# QM Notes – Momentum Operator

The probability of finding a particle between a and b at time t, is,…
$\int\limits_a^b \left |\psi(x,t)\right |^2 dx$
Since the particle must be somewhere the statistical interpretation of QM requires the normalization condition,…
$\int\limits_\infty^\infty \left |\psi(x,t)\right |^2 dx = 1$
In general the average value of some function f() is (called the ‘expectation value’ in QM),…
$\textless f(j) \textgreater = \sum\limits_{j=0}^\infty f(j)p(j)$ ,….. where p(j) is the probability of f(j).
So, for example, the expectation value for position, x, would be,…
$\textless x \textgreater = \int\limits_{-\infty}^\infty x \left |\psi(x,t)\right |^2 dx \text{ ....or .... } = \int\limits_{-\infty}^\infty x \psi^* \psi dx$
Observables in quantum mechanics are represented by Hermitian operators, self-adjoint operators or matricies that are equal to their own transpose-conjugate. (i.e. position, momentum, Hamiltonian, etc). This is required so that eiginvalues are Real numbers, and to facilitate the association of these operators with the orthonormal basis underlying Hilbert space. On account of this, the x-operator above, can be moved over to operate on the second factor of the inner product…
$\textless x \psi^* , \psi \textgreater = \textless \psi^*, x \psi \textgreater$
$\textless x \textgreater = \int\limits_{-\infty}^\infty \psi^* x \psi \text{ }dx$

Now, this position expectation value changes as the wavefunction $\psi$ evolves in time. The expectation value of this rate of change, the “velocity”, would be,…
$\frac{d\textless x \textgreater}{dt} = \int\limits_{-\infty}^\infty x \frac{\partial}{\partial t} (\psi^* \psi) \text{ } dx$ , ….. or via Leibniz rule ,….

$\frac{d\textless x \textgreater}{dt} = \int\limits_{-\infty}^\infty x (\psi^* \frac{\partial\psi} {\partial t} + \frac{\partial\psi^*} {\partial t}\psi) \text{ } dx$ ………………..(1)

The Schrödinger equation shows what the rate of change of $\psi$ with respect to time must be,…
$\frac{\partial\psi}{\partial t} = \frac{i\hbar}{2m}\frac{\partial^2\psi}{\partial x^2} - \frac{i}{\hbar} V \psi$ … and its complex conjugate,… $\frac{\partial\psi^*}{\partial t} = -\frac{i\hbar}{2m}\frac{\partial^2\psi^*}{\partial x^2} + \frac{i}{\hbar} V \psi^*$

Substituting these into equation (1) above, allows the potential terms, $\pm \frac{i}{\hbar} V\psi^*\psi$, to cancel out, and results in,….
$\frac{d\textless x \textgreater}{dt} = \int\limits_{-\infty}^\infty x \frac{i\hbar}{2m}\left (\psi^*\frac{\partial^2\psi}{\partial x^2} - \frac{\partial^2\psi^*}{\partial x^2}\psi \right ) \text{ } dx$
Factoring out a partial derivative operator and moving the constants outside the integral,…
$\frac{d\textless x \textgreater}{dt} = \frac{i\hbar}{2m} \int\limits_{-\infty}^\infty x \frac{\partial}{\partial x} \left (\psi^*\frac{\partial\psi}{\partial x} - \frac{\partial\psi^*}{\partial x}\psi \right ) \text{ } dx$ ………………..(2)

This equation can be simplified using ‘integration by parts’ which works as follows;
———————————————————-
Starting with a derivative of a product and expanding via Leibniz rule, ….
$\frac{d(fg)}{dx} = f \frac{dg}{dx} + \frac{df}{dx} g$ …. or,…. $\frac{d(fg)}{dx} - \frac{df}{dx} g = f \frac{dg}{dx}$
Integrating each side,….
$fg\bigg|_{-\infty}^\infty - \int\frac{df}{dx} g = \int f \frac{dg}{dx}$
The wavefunction will be zero at the boundary or at infinity, so that term can be dropped.
$\int f \frac{dg}{dx} = - \int\frac{df}{dx} g$
The result is equivalent to moving the f into the derivative and the g out, and then negating.
———————————————————-

Performing an ‘integration by parts’ on equation (2) above will move the $x$ into the $\frac{\partial}{\partial x}$, and the $\left (\psi^*\frac{\partial\psi}{\partial x} - \frac{\partial\psi^*}{\partial x}\psi \right )$ out, plus negating results in,…
$\frac{d\textless x \textgreater}{dt} = -\frac{i\hbar}{2m} \int\limits_{-\infty}^\infty \frac{\partial x}{\partial x} \left (\psi^*\frac{\partial\psi}{\partial x} - \frac{\partial\psi^*}{\partial x}\psi \right ) \text{ } dx$

Performing a integration by parts again, on the second term in parentheses, will move the $\psi$ into the partial derivative, and the $\psi^*$ out, and negating,….
$\frac{d\textless x \textgreater}{dt} = -\frac{i\hbar}{2m} \int\limits_{-\infty}^\infty \left (\psi^*\frac{\partial\psi}{\partial x} + \psi^*\frac{\partial\psi}{\partial x} \right ) \text{ } dx$ ,….

$\frac{d\textless x \textgreater}{dt} = -\frac{i\hbar}{2m} \int\limits_{-\infty}^\infty 2 \psi^*\frac{\partial\psi}{\partial x} \text{ } dx$ ,….

$\frac{d\textless x \textgreater}{dt} = -\frac{i\hbar}{m} \int\limits_{-\infty}^\infty \psi^*\frac{\partial\psi}{\partial x} \text{ } dx$ ,….

$\frac{d\textless x \textgreater}{dt} = \frac{1}{m} \int\limits_{-\infty}^\infty \psi^*\frac{\hbar}{i}\frac{\partial}{\partial x}\psi\text{ } dx$

Now, momentum is mass times velocity, so multiplying both sides by m, will result in the ‘expectation value’ of momentum,….
$m\frac{d\textless x \textgreater}{dt} = \int\limits_{-\infty}^\infty \psi^*\frac{\hbar}{i}\frac{\partial}{\partial x}\psi\text{ } dx$
This is similar in form to the expectation value equation for position above,….
$\textless x \textgreater = \int\limits_{-\infty}^\infty \psi^* x \psi \text{ }dx$
Where the position operator, x, was placed between $\psi^*$ and $\psi$. Therefore, momentum in quantum mechanics is associated with an operator,…
$p = -i\hbar\frac{\partial}{\partial x}$

The first term on the right side of the Schrödinger equation is the momentum form of kinetic energy….

$i\hbar\frac{\partial\psi}{\partial t} = \frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2} - V \psi$

kinetic energy $= \frac{1}{2}mv^2 = \frac{1}{2}\frac{m^2v^2}{m} = \frac{p^2}{2m} = \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}$

-mjg