Homework Help: Quantum Mechanics: The First Step in Proving the Constancy of a Normalization

1. Jan 31, 2013

QuantumBunnii

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

This is a much more general question regarding differential equations; however, since it was presented in a quantum mechanics text (and physicists often make appeals to empirical considerations in their mathematics), I thought it might be appropriate to post here.
The following is the first step in Griffith's endeavor to prove the constant normalization of an evolving wavefunction:

$\frac{d}{dt} \int |ψ(x, t)|^2|\,dx$ = $\int \frac{\partial }{\partial t}|ψ(x, t)|^2|\,dx$

He notes "that the integral is a function only of t, so I use a total derivative (d/dt) in the first expression, but the integrand is a function of x as well as t, so it's a partial derivative in the second one.

Firstly, I don't understand why he says that the intregral is a function only of t. To me, it seems that nothing in this equation is a function *only* of t, as the wavefunction is a function of x and t evaluated with respect to x.
Because of this, I also don't see what justifies going from a total derivative to a partial one. Consider the simple case of integrating the function $ψ= xt$:

$\frac{d}{dt} \int xt\,dx = \frac{d}{dt} \frac{1}{2} x^2 t + C = \frac{1}{2} (x^2 + \frac{dx^2}{dt}t)$

$\int \frac{\partial }{\partial t}xt\,dx = \int x\,dx = \frac{1}{2} x^2 + C$

The two cases clearly aren't the same.

2. Relevant equations

N/A

3. The attempt at a solution

My thoughts are:

(1) Perhaps $\frac{dx^2}{dt}$ is simply constant with respect to time, which would in fact make the two cases equal. However, this doesn't seem very reasonable, as is it a common feat of quantum mechanics to calculate the average velocity of the particle.

(2) Maybe my attempt at pursuing the integral in the first case is flawed, as t is itself a variable which should not be ignored in the integral. This mistake may be what deprives the equality (of course, I'm not sure).

This may seem a very elementary or otherwise intuitive concern, but it's been bothering me for quite some time. Any help is appreciated. :)

2. Jan 31, 2013

cepheid

Staff Emeritus
I think you're forgetting that the integral for normalization of the wavefunction is actually $$\int_{-\infty}^{\infty} |\psi(x,t)|^2\,dx$$ a definite integral, with limits. If you integrate this multivariable function over all possible values of x, you eliminate x as a variable, leaving only a variation with t. (Just like when you integrate a function of a single variable over some definite interval, you eliminate that variable, leaving you with a constant value). In the multivariable case, it's like you do a whole continuous sequence of such integrals with respect to x, one for each value of t. So what you're left with is a constant value at each value of t.

To visualize this: think about the domain of the function: the 2-dimensional x-t plane. The function |ψ|^2 can be represented as some 2D surface over top of this plane: the height of the surface above the point (x,t), being the value of |ψ|^2 at those values of x and t. Now, integrating over x, corresponds to taking slices of this surface along one axis and adding them all together. The end result is a 1D curve that goes only in the t-direction.

Last edited: Jan 31, 2013
3. Feb 1, 2013

QuantumBunnii

Thanks!

I'm not quite sure I can 'visualize' it too well, but-- in spite of this-- I think I understand it much better now.
The most important facet of this integral is the fact that it's, as you said, definite . This would force the x-variables to become constants after the integral is pursued, in which case we will be differentiating only t-variables. In this sense, it wouldn't matter whether we differentiate partially inside the integral (simply treating all x variables as constants before) or differentiating totally outside of it (treating all x variables as constants after). You would get the same result.

-- Thanks again for the help. :)

4. Feb 1, 2013

DimReg

The x variable in the wavefunction is just a coordinate, it's not the position of the particle or any other dynamical quantity. So the time derivative of x^2 is actually zero, and the two integrals are the same (the two you showed were "clearly not the same").

Specifically in this case, the "x" is a position operator, expressed in the x basis, and in the representation where operators are constant in time (I think it's the schrodinger picture?). While everything said about the integral being definite are true, this is what's "really" going on. By that I mean Griffiths is assuming you know that it is ψ(x,t) not ψ(x(t),t), which is why he doesn't spend much time explaining that step of the derivation to you.

This is actually an important point for understanding the physics of QM. It means two completely different things for the time derivative of the x operator to be non zero, and the time derivative of the expectation value of x to be non zero. The first is pure mathematics, the second is directly physical.

5. Feb 3, 2013

aim1732

Isn't knowing x(t) where x is the position of the particle beyond QM? I mean you know it's position at any time with certainty and you haven't even said that the wave fn. is a position eigenfunction. Plus,differentiating x(t) will give p(t)?I am not sure what's going on.

Also I do remember somewhere in the Ehrenfest theorem we explicitly say ∂x/∂t is zero.

6. Feb 3, 2013

cepheid

Staff Emeritus
You're right, there is no such thing as x(t) (a deterministic particle trajectory). However, I don't think that anybody above claimed that it existed, or that we were differentiating it.