Homework Help: Derivation of the velocity of an expectation value

1. May 31, 2010

mtmn

1. The problem statement, all variables and given/known data
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. Relevant Equations.

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

3. 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??

Matt

2. May 31, 2010

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.

3. May 31, 2010

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!! =)

4. May 31, 2010

Dickfore

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

$$\int_{-\infty}^{\infty}{|\Psi(x, t)|^{2} \, dx} = 1$$

to converge.