Derivation of the velocity of an expectation value

[mtmn]

Homework Statement

I am trying to derive for myself the velocity of the expectation value from the information given, specifically that

<x> = \int_{-\infty}^{\infty}x|\Psi (x,t)|^2 dt (1)

Eq (1) can be transformed into,

\frac{d<x>}{dt} = -\frac{i\hbar}{m}\int\Psi^*\frac{\partial\Psi}{\partial x}dx (2)

I couldn't come up with Eq. (2). Below is my work.

2. Homework Equations .

Schr\ddot{o}dinger Equation
i\hbar\frac{\partial\Psi}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}+V\Psi

The Attempt at a Solution

Taking the derivative of both sides of Eq. (1) results in,

\frac{d<x>}{dt} = \int x\frac{\partial}{\partial t}|\Psi|^2 dx (3)

As \Psi^2 = \Psi^*\Psi taking the derivative of the r.h.s of Eq.(3) results in,

\int x\left(\Psi^*\frac{\partial\Psi}{\partial t}+\Psi\frac{\partial\Psi^*}{\partial t}\right) (4)

Manipulating the Schr\ddot{o}dinger Equation to get \frac{\partial\Psi}{\partial T}

-\frac{\partial\Psi}{\partial t} = \frac{\hbar}{2im}{\partial^2\Psi}{\partial x^2}-\frac{V\Psi}{i\hbar} (5)

Multiplying both sides of Eq. (5) by i^2,

\frac{\partial\Psi}{\partial t} = \frac{i\hbar}{2m}\frac{\partial^2\Psi}{\partial x^2}-\frac{iV\Psi}{\hbar} (6)

Taking the complex conjugate of Eq. (6),

\frac{\partial\Psi^*}{\partial t} = -\frac{i\hbar}{2m}\frac{\partial^2\Psi^*}{\partial x^2}+\frac{iV\Psi^*}{\hbar}(7)

Substituting Eq. (6) & (7) into Eq. (4) results in,

=\Psi^*\frac{i\hbar}{2m}\frac{\partial^2\Psi}{\partial x^2}-V\frac{i\Psi^*\Psi}{\hbar}-\Psi\frac{i\hbar}{2m}\frac{\partial^2\Psi^*}{\partial x^2}+V\frac{i\Psi^*\Psi}{\hbar}

=\Psi^*\frac{i\hbar}{2m}\frac{\partial^2\Psi}{\partial x^2}-\Psi\frac{i\hbar}{2m}\frac{\partial^2\Psi^*}{\partial x^2}

=\frac{i\hbar}{2m}\left(\Psi^*\frac{\partial^2\Psi}{\partial x^2}-\Psi\frac{\partial^2\Psi^*}{\partial x^2}\right)(8)

Eq. (8) is equivalent to,

\frac{i\hbar}{2m}\left[\frac{\partial}{\partial x}\left(\Psi^*\frac{\partial\Psi}{\partial x}-\Psi\frac{\partial\Psi^*}{\partial x}\right)\right](9)

(The inside can be confirmed just by taking the derivative, which I'm going to leave out for brevities sake)

So, the velocity of the expectation value now becomes,

\frac{d<x>}{dt} = \frac{i\hbar}{2m}\int x\frac{\partial}{\partial x}\left(\Psi^*\frac{\partial\Psi}{\partial x}-\Psi\frac{\partial\Psi^*}{\partial x}\right)dx(10)

Taking the constant out I solved the integrand by using integration by parts.

Let u=x
dv=\frac{\partial}{\partial x}\left(\Psi^*\frac{\partial\Psi}{\partial x}-\Psi\frac{\partial\Psi^*}{\partial x}\right)dx.

Then, du=dx
v=\left(\Psi^*\frac{\partial\Psi}{\partial x}-\Psi\frac{\partial\Psi^*}{\partial x}\right)

Substituting this into \int udv = \int uv-\int vdu

x\left(\Psi^*\frac{\partial\Psi}{\partial x}-\Psi\frac{\partial\Psi^*}{\partial x}\right)\bigg|_{\infty}^{\infty}-\int \left(\Psi^*\frac{\partial\Psi}{\partial x}-\Psi\frac{\partial\Psi^*}{\partial x}\right)dx (11)

Before the integral that term goes to 0 because (as I'm told, I don't actually know how to prove this) \Psi goes to zero must quicker then x goes to infinity. So the velocity becomes,

\frac{d<x>}{dt} = -\frac{i\hbar}{2m}\int \left(\Psi^*\frac{\partial\Psi}{\partial x}-\Psi\frac{\partial\Psi^*}{\partial x}\right)dx(12)

I transformed,

\Psi\frac{\partial\Psi^*}{\partial x}=\frac{\partial\Psi^*\Psi}{\partial x}-\Psi^*\frac{\partial\Psi}{\partial x}(13)

Substituting Eq (13) into Eq (12),

\frac{d<x>}{dt} = -\frac{i\hbar}{2m}\int \left(\Psi^*\frac{\partial\Psi}{\partial x}-\frac{\partial\Psi^*\Psi}{\partial x}+\Psi^*\frac{\partial\Psi}{\partial x}\right)dx

=\frac{d<x>}{dt} = -\frac{i\hbar}{2m}\int \left(2\Psi^*\frac{\partial\Psi}{\partial x}-\frac{\partial\Psi^*\Psi}{\partial x}\right)dx(14)

Eq.(14) Becomes

\frac{d<x>}{dt} = -\frac{i\hbar}{m}\int \Psi^*\frac{\partial\Psi}{\partial x}dx+\frac{i\hbar}{2m}\int\frac{\partial\Psi^*\Psi}{\partial x}dx(15)

The left part of the r.h.s. of Eq.(15) is what I want but I don't know how to eliminate the right portion of it.

Anyone have any ideas??

Thanks in advance!

Matt

 

[Dickfore]

Eqn. (15) Looks incorrect. You should only have the first term on the rhs.

EDIT:
Of course, the second term is an integral of a derivative of |\Psi|^{2} = \Psi^{\ast} \, \Psi, so it can be integrated. But, the wavefunction disappears at infinity, so the contribution from that term is zero.

 

[mtmn]

So, I can assume again that \Psi^2\Psi goes to zero as it goes to infinity? If that's the case then thanks! =)

 

[Dickfore]

So, I can assume again that \Psi^2\Psi goes to zero as it goes to infinity? If that's the case then thanks! =)

Of course. This is a necessary (but not sufficient) condition for the probability normalization integral:

<br /> \int_{-\infty}^{\infty}{|\Psi(x, t)|^{2} \, dx} = 1<br />

to converge.