# Normalisation of quantum states

1. Oct 12, 2012

### bayners123

Hi,

Just a little thing that's been puzzling me:

Consider a state

$$\mid \psi \rangle = \frac{1}{\sqrt{2}} \mid A \rangle + \frac{1}{\sqrt{2}} \mid B \rangle$$

This is normalised since $$[\frac{1}{\sqrt{2}}]^2 + [\frac{1}{\sqrt{2}}]^2 = 1$$

Now let A = B:

$$\mid \psi \rangle = \frac{1}{\sqrt{2}} \mid A \rangle + \frac{1}{\sqrt{2}} \mid A \rangle = \frac{2}{\sqrt{2}} \mid A \rangle = {\sqrt{2}} \mid A \rangle$$

This isn't normalised anymore! What happened?

2. Oct 12, 2012

### Matterwave

A and B must be normalized and orthogonal for your normalization to work. A and A is obviously not orthogonal.

3. Oct 12, 2012

### Jazzdude

You implicitly assumed <A|B> = 0 in your first expression.

4. Oct 12, 2012

### bayners123

Ah ok.. The reason this came up is that I was looking at latter operators on paired states. So with 2 atoms in a 1,1 state you can get
$$\mid 2,2 \rangle = \mid 1,1 \rangle\mid 1,1 \rangle$$
And then you can use ladder operators to go down:
$$J_-\mid 2,2 \rangle = (J_-\mid 1,1 \rangle)\mid 1,1 \rangle + \mid 1,1 \rangle(J_- \mid 1,1 \rangle) \\ \mid 2,1 \rangle = \frac{1}{\sqrt{2}} \mid 1,0 \rangle\mid 1,1 \rangle + \frac{1}{\sqrt{2}} \mid 1,1 \rangle \mid 1,0 \rangle$$
Which is fine, but then the next one:
$$\sqrt{6} \mid 2,0 \rangle = \frac{1}{\sqrt{2}} \sqrt{2} \mid 1,-1 \rangle\mid 1,1 \rangle + \frac{1}{\sqrt{2}} \sqrt{2} \mid 1,0 \rangle\mid 1,0 \rangle + \frac{1}{\sqrt{2}} \sqrt{2} \mid 1,0 \rangle \mid 1,0 \rangle + \frac{1}{\sqrt{2}} \sqrt{2} \mid 1,0 \rangle \mid 1,-1 \rangle$$
is actually also fine now I look at it again.. Sorry for wasting your time!

5. Oct 13, 2012