Derivation of the Quantum Momentum
I’ve been working through Griffiths’ book on Quantum mechanics, and have been confused by his derivation of the quantum momentum operator. Here I attempt to derive it following the same steps as Griffiths, but with more detail where I got lost. The equation numbers correspond to the same equations within the book.
We start with the expectation value of the position of a particle (given its wave function): $$ \left<x\right> = \int_{-\infty}^{\infty} x \ \left|\Psi \left( x, t \right) \right|^2 dx \tag{1.28}$$
As this is the expectation value of position, we can then find how it changes over time - lets call this value the expectation value of velocity (which is intuiviely related to momentum): $$ \left<v\right> = \frac{d \left<x\right>}{dt} \tag{1.32}$$
Now we can try to expand the RHS, taking note of how the total derivative $\frac{d}{dt}$ becomes a partial derivative $\frac{\partial}{\partial t}$ as it acts on $\Psi(x, t)$: $$ \frac{d \left<x\right>}{dt} = \frac{d}{dt} \int_{-\infty}^{\infty} x \ \left|\Psi \left( x, t \right) \right|^2 dx = \int_{-\infty}^{\infty} x \ \frac{\partial }{\partial t} |\Psi(x, t)|^2 dx $$
Finding $\frac{\partial}{\partial t} |\Psi|^2$
Here we turn to calculate the inner term $\frac{\partial }{\partial t} |\Psi(x, t)|^2$. Remember that $\Psi$ is complex, and as such $|\Psi|^2 = \Psi^* \Psi$, where $\Psi^*$ is the complex congujate of $\Psi$. Using the product rule: $$ \frac{\partial}{\partial t} |\Psi|^2 = \frac{\partial}{\partial t} (\Psi^* \Psi) = \Psi^*\frac{\partial \Psi}{\partial t} + \frac{\partial \Psi^*}{\partial t}\Psi \tag{1.22}$$
Given a rearranged Schrödinger equation and its complex conjugate: $$ \frac{\partial \Psi}{\partial t} = \frac{i \hbar}{2m}\frac{\partial^2 \Psi}{\partial x^2} - \frac{i}{\hbar}V\Psi \tag{1.23}$$ $$ \frac{\partial \Psi^*}{\partial t} = -\frac{i \hbar}{2m}\frac{\partial^2 \Psi^*}{\partial x^2} + \frac{i}{\hbar}V\Psi^* \tag{1.24}$$
We can plug these equations back into equation 1.22 resulting in: $$ \frac{\partial}{\partial t} |\Psi|^2 = \Psi^*\left( \frac{i \hbar}{2m}\frac{\partial^2 \Psi}{\partial x^2} - \frac{i}{\hbar}V\Psi \right) + \left( -\frac{i \hbar}{2m}\frac{\partial^2 \Psi^*}{\partial x^2} + \frac{i}{\hbar}V\Psi^* \right)\Psi $$ Note that the $ \frac{i}{h}V\Psi^*\Psi $ term cancels out leaving: $$ \frac{\partial}{\partial t} |\Psi|^2 = \Psi^*\left( \frac{i \hbar}{2m}\frac{\partial^2 \Psi}{\partial x^2} \right) - \left( -\frac{i \hbar}{2m}\frac{\partial^2 \Psi^*}{\partial x^2} \right) \Psi = \frac{i\hbar}{2m} \left( \Psi^* \frac{\partial^2 \Psi}{\partial x^2} - \frac{\partial^2 \Psi^*}{\partial x^2} \Psi \right)$$
Now while we could stop here, we can actually futher simplfy this by “extacting” a $\frac{\partial}{\partial x}$ from the RHS. I’ll do this by first giving you the simplified form and proving the equality. We will see that: $$ \boxed{ \frac{\partial}{\partial t} |\Psi|^2 = \frac{i\hbar}{2m} \left( \Psi^* \frac{\partial^2 \Psi}{\partial x^2} - \frac{\partial^2 \Psi^*}{\partial x^2} \Psi \right) = \frac{\partial}{\partial x} \left[ \frac{i\hbar}{2m} \left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right)\right] } \tag{1.25}$$
This is important as the RHS of 1.25 is what we will return and subsitute in to find $\frac{d\left<x\right>}{dt}$.
