Derivation of the velocity of an expectation value

AI Thread Summary
The discussion centers on deriving the velocity of the expectation value from the given equation for the expectation value of position. The user initially struggles to derive the equation for the velocity, leading to a series of manipulations involving the Schrödinger equation and integration by parts. Ultimately, they arrive at an expression for the velocity, but are unsure about eliminating a term that appears in their final equation. Other participants confirm that the additional term can be disregarded since it integrates to zero due to the wavefunction's behavior at infinity, reinforcing the necessity for the wavefunction to vanish at infinity for normalization. The conversation concludes with a consensus on the correctness of this assumption.
mtmn
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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
 
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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.
 
So, I can assume again that \Psi^2\Psi goes to zero as it goes to infinity? If that's the case then thanks! =)
 
mtmn said:
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.
 
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