Commutation Relation between position and linear momentum

Commutation Relation between position and linear momentum

$$[\hat y, \hat P_y]=i\hbar$$

$$[\hat Z, \hat P_z]= i\hbar$$

$$[\hat P_x, \hat x]= -\hbar $$ and so on.

Physical Significance:

1. One component of position and corresponding component of linear momentum of a particle can not be measured.

2. \(\hat x\) and \(\hat P_x\) can not have simultaneous eigen state.

3. The order of measurement is important in this case.

4. The measurement of x disturb the system for the measurement of \(\P_x\). So we can not measured \(\hat x\) and \(\hat P_x\) simultaneously and accurately.

Here \(\hat x\) and \(\hat P_x\) are said to be non-commutating operators. This is also consistent with Heisenberg uncertainty principle.

Q. \([\hat x,\hat P_y]\psi=?\)

Proof: $$[\hat x, \hat P_y]\psi= \hat x \hat P_y \psi-\hat P_y \hat x \psi$$

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

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

$$=0\psi$$

$$=0$$

Here, \(\hat x\) and \(\hat P_y\) can be measured accurately and simultaneously.

Similarly, $$[\hat x, \hat P_z]=0[\hat y, \hat P_x]=0, [\hat Z, \hat P_x]=0 \; and \; so \;on$$

Show that $$ (i) [\hat x, \hat P_x^n]= ni\hbar \hat P_x^{n-1}$$

$$(ii) [\hat x^n , \hat P_x]= ni\hbar \hat x^{n-1}$$

\(\Rightarrow\) Solution:

$$[\hat x, \hat P_x^n]=[\hat X, \hat P_x\cdot \hat P_x^{n-1}]$$

$$=\hat P_x[\hat x, \hat P_x^{n-1}]+ [\hat x, \hat P_x]\hat P_x^{n-1}$$

$$=\hat P_x[\hat x, \hat P_x^{n-1}]+i\hbar \hat P_x^{n-1}\dotsm(1)$$

$$=\hat P_x[\hat x, \hat P_x\cdot \hat P_x^{n-2}]+ [\hat x, \hat P_x]\hat P_x^{n-2}]+ i\hbar \hat P_x^{n-1}$$

$$=\hat P_x^2[\hat x, \hat P_x^{n-2}]+ i\hbar \hat P_x^{n-1}+i\hbar \hat P_x^{n-1}$$

$$=\hat P_x^2[\hat x, \hat P_x^{n-2}]++ 2i\hbar P_x^{n-1}\dotsm(2)$$

Repeating above process upto r-times [ iteration ]

$$=\hat P_x[\hat x, hat P_x^{n-r}]+ ri\hbar \hat P_x^{n-1}$$

Put r= n-1, we get

$$=\hat P_x^{n-1}[\hat x, \hat P_x]+ (n-1)i\hbar\hat P_x^{n-1}$$

$$=\hat P_x^{n-1}(i\hbar)+ (n-1)i\hbar \hat P_x^{n-1}$$

$$=i\hbar \hat P_x^{n-1}+ ni\hbar P_x^{n-1}-i\hbar P_x^{n-1}$$

$$[\hat x, \hat P_x^n]= ni\hbar P_x^{n-1}$$

$$Proved$$

Q. \([\hat x^n, \hat P_x]= ni\hbar x^{n-1}\)

Let \(\psi\) be the wave function associated with commutator \([\hat x^n, \hat P_x]\). Then,

$$[\hat x^n, \hat P_x]\psi= [\hat x^n \hat P_x- \hat P_x \hat x^n ]\psi$$

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

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

$$=(-i\hbar)\biggl[\hat x^n \frac{\partial \psi}{\partial x}- \hat x^n \frac{\partial x}{\partial x}- n\hat x^{n-1}\psi\biggr]$$

$$=i\hbar n\hat x^{n-1}\psi$$

$$[\hat x^n, \hat P_x]= i\hbar n \hat x^{n-1}$$

$$Proved$$

Q. 1 $$[\hat x, \hat P_x^2] = 2i\hbar \hat P_x$$

Proof:

$$[\hat x, \hat P_x^2]= [\hat x \hat P_x,\cdot \hat P_x]$$

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

$$=\hat P_x(i\hbar )+ (i\hbar ) \hat P_x$$

$$=2i\hbar \hat P_x$$

$$Proved$$

Q.2 $$[\hat x, \hat P_x^3]= 3i\hbar \hat P_x^2$$

Proof:

$$[\hat x, \hat P_x^3]= [\hat x, \hat P_x^2\cdot \hat P_x]$$

$$=\hat P_x^2(i\hbar)+ [\hat x, \hat P_x\hat P_x]\hat P_x$$

$$=i\hbar \hat P_x^2+ \biggl[\hat P_x[\hat x, \hat P_x]+[\hat x \hat P_x]\hat P_x\biggr]\hat P_x$$

$$=i\hbar \hat P_x^2 + [ \hat P_x i\hbar + i\hbar \hat P_x]\hat P_x$$

$$=i\hbar \hat P_x^2+ 2i\hbar \hat P_x^2$$

$$=3i\hbar \hat P_X^2$$

$$Proved$$

Q. $$[\hat r, \hat p]= ?$$

Let \(\psi(r)\) be the wave function associated with commutator \([\hat r, \hat P]\).

Now, $$[\hat r, \hat P] \psi= (\hat r\hat P- \hat P \hat r)\psi$$

$$=[r(-i\hbar\nabla)- (-i\hbar \nabla ) r]\psi$$

$$=(-i\hbar)[r\nabla \psi- \nabla ( r\psi)]$$

$$= -(i\hbar) [ r\nabla \psi- (\nabla r) \psi- r\nabla \psi]$$

$$=-i\hbar [ -(\nabla\cdot r)]\psi$$

$$= 3i\hbar \psi$$

$$\therefore\ ; [\hat r, \hat P] = 3i \hbar $$

$$proved$$

Q. $$[\hat P_x,\hat F(x)]= ?$$

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

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

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

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

$$=(-i\hbar)\frac{\partial F(x)}{\partial x}\psi$$

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

$$Proved$$

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.