Question

Show that

𝑑⟨𝑝⟩

𝑑𝑡

= ⟨−

𝜕𝑉

𝜕𝑥⟩

i.e. expectation values follow Newton’s law.

Answer 1

• As indicated, let us calculate $\frac{d}{dt}\int |\psi(x,t)|^2 dx$, using that we can differentiate under the integral sign :

$\frac{d}{dt}\int |\psi(x,t)|^2dx = \int(\frac{\partial}{\partial t}|\psi(x,t)|^2)dx$

Now we use that $|\psi|^2=\psi \bar{\psi}$ :

$\int \frac{\partial}{\partial t}(\psi\bar{\psi})dx =\int ((\frac{\partial}{\partial t}\psi)\bar{\psi}+(\frac{\partial}{\partial t}\bar{\psi})\psi)dx$

Now we will use the fact that $\psi$ satisfies the Schrodinger equation

$\hat{H}\psi = i\hbar\partial_t \psi$

And as hamoltonian is a real operator

$\bar{\hat{H}}=\hat{H}$ and thus

$\hat H \bar{\psi}=-i\hbar \partial_t \bar{\psi}$

Using these expressions gives us

$\int \left((\frac{-i}{\hbar}\hat H\psi)\bar{\psi}+(\frac{i}{\hbar}\hat H \bar\psi)\psi\right)dx$

But we know that as $\hat H$ is a hermitian operator, we have $\int \bar \psi \hat H \psi dx = \int\psi\hat H \bar \psi dx$ and thus the integral is zero, $\frac{d}{dt}\int |\psi(x,t)|^2dx=0$, $\int |\psi(x,t)|^2dx=||\psi||^2=const$ and as initially it is 1, it is always normalized.

• First let's write the expression for the expectation value of $p$

$<p>=\int \bar\psi (-i\hbar\frac{\partial}{\partial x})\psi dx$

Deriving with respect to time gives us

$-i\hbar \left( \int(\partial_t\bar \psi)(\partial_x\psi)dx+\int(\bar\psi) (\partial_t\partial_x\psi) dx \right)$

As previously, we will use the Schrodinger equation to replace $\partial \psi$ :

$-i\hbar\left( \int\frac{i}{\hbar}(\hat H \bar{\psi})(\partial_x\psi)dx+\int(\bar\psi) \frac{-i}{\hbar}(\partial_x\hat H \psi) \right)$

Now, using that hamiltonian is a hermitian operator we can write it as

$\int \bar \psi (\hat H \partial_x\psi - \partial_x \hat H \psi)dx$

Now using that $\hat H = -\frac{\hbar^2}{2m}\partial_{xx}+V$ we find

$\int \bar\psi (-\frac{\hbar^2}{2m}\partial_{xxx}\psi +V\partial_x \psi +\frac{\hbar^2}{2m}\partial_{xxx}\psi-(\partial_xV)\psi -V\partial_x\psi)dx$

$\int \bar\psi (-\partial_xV)\psi dx=<-\frac{\partial}{\partial x}V>$

$\frac{d}{dt}<p>=<-\frac{\partial V}{\partial x}>$