Square of angular momentum operator, Ladder operator and some numerical examples

Square of angular momentum operator\(\hat L^2\):

In cartesian co-ordinate system:

$$\hat L^2= \hat L\cdot \hat L = (\hat i \hat L_x+ \hat J \hat L_y+ \hat K \hat L_z)\cdot (\hat I \hat L_x+ \hat J \hat L_Y+\hat L_Y+ \hat K \hat L_Z)$$

$$\hat L^2= \hat L_x^2+ \hat L_y^2+ \hat L_z^2\dotsm(1)$$

Commutation relation between \(\hat L^2\) and component of \(\hat L\).

$$[\hat L_x, \hat L^2]\)= ?

$$[\hat L^2, \hat L_x]=0$$

$$[\hat L^2, \hat L_Y]=0$$

$$[\hat L^2, \hat L_z] =0$$

Hence square of angular momentum and component of angular momentum of a particle can be measured accurately and simultaneously.

Ladder Operator:

The operator of the form

$$\hat L_+= \hat L_x+ i\hat L_y$$

and

$$\hat L_- = \hat L_x - i\hat L_y $$ are called ladder operator.

Where \(\hat L_x\) and \(\hat L_y\) are x and y-component of angular momentum operator.

\(\hat L_+\) is known as raising operator because it raises the m^{th} level wave function to (m+1)^{th} level. \(\hat L_-\) is known as lowering operator because it lowers the \(m^{th}\) level wavefunction to \((m-1)^{th}\) level.

Q. Show that

$$(a) [\hat L_x, \hat L_y] = i\hat h \hat L_z$$

$$(b) [\hat L\times \hat L] = i\hbar\hat L$$ where \(\hat L\) is angular momentum operators.

$$(c) [\hat L_i, \hat L_j] = \sum_{ijk}i\hbar \hat L_k$$

Where, \(\sum_{ijk}= +1\) For even permutation

= -1 for odd permutation

=0 , i=j

$$[ \hat i = x \hat j= y , \hat k = \hat z ]$$

Proof: (b) \([ \hat L \times \hat L\)]

$$= [ \hat i \hat L_x+ \hat j \hat L_y+ \hat k \hat L_z] \times [ \hat i \hat L_x+ \hat j \hat L_y+ \hat k \hat L_z]$$

$$=\hat k \hat L_x \hat L_y- \hat j \hat L_x \hat L_z- \hat k \hat L_y \hat L_x + \hat i \hat L_y \hat L_z+ \hat j \hat L_z \hat L_x - i\hat L_z \hat L_y$$

$$= \hat i[ \hat L_y \hat L_z- \hat L_z \hat L_y)+ \hat j ( \hat L_z \hat L_x - \hat L_x \hat L_z ) + \hat k ( \hat L_x \hat L_y- \hat L_y \hat L_x)$$

$$= \hat i [ \hat L_y, \hat L_z] + \hat j [ \hat L_z, \hat L_x] + \hat k [ \hat L_x, \hat L_y]$$

$$= \hat i i\hbar \hat L_x+ \hat j i\hbar L_y+ \hat k i\hbar \hat L_z$$

$$= i\hbar ( \hat i \hat L_x+ \hat j \hat L_y+ \hat k \hat L_z)$$

$$= i\hbar \hat L $$

$$Proved$$

Again, (a) Alternatively, \( [ \hat L_x, \hat L_y] \psi\)

$$= \hat L_x \hat L_y \psi- \hat L_y \hat L_x \psi(1)$$

Again, $$\hat L_x \hat L_y \psi= (-i\hbar)\biggl(y\frac{\partial}{\partial z}- z\frac{\partial}{\partial y}\biggr) (-i\hbar)\biggl(z\frac{\partial }{\partial x} - x\frac{\partial}{\partial z}\biggr)\psi$$

$$= (-i\hbar)^2 \biggl(y\frac{\partial }{\partial z}- z \frac{\partial}{\partial Y}\biggr)\biggl[z\frac{\partial \psi}{\partial x}- x\frac{\partial \psi}{\partial z}\biggr]$$

