Chapter 7, Problem 7C.15P

(a)

Interpretation Introduction

Interpretation:

The commutator for $\left[\widehat{H},{\widehat{p}}_x\right]$ has to be calculated using (i) $V(x)=V_0$ and (ii) $V(x)=\frac{1}{2}{k}_t{x}^2$.

Concept introduction:

The commutators are the operators obtained from the operators which commute with each other.  The commutator of the operators like position and momentum operators is given by the formula shown below.

    $\left[\widehat{x},{\widehat{p}}_x\right]=\left(\widehat{x}\cdot {\widehat{p}}_x-{\widehat{p}}_x\cdot \widehat{x}\right)$

Answer to Problem 7C.15P

The commutator for $\left[\widehat{H},{\widehat{p}}_x\right]$ using $V(x)=V_0$ and $V(x)=\frac{1}{2}{k}_t{x}^2$ is zero and $xi\hslash k_t$ respectively.

Explanation of Solution

The commutator of the operators is given by the formula shown below.

    $\left[\widehat{H},{\widehat{p}}_x\right]=\left(\widehat{H}\cdot {\widehat{p}}_x-{\widehat{p}}_x\cdot \widehat{H}\right)$        (1)

The value of operator $\widehat{H}$ is given as shown below.

    $\widehat{H}=\frac{{\widehat{p}}_x^2}{2m}+\widehat{V}\left(x\right)$

Substitute the values of operators in the equation (1) for $V(x)=V_0$ as shown below.    $\begin{array}{c}\left[\widehat{H},{\widehat{p}}_x\right]\psi =\left(\widehat{H}\cdot {\widehat{p}}_x-{\widehat{p}}_x\cdot \widehat{H}\right)\psi \\ =\left(\left(\frac{{\widehat{p}}_x^2}{2m}+V_0\right){\widehat{p}}_x-{\widehat{p}}_x\left(\frac{{\widehat{p}}_x^2}{2m}+V_0\right)\right)\psi \end{array}$

The above is equation is simplified by substituting the value of momentum operator $\left(-i\hslash \frac{\text{d}}{\text{d}x}\right)$ as shown below.

    $\begin{array}{c}\left[\widehat{H},{\widehat{p}}_x\right]\psi =\left(\left(\frac{{\widehat{p}}_x^2}{2m}+V_0\right){\widehat{p}}_x-{\widehat{p}}_x\left(\frac{{\widehat{p}}_x^2}{2m}+V_0\right)\right)\psi \\ =\left(\left(\frac{1}{2m}{\left(-i\hslash \frac{\text{d}}{\text{d}x}\right)}^2+V_0\right)\left(-i\hslash \frac{\text{d}}{\text{d}x}\right)\psi -\left(-i\hslash \frac{\text{d}}{\text{d}x}\right)\left(\frac{1}{2m}{\left(-i\hslash \frac{\text{d}}{\text{d}x}\right)}^2+V_0\right)\psi \right)\\ =\left(\left(\frac{1}{2m}\left(-{\hslash }^2\frac{{\text{d}}^2}{\text{d}{x}^2}\right)+V_0\right)\left(-i\hslash \frac{\text{d}\psi }{\text{d}x}\right)-\left(-i\hslash \frac{\text{d}}{\text{d}x}\right)\left(\frac{1}{2m}\left(-{\hslash }^2\frac{{\text{d}}^2\psi }{\text{d}{x}^2}\right)+V_0\psi \right)\right)\\ =\left(\frac{1}{2m}\left(i{\hslash }^3\frac{{\text{d}}^3\psi }{\text{d}{x}^3}\right)+V_0\left(-i\hslash \frac{\text{d}\psi }{\text{d}x}\right)\right)-\left(\frac{1}{2m}\left(i{\hslash }^3\frac{{\text{d}}^3\psi }{\text{d}{x}^3}\right)+V_0\left(-i\hslash \frac{\text{d}\psi }{\text{d}x}\right)\right)\end{array}$

The above equation on solving gives the value of zero.

