What Is the Normalization Constant for a 2-Electron Antisymmetric Spin Function?

[physgirl]

Homework Statement

Given that the antisymmetric spin function for a 2 electron system is N[a1b2-a2b1], find the normalization constant N. (and by a and b I mean the alpha and beta spin states and by 1 and 2, I mean the labels on the two electrons...

Homework Equations

Normalization: 1=integral over all relevant space of (wavefunction*wavefunction)

The Attempt at a Solution

So I tried to square the spin function given, set it equal to 1, and solve for N. However, as the squared value of [a1b2-a2b1]... or what I THINK is the squared value of that, I kept on getting zero... what am I doing wrong in doing:

square of spin function=(N[a1b2-a2b1])^2
=(N^2)<a1b2-a2b1|a1b2-a2b1>
=(N^2)[<a1|a1><b2|b2>-<a1|a2><b2|b1>-<a2|a1><b1|b2>+<a2|a2><b1|b1>]

and because any sort of <a|a> is 1 and so is <b|b>, all those braket stuff are equal to 1, which overall makes the equation 0... I'm confused :(

 

[mjsd]

no, while <a1 b2|a1 b2> = 1, <a1 b2|a2 b1>=0
therefore you get
1= N^2 (1+0+0+1)
1= 2N^2
N = 1/sqrt(2) which is what one would expect naively.

note: the 2e- system has 4 states:
|a1 a2>,|a1 b2>,|a2 b1>,|a2 b2>
these are tensor products. eg. |a1 a2> = |a1>|a2>

 

[physgirl]

how did you get (1+0+0+1)? I kept on getitng something like 1-1+1-1 or something that kept on cancelling all out to 0 :(

 

[jostpuur]

square of spin function=(N[a1b2-a2b1])^2
=(N^2)<a1b2-a2b1|a1b2-a2b1>
=(N^2)[<a1|a1><b2|b2>-<a1|a2><b2|b1>-<a2|a1><b1|b2>+<a2|a2><b1|b1>]

and because any sort of <a|a> is 1 and so is <b|b>, all those braket stuff are equal to 1, which overall makes the equation 0... I'm confused :(

Your state

$$|\psi\rangle = N|\alpha_1\beta_2-\alpha_2\beta_1\rangle$$

looks bad. Use this kind of notation instead:

$$|\psi\rangle = N\big(|\alpha_1\beta_2\rangle-|\alpha_2\beta_1\rangle\big)$$

When you compute

$$\langle\psi|\psi\rangle$$

don't start splitting states $|\alpha_i\beta_j\rangle$ into sums of states $|\alpha_i\rangle$ and $|\beta_j\rangle$, because that is wrong.

 

[physgirl]

Ok, I got it now : ) but now my question is, I got to the point where N=sqrt(1/2). So can N be both positive AND negative of sqrt(1/2)?

 

[jostpuur]

Ok, I got it now : ) but now my question is, I got to the point where N=sqrt(1/2). So can N be both positive AND negative of sqrt(1/2)?

In fact

$$N=e^{i\theta}\sqrt{1/2}$$

are all valid normalization constants, where theta is some arbitrary constant, but the simplest $\sqrt{1/2}$ is usually preferred.