# Bra ket notation for magnitude of two vectors

1. Apr 11, 2017

### Vitani11

1. The problem statement, all variables and given/known data
If I had two vectors say ⟨em|f⟩⟨f|em⟩ does this equal |⟨em|f⟩|2? e is a basis and f is some arbitrary function. I ask this because I have a problem which is to show the following: Show that for the fourier expansion of |f⟩ in terms of Fourier basis vectors |em⟩ is ρ2(|f⟩,|fn⟩) = {⟨f|f⟩-∑|⟨em|f⟩|2+∑|amn-|⟨em|f⟩|2.

2. Relevant equations
a = linear combinations of a(ij)m
ρ2(|f⟩,|fn⟩) fourier expansion of |f⟩ in terms of fourier basis vectors |em
f is some function
fn is the nth function
All summations are from m=-n to n
3. The attempt at a solution
Here is what I have done:
ρ2(|f⟩,|fn⟩)=(⟨f|-⟨fn|)(|f⟩-|fn⟩)
=⟨f|f⟩-⟨f|fn⟩-⟨fn|f⟩+⟨fn|fn
= ⟨f|f⟩-⟨f|∑amn|em⟩-⟨fn|f⟩+⟨fn|∑amn|em
=⟨f|f⟩-⟨f|∑amn|em⟩-⟨fn|∑|⟨em|f⟩|em
= ⟨f|f⟩-⟨fn|∑⟨em|f⟩em⟩+⟨fn|∑amn|em⟩-⟨f|∑amn|em
= ⟨f|f⟩-∑⟨em|f⟩⟨fn|em⟩+Σ(amn⟨fn|em⟩-amn⟨f|em⟩)

Now what do I do? I see that I need to get rid of fn, but even with a list of bra-ket rules I can't seem to figure it out.

Last edited: Apr 11, 2017
2. Apr 11, 2017

### andrewkirk

It is correct that $\langle e_m|f\rangle\langle f|e_m\rangle=|\langle e_m|f\rangle|^2$.

To prove it, just write $\langle e_m|f\rangle$ as $a+bi$ and use the facts that $\langle e_m|f\rangle=\langle f|e_m\rangle^*$ and $|a+bi|=\sqrt{a^2+b^2}$.

3. Apr 11, 2017

### Vitani11

Awesome. That helps a lot. Can I write ∑⟨em|f⟩⟨fn|em⟩ = ∑⟨em|f⟩⟨f|em⟩ = ∑|⟨em|f⟩|2? At least if I can make an argument for fn = f. I think that is what needs to be done at least because that is the final answer for the first part of the equation.

4. Apr 11, 2017

### Vitani11

⟨fn| = Σamn|em⟩ this I know. So therefore is its conjugate |fn⟩ equal to Σamn⟨em|?

Last edited: Apr 11, 2017
5. Apr 11, 2017

### andrewkirk

I would recommend against using the word 'conjugate' in relation to the vector $|f_n\rangle$. It is natural to talk about the conjugate of the function $f$, which we denote by $f^*$, but one doesn't normally talk about the conjugates of vectors, and in most cases the idea of the 'conjugate of a vector' is either meaningless, or means something different from what is intended here. With vectors, one talks about duals. The dual of $|f\rangle$ is $\langle f|$.

This may seem picky, as in this context it is natural to identify the function $f$ with the vector $|f\rangle$, but the differences can become important as one goes deeper into Hilbert Space territory. Also, since we are going to all the trouble of treating it as a vector by writing the bar and ket symbols around it, we might as well use vector terminology to refer to it.

Having got that preamble out of the way' let's look at your rephrased question: 'is $\sum_m a^m{}_n|e_m\rangle$ the dual of $\sum_m a^m{}_n\langle e_m|$?'

$$\sum_m a^m{}_n|e_m\rangle =\sum_m |a^m{}_n\,e_m\rangle$$
and the dual of this is
$$\sum_m \langle a^m{}_n\,e_m|$$
What happens to the coefficients $a^m{}_n$ when we take them outside of that bra?

6. Apr 11, 2017

### Vitani11

If you take the amn out of the bra you will only have a basis and some scalar... so it will give you the sum over some scalar from n to m, but if it is a coefficient it doesn't change. Therefore amn is simply a and would then be able to be taken out of the whole summation. I don't know - can you do that? but then you would get a sum over all basis vectors which just sounds funny without components until you introduce the amn again. Sorry, this might have sounded ridiculous.

7. Apr 11, 2017

### andrewkirk

That depends on whether the vector space is over the Real or the Complex numbers. Which one is it in this case?

8. Apr 11, 2017

### Vitani11

Okay, so if the vector space is over real numbers it does not change but for a vector space over complex numbers the coefficient changes? Can you explain how this is? If I sum over all components of a vector in the complex plane the coefficient changes because the coefficient is itself a basis... As in there is one component which is real and one which is imaginary for the basis of the complex plane

9. Apr 11, 2017

### andrewkirk

Using $*$ to denote conjugate for scalars and dual for vectors, we have
$$|\alpha v\rangle^*=\Big(\alpha|v\rangle\Big)^*=\alpha^*|v\rangle^*=\alpha^*\langle v|$$