Dealing with normalized quantum functions

  • Thread starter Thread starter Jimmy25
  • Start date Start date
  • Tags Tags
    Functions Quantum
Jimmy25
Messages
69
Reaction score
0

Homework Statement



If <∅| is normalized, show that:

<∅|∅>=1=<∅|n><n|∅>

(where ∅ is a non-eigenfunction wave function composed of Ʃc(n)ψ(n).

Homework Equations





The Attempt at a Solution



I can show that <∅|∅>=Ʃc*(n)c(n) (=1). But the next part of the question asks to use your proof to show Ʃc*(n)c(n)=1 so that's not the way it is supposed to be done. I feel like I'm missing something very simple.
 
Physics news on Phys.org
Jimmy25 said:

Homework Statement



If <∅| is normalized, show that:

<∅|∅>=1=<∅|n><n|∅>

(where ∅ is a non-eigenfunction wave function composed of Ʃc(n)ψ(n).

It seems to me that the problem statement must be


<∅|∅>=1=Ʃn<∅|n><n|∅>.

This is just the usual expansion over basis states. What you wrote can't be true for arbitrary n; it only works when you sum over all the basis states.

In matrix language, Ʃn|n><n| is just the identity matrix.

BBB
 
Aha that was what I suspected.

With the correction, is there a way to prove that it is equal to one without having to use that Ʃc*(n)c(n) (=1)?
 
Jimmy25 said:
Aha that was what I suspected.

With the correction, is there a way to prove that it is equal to one without having to use that Ʃc*(n)c(n) (=1)?

I would just say Ʃn|n><n| = I, the identity matrix, because that is the definition of a complete set of states {|n>}. Then

<θ|I|θ> = <θ|θ> =1

because |θ> is normalized. This is not the complicated part of quantum mechanics.

BBB
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top