0

The most general solution for the Schrödinger Equation with $V = 0$ :

$$\Psi(x,t)=\int_{-\infty}^{\infty} A(k) e^{ikx} e^{−i\hbar k^2t/2m} dk$$

To normalize this, what constraints will be placed on $A(k)$?

What is implied about $A(k)$ from:

$$\int_{-\infty}^{\infty}\left|\;\int_{-\infty}^{\infty} A(k) e^{ikx} e^{−i\hbar k^2t/2m} dk\;\right|^2\;dx = 1$$

Community
  • 1
Truth-seek
  • 1,427

1 Answers1

1

The constraint is,

$$\int \psi_p(x,t)^* \psi_q(x,t) dx = \delta(p-q),$$

Where $\delta$ is the Dirac delta function.

You also need to understand how to properly square an integral.

$$ \Big| \int f(k) dk \Big|^2 = \Big( \int f(k) \ dk \Big)^* \Big( \int f(q) \ dq\Big) = \int \int f^*(k) f(q) \ dk dq$$

Spencer
  • 12,271
  • What is $p$ and $q$ representing? Can you plz elaborate – Truth-seek Mar 30 '18 at 07:55
  • Sure! The variables $p$ and $q$ are momenta corresponding to the momentum eigen states $\psi_p$ and $\psi_q$. The eigenstates are normalized according to the formula I have given above. Your formula has three integrals, you can evaluate the x integration using my formula and then the delta function kills one of the two momentum integrals. – Spencer Mar 30 '18 at 14:16
  • Am sorry if this is dumb: but why do we need momentum eigenstates $\psi_p$'s, we have \Psi and we have to perform the integral to normalize it ryt? – Truth-seek Mar 30 '18 at 14:25
  • The eigenstates are implicitly in your problem in the form of the complex exponentials. You need to use this property of then to simplify it. – Spencer Mar 30 '18 at 14:29
  • Oh sorry you mean to say that $\psi_p$'s are the $e^{ikx}$s? But where does the product $e^{-ik_1x}e^{ik_2x}$ appear in the integral? – Truth-seek Mar 30 '18 at 14:35
  • Sorry for asking these (dumb, maybe) questions. But It would be nice if you could plz explain in detail what we can deduce about the coefficients $A(k)$ – Truth-seek Mar 30 '18 at 14:49