Derivation of Schrodinger's equation in momentum space

[Happiness]

Homework Statement

How do you get from (3.171) to (3.172)? In particular, why is

$\int e^{-ip.r/{\hbar}}\frac{p_{op}^2}{2m}\Psi(r,t)\,dr=\int\frac{p_{op}^2}{2m}[e^{-ip.r/{\hbar}}\Psi(r,t)]\,dr$? $\,\,\,\,\,$-- (1)

Homework Equations

The Attempt at a Solution

For (1) to be true, it must be true that
$\int\frac{\partial ^2}{\partial x^2}[e^{-ip.x/{\hbar}}\Psi(r,t)]\,dr=\int e^{-ip.x/{\hbar}}\frac{\partial ^2}{\partial x^2}\Psi(r,t)\,dr$. $\,\,\,\,\,$-- (2)

Using integration by parts and the fact that $\Psi(r,t)=0$ at infinity, I am able to show that (2) is true provided

$\int\frac{-ip}{\hbar}e^{-ipx/\hbar}\frac{\partial}{\partial x}\Psi(r,t)\,dr=0$ $\,\,\,\,\,$-- (3) or

$\int\Psi(r,t)\,(\frac{-ip}{\hbar})^2e^{-ipx/\hbar}\,dr=0$.$\,\,\,\,\,$-- (4)

But I don't know how (3) or (4) is true?

 

[DrClaude]

The author has left out steps leading to (3.171). The way it is written there, p is already an operator in momentum space, meaning that it is a simple multiplication.

Edit: Maybe it is not missing steps, but the author has taken the problem backward. I would start with the Schrödinger equation for $\Psi(\mathbf{r},t)$, write $\Psi(\mathbf{r},t)$ in terms of $\Phi(\mathbf{p},t)$, and get the Schrödinger equation for $\Phi(\mathbf{p},t)$.

 

[Happiness]

The author has left out steps leading to (3.171). The way it is written there, p is already an operator in momentum space, meaning that it is a simple multiplication.

Edit: Maybe it is not missing steps, but the author has taken the problem backward. I would start with the Schrödinger equation for $\Psi(\mathbf{r},t)$, write $\Psi(\mathbf{r},t)$ in terms of $\Phi(\mathbf{p},t)$, and get the Schrödinger equation for $\Phi(\mathbf{p},t)$.

Since the integration in (3.171) is done with respect to r, shouldn't p still be an operator in configuration space?

How I see the steps leading to (3.171) is as follows:

Since $\Psi(r,t)$ satisfies the Schrodinger equation, any linear combination of $\Psi$'s satisfies the Schrodinger equation too by the principle of superposition (or by the linearity of the operators in the Schrodinger equation).

 

[Happiness]

Oh I got it now. The author used

$H\Psi(r,t)=E\Psi(r,t)=[\frac{p^2}{2m}+V(r,t)]\Psi(r,t)$.

So the p$^2$ in (3.171) is a constant and not an operator.

 

[DrClaude]

Oh I got it now. The author used

$H\Psi(r,t)=E\Psi(r,t)=[\frac{p^2}{2m}+V(r,t)]\Psi(r,t)$.

So the p$^2$ in (3.171) is a constant and not an operator.

In the equation you wrote, p² has got to be an operator. It is in momentum space that it is a number.

I'm still trying to figure how to best approach the way it is presented in your book.

 

[Happiness]

I think I got it.

$(2\pi\hbar)^{-3/2}\int e^{-ip.r/\hbar}\frac{p_{op}^2}{2m}\Psi(r,t)\,dr\\$

first substitute $\Psi(r,t)$ in terms of $\Phi(p',t)$
$=(2\pi\hbar)^{-3/2}\int e^{-ip.r/\hbar}\frac{p_{op}^2}{2m}(2\pi\hbar)^{-3/2}\int e^{ip'.r/\hbar}\Phi(p',t)\,dp'\,dr\\$

then group the exponential terms together
$=\int\frac{p_{op}^2}{2m}(2\pi\hbar)^{-3}\int e^{i(p'-p).r/\hbar}\Phi(p',t)\,dp'\,dr\\$

since $p_{op}$ and the r integration commute, we have
$=\frac{p_{op}^2}{2m}\int(2\pi\hbar)^{-3}\int e^{i(p'-p).r/\hbar}\Phi(p',t)\,dp'\,dr\\$

since the r integration and the p' integration commute, we have
$=\frac{p_{op}^2}{2m}\int(2\pi\hbar)^{-3}\int e^{i(p'-p).r/\hbar}\,dr\,\Phi(p',t)\,dp'\\$

next perform the r integration
$=\frac{p_{op}^2}{2m}\int\delta(p'-p)\Phi(p',t)\,dp'\\ =\frac{p_{op}^2}{2m}\Phi(p,t)\\$

 

[Happiness]

I still have one unresolved question:

How does $p_{op}^2$ change from being $-\hbar^2\nabla^2$ to being $p^2$, where $p$ is a number? At which step does it occur and why?

 

[nrqed]

I think I got it.