“Extraction” of $\frac{\partial}{\partial x}$
How did we move $\frac{\partial}{\partial x}$? Let’s “distribute” the $\frac{\partial}{\partial x}$ using the product rule and see. Ignoring the constants: $$ \frac{\partial}{\partial x} \left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) = \left(\frac{\partial \Psi^*}{\partial x}\frac{\partial \Psi}{\partial x} + \Psi^*\frac{\partial^2 \Psi}{\partial x^2}\right) - \left(\frac{\partial^2 \Psi^*}{\partial x^2}\Psi + \frac{\partial \Psi^*}{\partial x}\frac{\partial \Psi}{\partial x}\right) $$
Note the terms that cancel: The first term $\frac{\partial \Psi^*}{\partial x}\frac{\partial \Psi}{\partial x}$ and the last term $\frac{\partial \Psi^*}{\partial x}\frac{\partial \Psi}{\partial x}$! We see that the expanded form of the expression ends up being: $$ \frac{\partial}{\partial x} \left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) = \Psi^* \frac{\partial^2 \Psi}{\partial x^2} - \frac{\partial^2 \Psi^*}{\partial x^2} \Psi $$ Which proves Equation 1.25.
Returning to find $\frac{d\left<x\right>}{dt}$
We take the results from Equation 1.25 and see that: $$ \frac{d \left<x\right>}{dt} = \int x \ \frac{\partial}{\partial t}|\Psi|^2 dx = \int x \ \frac{\partial}{\partial x} \left[ \frac{i\hbar}{2m} \left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right)\right] dx \tag{1.29}$$
This is another spot where I needed more verbose derivations that what was given in Griffiths. We can actually simplify this further with an “obvious” integration-by-parts. Integration-by-parts is often written as: $$ \int u dv = uv - \int v du $$
But this terse form omits the boundary terms and the common variable. It is more helpful in this form: $$ \int_{a}^{b} u \frac{dv}{dx} dx = uv \ \Biggr|_{a}^{b} - \int_{a}^{b} \frac{du}{dx} v dx $$
Now when we look at Equation 1.29, we can see that a smart choice for our smaller functions is: $$ u(x) = x $$ $$ v(x) = \left[ \frac{i\hbar}{2m} \left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right)\right] $$ We can see that $v(x)$ was inside of the $\frac{\partial}{\partial x}$ in Equation 1.29.
After putting the bounds back in, we can replace 1.29 using the integration-by-parts. For readability lets have $ \frac{dv}{dx} = \frac{\partial}{\partial x} \left[ … \right]$: $$\begin{align} \int_{-\infty}^{\infty} x \ \frac{\partial}{\partial x} \left[ … \right] dx & = uv \ \Biggr|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \frac{du}{dx} v dx \\ & = x v \ \Biggr|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} 1 \ v dx \\ & = x \frac{i\hbar}{2m} \left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) \ \Biggr|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \frac{i\hbar}{2m} \left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) dx \end{align}$$
As the $\Psi$ & $\Psi^*$ go to 0 as we approach $\infty$, we can get rid of the first term. This leaves:
$$ \int_{-\infty}^{\infty} x \ \frac{\partial}{\partial x} \left[ … \right] dx = - \frac{i\hbar}{2m} \int_{-\infty}^{\infty} \left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) dx $$
This can be even further simplified using another integration-by-parts as the second term has a similar form to to the last equation. We allow $u = \Psi$ and $v = \Psi^*$: $$ \int_{-\infty}^{\infty} \frac{\partial \Psi^*}{\partial x} \Psi \ dx = \Psi \Psi^* \Biggr|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \frac{\partial \Psi}{\partial x} \Psi^* dx $$
Again, the first term goes to 0 and we see this is the same term.
$$\begin{align} \int_{-\infty}^{\infty} x \ \frac{\partial}{\partial x} \left[ … \right] dx & = - \frac{i\hbar}{2m} \int_{-\infty}^{\infty} \left( \Psi^* \frac{\partial \Psi}{\partial x} - \left[ - \frac{\partial \Psi}{\partial x} \Psi^* dx \right] \right) \\ & = - \frac{i\hbar}{2m} \int_{-\infty}^{\infty} \left( \Psi^* \frac{\partial \Psi}{\partial x} + \frac{\partial \Psi}{\partial x} \Psi^* \right) dx \\ & = - \frac{i\hbar}{m} \int_{-\infty}^{\infty} \left( \Psi^* \frac{\partial \Psi}{\partial x} \right) dx \end{align}$$
Looking at Equation 1.29, we finally see that: $$ \frac{d \left<x\right>}{dt} = - \frac{i\hbar}{m} \int_{-\infty}^{\infty} \left( \Psi^* \frac{\partial \Psi}{\partial x} \right) dx \tag{1.31}$$
Now as we defined the expectation value of velocity (Equation 1.32), we can multiply by $m$ to get the expectation value of momentum. $$ \boxed{ \left<p\right> = m \frac{d\left<x\right>}{dt} = -i\hbar \int \left( \Psi^* \frac{\partial \Psi}{\partial x} \right) dx } \tag{1.33}$$