y) [ \hat L_y, \hat L_z] = ( i\hbar ) \hat L_x[ \hat L_z, \hat L_z] =0\)

Now, $$[\hat L_x, \hat L_y] = [\hat y \hat P_z- Z \hat P_y , \hat L_y]$$

$$= [ \hat y \hat P_z, \hat L_ y ] - [ \hat Z \hat P_y , \hat l_y]$$

$$= [ \hat y \hat P_z, \hat z \hat P_x- \hat x \hat P_z] - [ \hat z \hat P_y , \hat z \hat P_x- \hat x \hat P_z]$$

$$= [\hat y \hat P_z, \hat z \hat P_x] - [ \hat y \hat P_z, \hat x \hat P_z] - [ \hat z \hat P_y, \hat z \hat P_z] + [ \hat z \hat P_y, \hat x \hat P_z]$$

$$= [ \hat y \hat P_z, \hat z \hat P_z] + [ \hat z \hat P_y, \hat x \hat P_z]$$

$$= \hat x [ \hat y \hat P_z, \hat P_x] + [ \hat y \hat P_z, \hat Z_x] \hat P_x + \hat x [ \hat z \hat P_y, \hat P_z] + [ \hat Z \hat P_y \hat H] \hat P_z$$

$$= \biggl[ \hat y [ \hat P_z, \hat z ] + [ \hat y , \hat z ] \hat P_z \biggr] \hat P_x + \hat x \biggl[ \hat z [ \hat P_y, \hat P_z] + [ \hat z , \hat P_z ] \hat P_y\biggr]$$

$$= i\hbar \hat y \hat P_x+ \hat x ( i \hbar ) \hat P_y$$

$$=(-i\hbar) [ \hat x \hat P_y- \hat y \hat P_x] $$

$$= i\hbar \hat L_z$$

$$\therefore\;\; [ \hat L_x, \hat L_y] = i\hbar \hat L_z$$

Hence the different components of angular momentum o f particle can not be measured accurately and simultaneously. We could not have knowledge of a system by measuring individual component of angular momentum accurately and simultaneously.

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.