Chapter 11, Problem 11.23P

(a)

To determine

Show that if $\left[\widehat{G},\widehat{H}\right]=0$, then $\Psi$ satisfies the Schrodinger equation.

Answer to Problem 11.23P

It is showed that if $\left[\widehat{G},\widehat{H}\right]=0$, then $\Psi$ satisfies the Schrodinger equation.

Explanation of Solution

Write the expression for the Schrodinger equation using the given equation 11.108.

    $\begin{array}{c}i\hslash \frac{\partial \psi }{\partial t}=i\hslash \frac{\partial }{\partial t}{e}^{\widehat{G}\left(t\right)}\Psi \left(0\right)\\ =i\hslash \frac{\partial }{\partial t}\left[1+\widehat{G}\left(t\right)+\frac{1}{2}\widehat{G}\left(t\right)+\frac{1}{3!}\widehat{G}\left(t\right)\widehat{G}\left(t\right)+\cdot \cdot \cdot \right]\Psi \left(0\right)\end{array}$        (I)

Write the expression for the Hamiltonian.

    $H=i\hslash \dot{G}$        (II)

Use equation (II) in (I) and it can be written as

$\begin{array}{c}i\hslash \frac{\partial \psi }{\partial t}=i\hslash \left[\begin{array}{l}0+\stackrel{\cdot }{\widehat{G}}\left(t\right)\widehat{G}\left(t\right)+\frac{1}{2}\stackrel{\cdot }{\widehat{G}}\left(t\right)\widehat{G}\left(t\right)+\frac{1}{2}\widehat{G}\left(t\right)\stackrel{\cdot }{\widehat{G}}\left(t\right)\\ +\frac{1}{3!}\stackrel{\cdot }{\widehat{G}}\left(t\right)\widehat{G}\left(t\right)\widehat{G}\left(t\right)+\frac{1}{3!}\widehat{G}\left(t\right)\stackrel{\cdot }{\widehat{G}}\left(t\right)\widehat{G}\left(t\right)+\frac{1}{3!}\widehat{G}\left(t\right)\widehat{G}\left(t\right)\stackrel{\cdot }{\widehat{G}}\left(t\right)+\cdot \cdot \cdot \end{array}\right]\Psi \left(0\right)\\ =i\hslash \left[\begin{array}{l}\widehat{H}\left(t\right)+\frac{1}{2}\widehat{H}\left(t\right)\widehat{G}\left(t\right)+\frac{1}{2}\widehat{G}\left(t\right)\widehat{H}\left(t\right)+\frac{1}{3!}\widehat{H}\left(t\right)\widehat{G}\left(t\right)\widehat{G}\left(t\right)\\ +\frac{1}{3!}\widehat{G}\left(t\right)\widehat{H}\left(t\right)\widehat{G}\left(t\right)+\frac{1}{3!}\widehat{G}\left(t\right)\widehat{G}\left(t\right)\widehat{H}\left(t\right)+\cdot \cdot \cdot \end{array}\right]\Psi \left(0\right)\end{array}$        (III)

If the $\left[\widehat{G},\widehat{H}\right]=0$, the equation (III) becomes,

$\begin{array}{c}i\hslash \frac{\partial \psi }{\partial t}=\widehat{H}\left(t\right)\left[1+\widehat{G}\left(t\right)+\frac{1}{2}\widehat{G}\left(t\right)\widehat{G}\left(t\right)+\cdot \cdot \cdot \right]\Psi \left(0\right)\\ =\widehat{H}\left(t\right)e^{\widehat{G}\left(t\right)}\Psi \left(0\right)\\ =\widehat{H}\left(t\right)\Psi \left(t\right)\end{array}$        (IV)

So if equation 11.108 is a solution to the Schrodinger equation if $G$ and $H$ commute which is not typically cause for time-dependent Hamiltonians.

Conclusion:

Therefore, it is showed that if $\left[\widehat{G},\widehat{H}\right]=0$, then $\Psi$ satisfies the Schrodinger equation.

(b)

To determine

Check the correct solution in the general case $\left(\left[\widehat{G},\widehat{H}\right]\ne 0\right)$ is given in equation 11.109.

Answer to Problem 11.23P

It is showed the correct solution in the general case $\left(\left[\widehat{G},\widehat{H}\right]\ne 0\right)$ is given in equation 11.109.

Explanation of Solution

Take the derivation of equation 11.109 and recalling that,

$\begin{array}{c}\frac{d}{dx}{\displaystyle {\int }_a^{x}f\left(u\right)du}=f\left(x\right)\\ i\hslash \frac{\partial \Psi }{\partial t}=\left\{\begin{array}{l}0+\widehat{H}\left(t\right)+\left(-\frac{i}{\hslash }\right)\widehat{H}\left(t\right){\displaystyle {\int }_0^t\widehat{H}\left(t_2\right)d{t}_2}\\ +{\left(-\frac{i}{\hslash }\right)}^2\widehat{H}\left(t\right){\displaystyle {\int }_0^t\widehat{H}\left(t_2\right)\left({\displaystyle {\int }_0^{t_2}\widehat{H}\left(t_3\right)d{t}_3}\right)d{t}_2}\end{array}\right\}\Psi \left(0\right)\end{array}$        (V)

