kokoszzm

2022-06-27

I found in many places that the average momentum of a particle is given by:
$⟨p⟩={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\psi }^{\ast }\left(\frac{\hslash }{i}\right)\frac{\mathrm{\partial }\psi }{\mathrm{\partial }x}\phantom{\rule{mediummathspace}{0ex}}\mathrm{d}x$
I think that it comes from considering the classical momentum:
$⟨p⟩=m\frac{\mathrm{d}⟨x⟩}{\mathrm{d}t}$
and that the expected value of the position is given by:
$⟨x⟩={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x\phantom{\rule{mediummathspace}{0ex}}{|\psi \left(x,t\right)|}^{2}\phantom{\rule{mediummathspace}{0ex}}\mathrm{d}x$
But when replacing $⟨x⟩$ and differentiating inside the integral I don't know how to handle the derivatives of $\psi$ for getting the average momentum formula. Any suggestion?

laure6237ma

Use product rule to get the above into the form:

The second term contains $\frac{\mathrm{\partial }x}{\mathrm{\partial }t}$, which is 0.
This is just calculus. Now comes the crucial step of imposing physics: the Schrodinger equation: $i\hslash \frac{\mathrm{\partial }\mathrm{\Psi }}{\mathrm{\partial }t}=\stackrel{^}{H}\mathrm{\Psi }$ (and also $-i\hslash \frac{\mathrm{\partial }{\mathrm{\Psi }}^{\ast }}{\mathrm{\partial }t}=\stackrel{^}{H}{\mathrm{\Psi }}^{\ast }$)
Write the operator $\stackrel{^}{H}$ in terms of second order spatial derivative (acting on $\mathrm{\Psi }$ and ${\mathrm{\Psi }}^{\ast }$). Through Schrodinger's equation, you get a relation between second order spatial derivative and first order time derivative. Replace the first order time derivatives in (1) with second order spatial derivatives. And then integrate by parts, to reduce the second order spatial derivative to first order spatial derivative, using the boundary condition that $\frac{d\mathrm{\Psi }}{dx}\left(\text{and}\frac{d{\mathrm{\Psi }}^{\ast }}{dx}\right)$ go to zero at infinity. After a few steps of algebra, and you get it to the form , which is what you wanted: $⟨\stackrel{^}{p}⟩$
This (or a similar) calculation is usually given in various resources. Can you take it from here and do it yourself?

Roland Waters

Since your question is about the single-particle momentum operator, I will assume that we are dealing with a single particle moving in an arbitrary scalar potential. Suppose we start with
$⟨p⟩=m\frac{d⟨x⟩}{dt}$
and
$⟨x⟩=\int x|\psi {|}^{2}dx=\int {\psi }^{\ast }x\psi \phantom{\rule{thickmathspace}{0ex}}dx.$
Applying the time derivative to the above yields
$\frac{d⟨x⟩}{dt}=\int \frac{\mathrm{\partial }{\psi }^{\ast }}{\mathrm{\partial }t}x\psi +{\psi }^{\ast }\frac{\mathrm{\partial }x}{\mathrm{\partial }t}\psi +{\psi }^{\ast }x\frac{\mathrm{\partial }\psi }{\mathrm{\partial }t}\phantom{\rule{thickmathspace}{0ex}}dx$
Since the operator x has no explicit time dependence, the middle term is zero. In addition, the Schrodinger equation states
$\frac{\mathrm{\partial }\psi }{\mathrm{\partial }t}=\frac{1}{i\hslash }H\psi .$
Applying this to ${\psi }^{\ast }$, we have
$\frac{\mathrm{\partial }{\psi }^{\ast }}{\mathrm{\partial }t}=\frac{-1}{i\hslash }{\psi }^{\ast }H$
and substituting both of these in for the appropriate quantities in $\frac{d⟨x⟩}{dt}$:
$\frac{d⟨x⟩}{dt}=\int \frac{-1}{i\hslash }{\psi }^{\ast }Hx\psi +\frac{1}{i\hslash }{\psi }^{\ast }xH\psi \phantom{\rule{thickmathspace}{0ex}}dx=\frac{1}{i\hslash }\int {\psi }^{\ast }\left[x,H\right]\psi \phantom{\rule{thickmathspace}{0ex}}dx.$
Now all that is left is to calculate the commutator $\left[x,H\right]$. Since we're dealing with a single particle in an arbitrary scalar potential, we can write $H=\frac{{p}^{2}}{2m}+V\left(x,t\right)$ so that:
$\left[x,H\right]=\left[x,\frac{{p}^{2}}{2m}+V\left(x,t\right)\right]=\left[x,\frac{{p}^{2}}{2m}\right]+\left[x,V\left(x,t\right)\right].$
Since V(x,t) is a function of x alone*, we have that $\left[x,V\left(x,t\right)\right]=0$. We then use the commutator identity
$\left[A,BC\right]=\left[A,B\right]C+B\left[A,C\right]$
to write
$\left[x,\frac{{p}^{2}}{2m}\right]=\frac{1}{2m}\left(\left[x,p\right]p+p\left[x,p\right]\right)=\frac{1}{2m}\left(i\hslash p+pi\hslash \right)=\frac{i\hslash }{m}p$
since $\left[x,p\right]=i\hslash$. The representation of p in the position basis is $-i\hslash \frac{\mathrm{\partial }}{\mathrm{\partial }x}$, so in the position basis
$\left[x,H\right]=\frac{{\hslash }^{2}}{m}\frac{\mathrm{\partial }}{\mathrm{\partial }x}.$
Finally, substituting, we have that
$\frac{d⟨x⟩}{dt}=\frac{1}{i\hslash }\int {\psi }^{\ast }\left[x,H\right]\psi \phantom{\rule{thickmathspace}{0ex}}dx=\int {\psi }^{\ast }\left(\frac{\hslash }{im}\right)\frac{\mathrm{\partial }\psi }{\mathrm{\partial }x}\phantom{\rule{thickmathspace}{0ex}}dx$
so that
$⟨p⟩=m\frac{d⟨x⟩}{dt}=\int {\psi }^{\ast }\left(\frac{\hslash }{i}\right)\frac{\mathrm{\partial }\psi }{\mathrm{\partial }x}\phantom{\rule{thickmathspace}{0ex}}dx.$
*What I mean here is that V is not a function of any other operators, and t isn't an operator in quantum mechanics, only a parameter.

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