Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.4.

The square modulus of the wave function which is the solution to the Schrödinger equation is interpreted as a probability density. As an example consider the wave function given by

We can normalize by requiring

Plugging in the formula and doing the integral gives

where we’ve taken the positive real root for . Note that could also be multiplied by a phase factor for any real without affecting normalization. This can be important in some applications where we need to add together wave functions.

Given this value for , we can plot 1. Here, we’ve taken and :

Since has its maximum at , that is where the particle is most likely to be found. The probability of the particle being found to the left of is

If , then drops to zero at so . If , then is an isosceles triangle symmetric about so .

The expectation value of is

where we used Maple to simplify the integration. If , then as expected.

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DeniskaOh the wave function given in the question is the solution of schrodinger equation ( which may or may not be the actual wave function of the particle) but after normalizing we get the wave function(actual/correct), is my reasoning correct?

gwrowePost authorEssentially, yes. The normalization is required to make the wave function consistent with the probabilistic interpretation, that is, equation 2 says that the probability of finding the particle somewhere in space must add up to 1.

TrishnaCan someone explain how we get a/2 in equation 7

gwrowePost authorIf you plug equation 1 into the integral in 7, the gory details are:

Now you can see why I use Maple to do the integrals and algebra(!)

pavithraWt expansion is used in 2nd step of previous comment??

gwrowePost authorYou just have to multiply the integrands out and then do the integrals. It’s a lot of work, but there’s nothing fancy about it.

HridiHow did you get Sqrt3/b? Will you please show the solution?

gwrowePost authorSolve eqn 4 for A.