(a) For a function $\mathit{f}\left(x\right)$that can be expanded in a Taylor series, show that $\mathit{f}\left(x+x_0\right)\mathbf{=}{\mathit{e}}^{\mathbf{i}\hat{\mathbf{p}}{\mathbf{x}}_{\mathbf{0}}\mathbf{I}\mathbf{h}}\mathit{f}\left(x\right)$

wherex_{0}

is any constant distance). For this reason, $\hat{\mathbf{p}}\mathbf{/}\sout{\mathbf{h}}$is called the generator of translations in space. Note: The exponential of an operator is defined by the power series expansion: ${\mathit{e}}^{\hat{\mathbf{Q}}}\mathbf{\equiv }\mathbf{1}\mathbf{+}\hat{\mathbf{Q}}\mathbf{+}\left(1/2\right){\hat{\mathbf{Q}}}^{\mathbf{2}}\mathbf{+}\left(1/3!\right){\hat{\mathbf{Q}}}^{\mathbf{3}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}$

(b) If $\mathit{\psi }\left(x,t\right)$satisfies the (time-dependent) Schrödinger equation, show that $\mathit{\psi }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{+}{\mathbf{t}}_{\mathbf{0}}\mathbf{)}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{i}\hat{\mathbf{H}}{\mathbf{t}}_{\mathbf{0}}\mathbf{/}\sout{\mathbf{h}}}\mathit{\psi }\left(x,t\right)$

where $\mathit{t}\mathit{\_}\left\{0\right\}$is any constant time); $\mathbf{-}\hat{\mathbf{H}}\mathbf{/}\sout{\mathbf{h}}$is called the generator of translations in time.

(c) Show that the expectation value of a dynamical variable$\mathit{Q}\left(x,p,t\right)$, at time , $\mathit{t}\mathbf{+}{\mathit{t}}_{\mathbf{0}}$can be written${}^{34}$

${<Q>}_{\mathbf{t}\mathbf{+}{\mathbf{t}}_{\mathbf{0}}}\mathbf{=}<\psi \left(x,t\right)\left|e^{i\hat{H}{t}_0/\sout{h}}\hat{Q}\left(\hat{x},\hat{p},t+t_0\right)e^{-i\hat{H}{t}_0/\sout{h}}\right|\psi \left(x,t\right)>$

Use this to recover Equation 3.71. Hint: Let${\mathit{t}}_{\mathbf{0}}\mathbf{=}\mathit{d}\mathit{t}$, and expand to first order in dt.

Taylor series about ${\mathit{x}}_{\mathbf{0}}$

Consider the function of$f\left(x\right)$which can be expanded using Taylor series. By expanding in a Taylor series about$x_0$, we get:

$f\left(x+x_0\right)=\sum _{n=0}^{\infty }\frac{1}{n!}{x}_0^n{\left(\frac{d}{dx}\right)}^{n}f\left(x\right)$

momentum operator is given by:

$p=\frac{\sout{h}}{i}\frac{d}{dx}\to \frac{d}{dx}=\frac{ip}{\sout{h}}$

$$\begin{gathered} f\left(x+x_0\right)=\sum _{n=0}^{\infty }\frac{1}{n!}{x}_0^n{\left(\frac{ip}{\sout{h}}\right)}^{n}f\left(x\right) \\ =\sum _{n=0}^{\infty }\frac{1}{n!}{\left(\frac{ip{x}_0}{\sout{h}}\right)}^{n}f\left(x\right) \end{gathered}$$

But the power series expansion is given by:

$e^{\hat{Q}}=1+Q+\frac{1}{2!}{\hat{Q}}^2+\frac{1}{3!}{\hat{Q}}^3+...=\sum _{n=0}^{\infty }\frac{1}{n!}{\left(\hat{Q}\right)}^n$

Using this expansion, the above equation becomes:

$f\left(x+x_0\right)=e^{ip{x}_0/\sout{h}}f\left(x\right)$

Now consider the function of$\psi \left(x,t\right)$which can be expanded using Taylor series. By expanding in a Taylor series about$t_0$, we get:

localid="1656049111899" $\psi \left(x,t+t_0\right)=\sum _{n=0}^{\infty }\frac{1}{n!}{x}_0^n{\left(\frac{d}{dt}\right)}^n\psi \left(x,t+t_0\right)$

But,

$i\sout{h}\frac{\partial \psi }{\partial t}=H\psi$

But this doesn't mean,role="math" localid="1656049008992" $i\sout{h}\frac{\partial }{\partial t}=H$ these two operators have the same effect only when they are acting on solutions to the time dependent Schrodinger equation. Apply this operator two times, we get:

$\left(i\sout{h}\frac{\partial }{\partial t}\right)\psi =i\sout{h}\left(H\psi \right)=H\left(i\sout{h}\frac{\partial }{\partial t}\right)=H^2\psi$

And so on, so we can write:

$\frac{\partial }{\partial t}=-\frac{i}{\sout{h}}H$

Substitute into the above equation we get:

$\psi \left(x,t+t_0\right)=\sum _{n=0}^{\infty }\frac{1}{n!}{x}_0^n{\left(-\frac{i}{\sout{h}}H\right)}^n\psi \left(x,t+t_0\right)$

using the power series expansion, we get

$\psi \left(x,t+t_0\right)=e^{-iH{t}_0/\sout{h}}\psi \left(x,t+t_0\right)$

Now we need to show that the expectation value of a dynamical variable$Q\left(x,p,t\right)$, at time$t+t_0$is:

we start with:

but $\psi \left(x,t+t_0\right)=e^{i\hat{H}{t}_0/\sout{h}}\psi \left(x,t\right)\mathrm{and}{\left(e^{i\hat{H}{t}_0/\sout{h}}\right)}^{\sout{\mathrm{I}}}=e^{i\hat{H}{t}_0/\sout{h}}$, so we can write:

now let $t_0=dt$be very small, then we can expand to the first order using the expansion of the exponential, as:

Where,