Therefore, the value of $\left[\widehat{H},{\widehat{p}}_x\right]$ for $V(x)=V_0$ is zero.

Substitute the values of operators in the equation (1) for $V(x)=\frac{1}{2}{k}_t{x}^2$ as shown below.    $\begin{array}{c}\left[\widehat{H},{\widehat{p}}_x\right]\psi =\left(\widehat{H}\cdot {\widehat{p}}_x-{\widehat{p}}_x\cdot \widehat{H}\right)\psi \\ =\left(\left(\frac{{\widehat{p}}_x^2}{2m}+\frac{1}{2}{k}_t{x}^2\right){\widehat{p}}_x-{\widehat{p}}_x\left(\frac{{\widehat{p}}_x^2}{2m}+\frac{1}{2}{k}_t{x}^2\right)\right)\psi \end{array}$

The above is equation is simplified by substituting the value of momentum operator $\left(-i\hslash \frac{\text{d}}{\text{d}x}\right)$ as shown below.

    $\begin{array}{c}\left[\widehat{H},{\widehat{p}}_x\right]\psi =\left(\left(\frac{{\widehat{p}}_x^2}{2m}+\frac{1}{2}{k}_t{x}^2\right){\widehat{p}}_x-{\widehat{p}}_x\left(\frac{{\widehat{p}}_x^2}{2m}+\frac{1}{2}{k}_t{x}^2\right)\right)\psi \\ =\left(\begin{array}{l}\left(\frac{1}{2m}{\left(-i\hslash \frac{\text{d}}{\text{d}x}\right)}^2+\frac{1}{2}{k}_t{x}^2\right)\left(-i\hslash \frac{\text{d}}{\text{d}x}\right)\psi -\\ \left(-i\hslash \frac{\text{d}}{\text{d}x}\right)\left(\frac{1}{2m}{\left(-i\hslash \frac{\text{d}}{\text{d}x}\right)}^2+\frac{1}{2}{k}_t{x}^2\right)\psi \end{array}\right)\\ =\left(\begin{array}{l}\left(\frac{1}{2m}\left(-{\hslash }^2\frac{{\text{d}}^2}{\text{d}{x}^2}\right)+\frac{1}{2}{k}_t{x}^2\right)\left(-i\hslash \frac{\text{d}\psi }{\text{d}x}\right)-\\ \left(-i\hslash \frac{\text{d}}{\text{d}x}\right)\left(\frac{1}{2m}\left(-{\hslash }^2\frac{{\text{d}}^2\psi }{\text{d}{x}^2}\right)+\frac{1}{2}{k}_t{x}^2\psi \right)\end{array}\right)\\ =\left(\frac{1}{2m}\left(i{\hslash }^3\frac{{\text{d}}^3\psi }{\text{d}{x}^3}\right)+\frac{1}{2}{k}_t{x}^2\left(-i\hslash \frac{\text{d}\psi }{\text{d}x}\right)\right)-\\ \left(\frac{1}{2m}\left(i{\hslash }^3\frac{{\text{d}}^3\psi }{\text{d}{x}^3}\right)+\frac{1}{2}{k}_t{x}^2\left(-i\hslash \frac{\text{d}\psi }{\text{d}x}\right)-i\hslash \frac{1}{2}{k}_t\frac{\text{d}{x}^2}{\text{d}x}\psi \right)\end{array}$

The above equation can be solved as shown below.

$\begin{array}{c}=\left(\frac{1}{2m}\left(i{\hslash }^3\frac{{\text{d}}^3\psi }{\text{d}{x}^3}\right)+\frac{1}{2}{k}_t{x}^2\left(-i\hslash \frac{\text{d}\psi }{\text{d}x}\right)\right)-\left(\frac{1}{2m}\left(i{\hslash }^3\frac{{\text{d}}^3\psi }{\text{d}{x}^3}\right)+\frac{1}{2}{k}_t{x}^2\left(-i\hslash \frac{\text{d}\psi }{\text{d}x}\right)-xi\hslash k_t\psi \right)\\ =\left(xi\hslash k_t\right)\psi \end{array}$

Therefore, the value of $\left[\widehat{H},{\widehat{p}}_x\right]$ for $V(x)=\frac{1}{2}{k}_t{x}^2$ is $xi\hslash k_t$.

