Let the Hamiltonian H, in one dimension, be
H = \(\rm \frac{p^2_x}{2m}\) +V(x).

The commutator of H with x, [H, x], is

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  1. \(\rm \frac{-ih}{m}px\)
  2. \(\rm \frac{-ih}{2m}p^2_x\)
  3. \(\rm \frac{ih}{m}px\)
  4. \(\rm \frac{ih}{2m}px\)

Answer (Detailed Solution Below)

Option 1 : \(\rm \frac{-ih}{m}px\)
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Detailed Solution

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Concept:

  • The commutator of two operators \({{\rm{\hat A}}}\) and \({\hat B}\) is denoted by,

\(\left[ {{\rm{\hat A,\hat B}}} \right]{\rm{ = }}\left[ {{\rm{\hat A\hat B - \hat B\hat A}}} \right]\)

  • To compute a given commutator, an arbitrary function(\(\Psi \)) is used so that the operators can operate as

\(\left[ {{\rm{\hat A,\hat B}}} \right]\Psi {\rm{ = }}\left[ {{\rm{\hat A\hat B - \hat B\hat A}}} \right]\Psi \)

  • Two operators are said to commute when their commutators are equal to zero, hence

​\(\hat A\hat B = \hat B\hat A\)

  • Any commutator (\({\hat A}\)) will commute by itself,

\(\left[ {{\rm{\hat A, \hat A}}} \right]{\rm{ = 0}}\)

Explanation:

  • The Hamiltonian operator (H) is given by,

H = \(\rm \frac{p^2_x}{2m}\) +V(x).

  • Now, the arbitrary function \(\Psi \) is used to compute the commutator of H with x.
  • The commutator of H with x is:

​\(\left[ {{\rm{\hat H,\hat x}}} \right]\Psi {\rm{ = }}\left[ {\hat H\hat x - \hat x\hat H} \right]\Psi \)

\( = \left[ {\left( { {{{P_x}^2} \over {2m}} + V(x)} \right)x - x\left( { {{{P_x}^2} \over {2m}} + V(x)} \right)} \right]\Psi \)

\( = \left( { {{{P_x}^2} \over {2m}} + V(x)} \right)x\Psi - x\left( { {{{P_x}^2} \over {2m}} + V(x)} \right)\Psi \)

\( = {{{P_x}^2} \over {2m}}\left( {x\Psi } \right) + V(x)x\Psi - x{{{P_x}^2} \over {2m}}\Psi - xV(x)\Psi \)

\( = {1 \over {2m}}{\left( { - i\hbar {\partial \over {\partial x}}} \right)^2}\left( {x\Psi } \right) - x{1 \over {2m}}{\left( { - i\hbar {\partial \over {\partial x}}} \right)^2}\Psi \), as\(\left[ {{P_x} = \left( { - i\hbar {{\partial \Psi } \over {\partial x}}} \right)} \right]\) \( = - {{{\hbar ^2}} \over {2m}}{\left( {{\partial \over {\partial x}}} \right)^2}\left( {x\Psi } \right) + {{x{\hbar ^2}} \over {2m}}{\left( {{\partial \over {\partial x}}} \right)^2}\Psi \)

\( = - {{{\hbar ^2}} \over {2m}}\left( {{\partial \over {\partial x}}} \right)\left( {\Psi + x{{\partial \Psi } \over {\partial x}}} \right) + {{x{\hbar ^2}} \over {2m}}\left( {{{{\partial ^2}\Psi } \over {\partial {x^2}}}} \right)\)

\( = - {{{\hbar ^2}} \over {2m}}\left( {{{\partial \Psi } \over {\partial x}} + x{{{\partial ^2}\Psi } \over {\partial {x^2}}} + {{\partial \Psi } \over {\partial x}}} \right) + {{x{\hbar ^2}} \over {2m}}\left( {{{{\partial ^2}\Psi } \over {\partial {x^2}}}} \right)\)

\( =- {{{\hbar ^2}} \over {2m}}\left( {2{{\partial \Psi } \over {\partial x}} + x{{{\partial ^2}\Psi } \over {\partial {x^2}}}} \right) + {{x{\hbar ^2}} \over {2m}}\left( {{{{\partial ^2}\Psi } \over {\partial {x^2}}}} \right)\)

\( = -{{2{\hbar ^2}} \over {2m}}\left( {{{\partial \Psi } \over {\partial x}}} \right) + {{x{\hbar ^2}} \over {2m}}\left( {{{{\partial ^2}\Psi } \over {\partial {x^2}}}} \right) + {{x{\hbar ^2}} \over {2m}}\left( {{{{\partial ^2}\Psi } \over {\partial {x^2}}}} \right)\)

\( = {{{i^2}{\hbar ^2}} \over m}\left( {{{\partial \Psi } \over {\partial x}}} \right)\) as, \({i^2} = - 1\)

\( = - {{i\hbar } \over m}\left( { - i\hbar {{\partial \Psi } \over {\partial x}}} \right)\)

\( = - {{i\hbar } \over m}{P_x}\)as \(\left[ {{P_x} = \left( { - i\hbar {{\partial \Psi } \over {\partial x}}} \right)} \right]\)​ 

Conclusion:

Hence, the commutator of H with x, \(\left[ {{\rm{\hat H,\hat x}}} \right]\Psi\), is

​\( - {{i\hbar } \over m}{P_x}\)

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