# Thread: Quantum Mechanics

1. ## Quantum Mechanics

This is for a physics course, but it's very intensive on math.

I have to prove:

$\displaystyle \int_{-\infty}^{\infty} \bar{\Psi} (x)\Psi (x)dx = 1$ is equivalent to:

$\displaystyle \int_{-\infty}^{\infty} \bar{a} (p) a(p)dp = 1$

This is all part of Schrodinger stuff. Obvious the bar over psi above is the conjugrate, and the same goes for a.

We are working on Fourier transforms. I'm not sure if anyone is familiar with this, but I think topsquark (?) might be.

Thanks.

2. Originally Posted by DiscreteW
This is for a physics course, but it's very intensive on math.

I have to prove:

$\displaystyle \int_{-\infty}^{\infty} \bar{\Psi} (x)\Psi (x)dx = 1$ is equivalent to:

$\displaystyle \int_{-\infty}^{\infty} \bar{a} (p) a(p)dp = 1$

This is all part of Schrodinger stuff. Obvious the bar over psi above is the conjugrate, and the same goes for a.

We are working on Fourier transforms. I'm not sure if anyone is familiar with this, but I think topsquark (?) might be.

Thanks.
In a nutshell, you go to momentum-space via a Fourier transform, set up the integral and use a property of the dirac delta function.

I have to go for a while. When I come back, I'll set out some details for you, unless I get beaten to the punch.

3. Originally Posted by DiscreteW
This is for a physics course, but it's very intensive on math.

I have to prove:

$\displaystyle \int_{-\infty}^{\infty} \bar{\Psi} (x)\Psi (x)dx = 1$ is equivalent to:

$\displaystyle \int_{-\infty}^{\infty} \bar{a} (p) a(p)dp = 1$

This is all part of Schrodinger stuff. Obvious the bar over psi above is the conjugrate, and the same goes for a.

We are working on Fourier transforms. I'm not sure if anyone is familiar with this, but I think topsquark (?) might be.

Thanks.
I don't like your notation but I'll play along.

I'm playing a bit loose with the $\displaystyle 2 \pi 's$ but this should see you right:

Moving from momentum space via a Fourier transform, your wave functions are:

$\displaystyle \Psi (x) = \int_{-\infty}^{\infty} a(p) e^{ipx} \, dp$.

$\displaystyle \bar{\Psi} (x) = \int_{-\infty}^{\infty} \bar{a}(p) e^{-ipx} \, dp = \int_{-\infty}^{\infty} \bar{a}(p') e^{-ip'x} \, dp'$. This change in dummy variable is essential to what follows so bear with it.

Then $\displaystyle \int_{-\infty}^{\infty} \bar{\Psi} (x)\Psi (x)dx = 1$ is equivalent to:

$\displaystyle 1 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} a(p) e^{ipx} \, dp \, \int_{-\infty}^{\infty} \bar{a}(p') e^{-ip'x} \, dp'\, \, dx$

$\displaystyle = \int_{-\infty}^{\infty} a(p) \, \int_{-\infty}^{\infty} \bar{a}(p')\, \int_{-\infty}^{\infty} e^{-ix(p' - p)} \, dx \, dp \, dp'$

$\displaystyle = \int_{-\infty}^{\infty} a(p) \, \int_{-\infty}^{\infty} \bar{a}(p')\, \delta(p' - p) \, dp \, dp'$

using the well known integral expression for the dirac delta-function: $\displaystyle \delta(p' - p) = \int_{-\infty}^{\infty} e^{-i(p' - p)x} \, dx$.

$\displaystyle = \int_{-\infty}^{\infty} a(p) \, \bar{a}(p) \, dp$

using the well known sifting property of the dirac delta-function: $\displaystyle \int_{-\infty}^{\infty} f(p') \delta(p' - p) \, dp' = f(p)$.

The result by the way is Parseval's Theorem.

4. Since I was mentioned early on I'll put in my two cents.

mr fantastic is exactly correct. There is likely a more precise derivation of the result, but this kind of "hand waving" approximation to a Mathematical derivation is typically used. We can get away with it for the following reasons:
1) $\displaystyle \psi$ is well behaved. ie. there are no "psychotic" points where the function or first and second derivatives don't exist, etc. (I believe the most precise way to state this is that $\displaystyle \psi(x)$ is a $\displaystyle C^1$ function: the function is continuous and the first derivative of the function is continuous. (We don't really need continuity in the second derivative, but most wavefunctions you'll see are continuous in the second derivative.)

2) $\displaystyle \lim_{x \to \pm \infty} \psi (x) = 0$

and, of course,
3) $\displaystyle \psi$ is square integrable.

There might be more restrictions on a wavefunction, but I think these are it.

-Dan