# Help!: Probability Amplitudes

1. Oct 21, 2012

### CrazyNeutrino

Im new to quantum mechanics and prof. James Binney writes the probability amplitude of going through two paths s and t, is the mod square of A(s)+A(t). Then he writes this as The mod square of A(s)+ mod square of A(t) + A(s)A(t)* + A(t)A(s). Why is it expanded like this? Could someone please prove this to me. Also he further confused me by rewriting this as mod square of A(t) + mod square of A(s) + twice the real part of A(s)A(t)*. Can you please explain this to me too.

2. Oct 21, 2012

### dextercioby

This is elementary algebra with complex numbers: |a+b|2 = (a+b) (a+b)* = ... ?

3. Oct 22, 2012

### CrazyNeutrino

So then mod square as + at =( as+at) (as+at)* = (as+at)(as-at)=as square + at square.
So where does mod square A(s)+ mod square of A(t) + A(s)A(t)* + A(t)*A(s) or A(t) + mod square of A(s) + twice the real part of A(s)A(t)* come from?

4. Oct 22, 2012

### Staff: Mentor

No, (as + at)* ≠ as - at.

Rather, (as + at)* = as* + at*.

5. Oct 22, 2012

### CrazyNeutrino

Ok..... Thanks it's beginning to make sense

6. Oct 22, 2012

### CrazyNeutrino

But where does the twice the real part come from

7. Oct 22, 2012

### Staff: Mentor

What you want to show is A(s)A(t)* + A(t)*A(s) = 2Re[A(s)A(t)*], right?

Do you know the rule for multiplying two complex numbers, in terms of their real and imaginary parts? Use it to expand both sides and write them in terms of real and imaginary parts.

At least that's the "brute force" way of showing it. There may be a "clever" way of doing it, but it doesn't come to my mind right now.

8. Oct 23, 2012

### MisterX

There's perhaps an easier way if you know that conjugation distributes across multiplication, and you know what happens when you add a complex number to its complex conjugate.

(xy)* = x*y*
I suppose you'd have to write out the real and imaginary parts to show this:

(xy)* = ((a + ib)(c + id))* = (ac + iad + ibc - bd)* = (ac - bd + i(ad + bc))* = ac - bd - i(ad + bc)
x*y* = (a - ib)(c - id) = ac -iad - ibc -bd = ac - bd - i(ad + bc)

But once you have that, you might let A(s)A(t)* = z, and it's obvious to me that z + z* = 2Re{z}, since the imaginary parts would cancel.