(b)

Interpretation Introduction

Interpretation:

The commutator for $\left[\widehat{H},\widehat{x}\right]$ has to be calculated using (i) $V(x)=V_0$ and (ii) $V(x)=\frac{1}{2}{k}_t{x}^2$.

Concept introduction:

The commutators are the operator obtained from the operators which commute with each other.  The commutator of the operators like position and momentum operators is given by the formula shown below.

    $\left[\widehat{x},{\widehat{p}}_x\right]=\left(\widehat{x}\cdot {\widehat{p}}_x-{\widehat{p}}_x\cdot \widehat{x}\right)$

Answer to Problem 7C.15P

The commutator for $\left[\widehat{H},\widehat{x}\right]$ using $V(x)=V_0$ and $V(x)=\frac{1}{2}{k}_t{x}^2$ is same and is equals to $\frac{1}{2m}\left(-{\hslash }^2\frac{{\text{d}}^{2}x}{\text{d}{x}^2}\right)$.

Explanation of Solution

The commutator of the operators is given by the formula shown below.

    $\left[\widehat{H},\widehat{x}\right]=\left(\widehat{H}\cdot \widehat{x}-\widehat{x}\cdot \widehat{H}\right)$        (2)

The value of operator $\widehat{H}$ is given as shown below.

    $\widehat{H}=\frac{{\widehat{p}}_x^2}{2m}+\widehat{V}\left(x\right)$

Substitute the values of operators in the equation (2) for $V(x)=V_0$ as shown below.    $\begin{array}{c}\left[\widehat{H},\widehat{x}\right]\psi =\left(\widehat{H}\cdot \widehat{x}-\widehat{x}\cdot \widehat{H}\right)\psi \\ =\left(\left(\frac{{\widehat{p}}_x^2}{2m}+V_0\right)\widehat{x}-\widehat{x}\left(\frac{{\widehat{p}}_x^2}{2m}+V_0\right)\right)\psi \end{array}$

The above is equation is simplified by substituting the value of momentum operator $\left(-i\hslash \frac{\text{d}}{\text{d}x}\right)$ as shown below.

    $\begin{array}{c}\left[\widehat{H},\widehat{x}\right]\psi =\left(\left(\frac{{\widehat{p}}_x^2}{2m}+V_0\right)\widehat{x}-\widehat{x}\left(\frac{{\widehat{p}}_x^2}{2m}+V_0\right)\right)\psi \\ =\left(\left(\frac{1}{2m}{\left(-i\hslash \frac{\text{d}}{\text{d}x}\right)}^2+V_0\right)x\psi -x\left(\frac{1}{2m}{\left(-i\hslash \frac{\text{d}}{\text{d}x}\right)}^2+V_0\right)\psi \right)\\ =\left(\begin{array}{l}\left(\frac{1}{2m}\left(-x{\hslash }^2\frac{{\text{d}}^2\psi }{\text{d}{x}^2}\right)+\frac{1}{2m}\left(-{\hslash }^2\frac{{\text{d}}^{2}x}{\text{d}{x}^2}\right)\psi +V_{0}x\psi \right)\\ -x\left(\frac{1}{2m}\left(-{\hslash }^2\frac{{\text{d}}^2\psi }{\text{d}{x}^2}\right)+V_0\psi \right)\end{array}\right)\\ =\frac{1}{2m}\left(-{\hslash }^2\frac{{\text{d}}^{2}x}{\text{d}{x}^2}\right)\psi \end{array}$

Therefore, the value of $\left[\widehat{H},\widehat{x}\right]$ for $V(x)=V_0$ is $\frac{1}{2m}\left(-{\hslash }^2\frac{{\text{d}}^{2}x}{\text{d}{x}^2}\right)$.