$$=(-i\hbar)^2\biggl[y\frac{\partial}{\partial z} \biggl(z\frac{\partial \psi}{\partial x}- x\frac{\partial \psi}{\partial z}\biggr)- z\frac{\partial}{\partial y}\biggl(z\frac{\partial \psi}{\partial x}- x \frac{\partial \psi}{\partial z}\biggr)\biggr]$$

$$(-i\hbar)^2\biggl[y\frac{\partial \psi}{\partial x}+ yz \frac{\partial^2\psi}{\partial z \partial x}- yx\frac{\partial^2 \psi}{\partial Z^2}- Z^2 \frac{\partial^2 \psi}{\partial y \partial x}+ zx \frac{\partial ^2\psi}{\partial y \partial z}\biggr]$$

And, $$\hat L_y \hat L_x \psi= (-i\hbar)\biggl(z\frac{\partial}{\partial x}- x\frac{\partial }{\partial z}\biggr)(-i\hbar)\biggl(y\frac{\partial}{\partial z}- z\frac{\partial }{\partial y}\biggr)\psi$$

$$=(-i\hbar)^2 \biggl( z\frac{\partial}{\partial x}- x \frac{\partial }{\partial z}\biggr)\biggl[ y\frac{\partial \psi}{\partial z}- z\frac{\partial \psi}{\partial y}\biggr]$$

$$= (-i\hbar)^2\biggl[z\frac{\partial }{\partial x}\biggl(y\frac{\partial\psi}{\partial z}- \frac{x\partial}{\partial z}\biggl(y\frac{\partial \psi}{\partial z}- z\frac{\partial \psi}{\partial y}\biggr)$$

$$=(-i\hbar)^2\biggl[z\frac{\partial y}{\partial x}\frac{\partial\psi}{\partial z}+ zy\frac{\partial ^2\psi}{\partial x\partial z}- z\frac{\partial z}{\partial x}\frac{\partial \psi}{\partial y}- z^2 \frac{\partial^2 \psi}{\partial x \partial y}- \frac{x \partial y \partial \psi}{\partial z \partial z}- xy\frac{\partial^2\psi}{\partial Z^2}+ x\frac{\partial z}{\partial z } \frac{\partial \psi}{\partial y}+ xz \frac{\partial^2\psi}{\partial z\partial y}\biggr]$$

$$= (-ih)^2\biggl[ zy\frac{\partial ^2\psi}{\partial x\partial z}- z^2\frac{\partial^2\psi}{\partial x \partial y}- xy\frac{\partial^2\psi}{\partial z^2}+ x \frac{\partial \psi}{\partial y}+ xz\frac{\partial^2 \psi}{\partial z\partial y}\biggr]$$

Now,

$$\hat L_x \hat L_y \psi- \hat L_y \hat L_x \psi= $$

$$(-i\hbar)^2\biggl[y\frac{\partial \psi}{\partial x}+ yz\frac{\partial^2\psi}{\partial z \partial x}- yx\frac{\partial^2 \psi}{\partial z^2}- z^2 \frac{\partial^2\psi}{\partial y\partial x}+ zx\frac{\partial^2\psi}{\partial y \partial z}- zy \frac{\partial^2\psi}{\partial x\partial z}$$

$$+ \frac{Z^2 \partial^2 \psi}{\partial x\partial y}+ xy\frac{\partial^2\psi}{\partial z^2}- x\frac{\partial\psi}{\partial y}- xz\frac{\partial^2 \psi}{\partial z \partial y}\biggr]$$

$$=(-i\hbar)^2\biggl[y\frac{\partial \psi}{\partial x}- x\frac{\partial \psi}{\partial y}\biggr]$$

$$= (-i\hbar)^2\biggl(y\frac{\partial}{\partial x}- x\frac{\partial }{\partial y}\biggr)\psi$$

$$=(-i\hbar)^2 (-)\biggl[x\frac{\partial}{\partial y}- y\frac{\partial}{\partial x}\biggr) (i\hbar)\psi$$

$$= i\hbar \hat L_z \psi$$

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

$$Proved$$

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.