$(2\pi\hbar)^{-3/2}\int e^{-ip.r/\hbar}\frac{p_{op}^2}{2m}\Psi(r,t)\,dr\\$

first substitute $\Psi(r,t)$ in terms of $\Phi(p',t)$
$=(2\pi\hbar)^{-3/2}\int e^{-ip.r/\hbar}\frac{p_{op}^2}{2m}(2\pi\hbar)^{-3/2}\int e^{ip'.r/\hbar}\Phi(p',t)\,dp'\,dr\\$

then group the exponential terms together
$=\int\frac{p_{op}^2}{2m}(2\pi\hbar)^{-3}\int e^{i(p'-p).r/\hbar}\Phi(p',t)\,dp'\,dr\\$

since $p_{op}$ and the r integration commute, we have
$=\frac{p_{op}^2}{2m}\int(2\pi\hbar)^{-3}\int e^{i(p'-p).r/\hbar}\Phi(p',t)\,dp'\,dr\\$

You cannot pull $p_{or}^2$ outside of the integral over r, since it is acting on that variable! You must apply it to the exponential and therefore it will generate a factor $p^2$ where now p is the variable, not an operator.

 

[Happiness]

I see the mistake.

$(2\pi\hbar)^{-3/2}\int e^{-ip.r/\hbar}\frac{p_{op}^2}{2m}\Psi(r,t)\,dr\\$

first substitute $\Psi(r,t)$ in terms of $\Phi(p',t)$
$=(2\pi\hbar)^{-3/2}\int e^{-ip.r/\hbar}\frac{p_{op}^2}{2m}(2\pi\hbar)^{-3/2}\int e^{ip'.r/\hbar}\Phi(p',t)\,dp'\,dr\\$

then group the exponential terms together (This step is wrong because $p_{op}^2$ is acting on $r$ but the exponential term has $r$ so they don't commute.)
$=\int\frac{p_{op}^2}{2m}(2\pi\hbar)^{-3}\int e^{i(p'-p).r/\hbar}\Phi(p',t)\,dp'\,dr\\$

instead it should be
$=(2\pi\hbar)^{-3/2}\int e^{-ip.r/\hbar}(2\pi\hbar)^{-3/2}\int\frac{p_{op}^2}{2m}e^{ip'.r/\hbar}\Phi(p',t)\,dp'\,dr\\$
$=(2\pi\hbar)^{-3/2}\int e^{-ip.r/\hbar}(2\pi\hbar)^{-3/2}\int\frac{p'^2}{2m}e^{ip'.r/\hbar}\Phi(p',t)\,dp'\,dr\\$

since the r integration and the p' integration commute, we have
$=\int\frac{p'^2}{2m}(2\pi\hbar)^{-3}\int e^{i(p'-p).r/\hbar}\,dr\,\Phi(p',t)\,dp'\\$

next perform the r integration
$=\int\frac{p'^2}{2m}\delta(p'-p)\Phi(p',t)\,dp'\\ =\frac{p^2}{2m}\Phi(p,t)\\$

 

[nrqed]

I see the mistake.

$(2\pi\hbar)^{-3/2}\int e^{-ip.r/\hbar}\frac{p_{op}^2}{2m}\Psi(r,t)\,dr\\$

first substitute $\Psi(r,t)$ in terms of $\Phi(p',t)$
$=(2\pi\hbar)^{-3/2}\int e^{-ip.r/\hbar}\frac{p_{op}^2}{2m}(2\pi\hbar)^{-3/2}\int e^{ip'.r/\hbar}\Phi(p',t)\,dp'\,dr\\$

then group the exponential terms together (This step is wrong because $p_{op}^2$ is acting on $r$ but the exponential term has $r$ so they don't commute.)
$=\int\frac{p_{op}^2}{2m}(2\pi\hbar)^{-3}\int e^{i(p'-p).r/\hbar}\Phi(p',t)\,dp'\,dr\\$

instead it should be
$=(2\pi\hbar)^{-3/2}\int e^{-ip.r/\hbar}(2\pi\hbar)^{-3/2}\int\frac{p_{op}^2}{2m}e^{ip'.r/\hbar}\Phi(p',t)\,dp'\,dr\\$
$=(2\pi\hbar)^{-3/2}\int e^{-ip.r/\hbar}(2\pi\hbar)^{-3/2}\int\frac{p'^2}{2m}e^{ip'.r/\hbar}\Phi(p',t)\,dp'\,dr\\$

since the r integration and the p' integration commute, we have
$=\int\frac{p'^2}{2m}(2\pi\hbar)^{-3}\int e^{i(p'-p).r/\hbar}\,dr\,\Phi(p',t)\,dp'\\$

next perform the r integration
$=\int\frac{p'^2}{2m}\delta(p'-p)\Phi(p',t)\,dp'\\ =\frac{p^2}{2m}\Phi(p,t)\\$

Perfect!

 

[rdhill13]

Hey I was wondering what textbook you were using for this question:And for the final solution, I had a convolution in there for the last integral in (3.172): V(p) * Psi(p,t)?

Cheers

 

[PeroK]

Hey I was wondering what textbook you were using for this question:And for the final solution, I had a convolution in there for the last integral in (3.172): V(p) * Psi(p,t)?

Cheers

The thread is four years old!