Substitute the values of operators in the equation (2) for $V(x)=\frac{1}{2}{k}_t{x}^2$ as shown below.    $\begin{array}{c}\left[\widehat{H},\widehat{x}\right]\psi =\left(\widehat{H}\cdot \widehat{x}-\widehat{x}\cdot \widehat{H}\right)\psi \\ =\left(\left(\frac{{\widehat{p}}_x^2}{2m}+\frac{1}{2}{k}_t{x}^2\right)\widehat{x}-\widehat{x}\left(\frac{{\widehat{p}}_x^2}{2m}+\frac{1}{2}{k}_t{x}^2\right)\right)\psi \end{array}$

The above is equation is simplified by substituting the value of momentum operator $\left(-i\hslash \frac{\text{d}}{\text{d}x}\right)$ as shown below.

    $\begin{array}{c}\left[\widehat{H},\widehat{x}\right]\psi =\left(\left(\frac{{\widehat{p}}_x^2}{2m}+\frac{1}{2}{k}_t{x}^2\right)\widehat{x}-\widehat{x}\left(\frac{{\widehat{p}}_x^2}{2m}+\frac{1}{2}{k}_t{x}^2\right)\right)\psi \\ =\left(\left(\frac{1}{2m}{\left(-i\hslash \frac{\text{d}}{\text{d}x}\right)}^2+\frac{1}{2}{k}_t{x}^2\right)x\psi -x\left(\frac{1}{2m}{\left(-i\hslash \frac{\text{d}}{\text{d}x}\right)}^2+\frac{1}{2}{k}_t{x}^2\right)\psi \right)\\ =\left(\left(\frac{1}{2m}\left(-{\hslash }^2\frac{{\text{d}}^2}{\text{d}{x}^2}\right)+\frac{1}{2}{k}_t{x}^2\right)x\psi -x\left(\frac{1}{2m}\left(-{\hslash }^2\frac{{\text{d}}^2\psi }{\text{d}{x}^2}\right)+\frac{1}{2}{k}_t{x}^2\psi \right)\right)\\ =\left(\frac{1}{2m}\left(-x{\hslash }^2\frac{{\text{d}}^2\psi }{\text{d}{x}^2}\right)+\frac{1}{2m}\left(-{\hslash }^2\frac{{\text{d}}^{2}x}{\text{d}{x}^2}\right)\psi +\frac{1}{2}{k}_t{x}^3\psi \right)\\ -\left(\frac{1}{2m}\left(-x{\hslash }^2\frac{{\text{d}}^2\psi }{\text{d}{x}^2}\right)+\frac{1}{2}{k}_t{x}^3\psi \right)\end{array}$

The above equation can be solved as shown below.

  $\begin{array}{c}=\left(\frac{1}{2m}\left(-x{\hslash }^2\frac{{\text{d}}^2\psi }{\text{d}{x}^2}\right)+\frac{1}{2m}\left(-{\hslash }^2\frac{{\text{d}}^{2}x}{\text{d}{x}^2}\right)\psi +\frac{1}{2}{k}_t{x}^3\psi \right)-\left(\frac{1}{2m}\left(-x{\hslash }^2\frac{{\text{d}}^2\psi }{\text{d}{x}^2}\right)+\frac{1}{2}{k}_t{x}^3\psi \right)\\ =\frac{1}{2m}\left(-{\hslash }^2\frac{{\text{d}}^{2}x}{\text{d}{x}^2}\right)\psi \end{array}$

Therefore, the value of $\left[\widehat{H},\widehat{x}\right]$ for $V(x)=\frac{1}{2}{k}_t{x}^2$ is $\frac{1}{2m}\left(-{\hslash }^2\frac{{\text{d}}^{2}x}{\text{d}{x}^2}\right)$.