What is the significance of normalising wavefunctions in quantum mechanics?

[Jimmy87]

Homework Statement

I am struggling to understand all the steps in a derivation involving a normalisation of a particular wavefunction. I get most of the steps. I have attached the derivation and put a star next to the steps I don't fully understand.

Homework Equations

Listed on attachment.

The Attempt at a Solution

For the first step I have starred, I get that the integral from - to + infinity for an even function is twice the integral from 0 to infinity but how do you know that this is an even function in the example?
For the next bit why does -2lambda appear on the denominator?
I don't get what the derivation is getting at in the final step - how do they know that A is real? and why do they write A*A - is it because the absolute square of A is A multiplied by its complex conjugate which in this case is 1 (as it disappears in the 4th step) so therefore you know A is real?

Thanks for any help offered!

 

[momoko]

You can easily check that e^-2lambda |x| is an even function.

 

[vela]

For the first step I have starred, I get that the integral from - to + infinity for an even function is twice the integral from 0 to infinity but how do you know that this is an even function in the example?

An even function satisfies f(x) = f(-x) for all x. Check if the wave function has this property.

For the next bit why does -2lambda appear on the denominator?

Have you taken calculus yet? This is basic integration. You should be able to figure it out yourself.

I don't get what the derivation is getting at in the final step - how do they know that A is real? and why do they write A*A - is it because the absolute square of A is A multiplied by its complex conjugate which in this case is 1 (as it disappears in the 4th step) so therefore you know A is real?

The normalization constant is taken to be real by convention. You could, in fact, use any complex $A$ that satisfies $\lvert A \rvert ^2 = \lambda$, but why make things more complicated than necessary?

 

[Jimmy87]

An even function satisfies f(x) = f(-x) for all x. Check if the wave function has this property.Have you taken calculus yet? This is basic integration. You should be able to figure it out yourself.The normalization constant is taken to be real by convention. You could, in fact, use any complex $A$ that satisfies $\lvert A \rvert ^2 = \lambda$, but why make things more complicated than necessary?

Thanks for your help. For the first point, I am not sure how you would check this in this case? For the 2nd point that is me being really stupid, that is just a simple integration of an exponential - I get that. What do you mean by your final point - what other complex A could satisfy the norm squared of A?

 

[Jimmy87]

I think I get the first point now. If I insert (-x) into the function then the whole function doesn't change sign so its even. I still don't get the last few steps though. I don't know what they are trying to show when they write A*A = lambda therefore A^2 = lambda?

 

[Jimmy87]

Wait I'm confused with part 1 again - if there was no absolute value brackets around the x then am I right in saying that this would be an odd integrand? Why does the x have an absolute value sign? Does this always appear in wavefunctions?