In all this case $\widehat{H}\left(t\right)\text{'}\text{s}$ are on the left, so no need to worry about the commuting operators. $t_2$ and $t_3$ are dummy variables and can be renamed them to $t_1$ and $t_2$ and it can be written as

$i\hslash \frac{\partial \Psi }{\partial t}=\widehat{H}\left(t\right)\left\{1+\left(-\frac{i}{\hslash }\right){\displaystyle {\int }_0^t\widehat{H}\left(t_1\right)d{t}_1+{\left(-\frac{i}{\hslash }\right)}^2{\displaystyle {\int }_0^t\widehat{H}\left(t_1\right)\left({\displaystyle {\int }_0^{t_1}\widehat{H}\left(t_2\right)d{t}_1}\right)d{t}_1}}\right\}\Psi \left(0\right)$        (VI)

Compare the equation (VI) with 11.109 and it can be written as

$i\hslash \frac{\partial \Psi }{\partial t}=\widehat{H}\left(t\right)+\Psi \left(t\right)$        (VII)

Conclusion:

Therefore, it is showed the correct solution in the general case $\left(\left[\widehat{G},\widehat{H}\right]\ne 0\right)$ is given in equation 11.109.

(c)

To determine

Show that $T\left[\widehat{G}\widehat{G}\right]=-\frac{2}{{\hslash }^2}{\displaystyle {\int }_0^t\widehat{H}\left(t_1\right)\left[{\displaystyle {\int }_0^{t_1}\widehat{H}\left(t_2\right)d{t}_2}\right]d{t}_1}$ and generalize to higher power of $\widehat{G}$.

Answer to Problem 11.23P

It is showed that $T\left[\widehat{G}\widehat{G}\right]=-\frac{2}{{\hslash }^2}{\displaystyle {\int }_0^t\widehat{H}\left(t_1\right)\left[{\displaystyle {\int }_0^{t_1}\widehat{H}\left(t_2\right)d{t}_2}\right]d{t}_1}$ and generalize to higher power of $\widehat{G}$.

Explanation of Solution

Write the expression for $T\left[\widehat{G}\left(t\right)\widehat{G}\left(t\right)\right]$.

$\begin{array}{c}T\left[\widehat{G}\left(t\right)\widehat{G}\left(t\right)\right]=T\left[{\left(-\frac{i}{\hslash }\right)}^2{\displaystyle {\int }_0^t\widehat{H}\left(t^{\prime }\right)d{t}^{\prime }{\displaystyle {\int }_0^t\widehat{H}\left(t^{″}\right)d{t}^{″}}}\right]\\ ={\left(-\frac{i}{\hslash }\right)}^2{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{t}T\left[\widehat{H}\left(t^{\prime }\right)\widehat{H}\left(t^{″}\right)\right]d{t}^{\prime }d{t}^{″}}}\\ ={\left(-\frac{i}{\hslash }\right)}^2\left[{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{t^{″}}\widehat{H}\left(t^{″}\right)\widehat{H}\left(t^{\prime }\right)d{t}^{\prime }d{t}^{″}}}+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{t^{\prime }}\widehat{H}\left(t^{\prime }\right)\widehat{H}\left(t^{″}\right)d{t}^{″}d{t}^{\prime }}}\right]\end{array}$        (VIII)

Replace the variables with $t^{\prime }\to t_2$ and $t^{″}\to t_1$ in first integral of equation (VIII) and $t^{\prime }\to t_1$ and $t^{″}\to t_2$ in second integral of equation (VIII), it can be written as

$T\left[\widehat{G}\left(t\right)\widehat{G}\left(t\right)\right]=2{\left(-\frac{i}{\hslash }\right)}^2{\displaystyle {\int }_0^t\widehat{H}\left(t_1\right)\left[{\displaystyle {\int }_0^{t_2}\widehat{H}\left(t_2\right)d{t}_2}\right]d{t}_1}$        (IX)

Use equation (IX) and write the expression to generalize to higher power of $\widehat{G}$.

$T\left[{\widehat{G}}^n\right]=n!{\left(-\frac{i}{\hslash }\right)}^n{\displaystyle {\int }_0^t\widehat{H}\left(t_1\right){\displaystyle {\int }_0^{t_1}\widehat{H}\left(t_2\right)\cdot \cdot \cdot {\displaystyle {\int }_0^{t_{n-1}}\widehat{H}\left(t_n\right)d{t}_{n}d{t}_{n-1}\cdot \cdot \cdot d{t}_{2}d{t}_1}}}$        (X)

Conclusion:

Therefore, it is showed that $T\left[\widehat{G}\widehat{G}\right]=-\frac{2}{{\hslash }^2}{\displaystyle {\int }_0^t\widehat{H}\left(t_1\right)\left[{\displaystyle {\int }_0^{t_1}\widehat{H}\left(t_2\right)d{t}_2}\right]d{t}_1}$ and generalize to higher power of $\widehat{G}$.