Bra ket notation for magnitude of two vectors

[Vitani11]

Homework Statement

If I had two vectors say ⟨e_m|f⟩⟨f|e_m⟩ does this equal |⟨e_m|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 |e_m⟩ is ρ²(|f⟩,|f_n⟩) = {⟨f|f⟩-∑|⟨e_m|f⟩|²+∑|a_mⁿ-|⟨e_m|f⟩|².

Homework Equations

a = linear combinations of a_(ij)^m
ρ²(|f⟩,|f_n⟩) Fourier expansion of |f⟩ in terms of Fourier basis vectors |e_m⟩
f is some function
f_n is the nth function
All summations are from m=-n to n

The Attempt at a Solution

Here is what I have done:
ρ²(|f⟩,|f_n⟩)=(⟨f|-⟨f_n|)(|f⟩-|f_n⟩)
=⟨f|f⟩-⟨f|f_n⟩-⟨f_n|f⟩+⟨f_n|f_n⟩
= ⟨f|f⟩-⟨f|∑a_mⁿ|e_m⟩-⟨f_n|f⟩+⟨f_n|∑a_mⁿ|e_m⟩
=⟨f|f⟩-⟨f|∑a_mⁿ|e_m⟩-⟨f_n|∑|⟨e_m|f⟩|e_m⟩
= ⟨f|f⟩-⟨f_n|∑⟨e_m|f⟩e_m⟩+⟨f_n|∑a_mⁿ|e_m⟩-⟨f|∑a_mⁿ|e_m⟩
= ⟨f|f⟩-∑⟨e_m|f⟩⟨f_n|e_m⟩+Σ(a_mⁿ⟨f_n|e_m⟩-a_mⁿ⟨f|e_m⟩)

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

 

[andrewkirk]

If I had two vectors say ⟨em|f⟩⟨f|em⟩ does this equal |⟨em|f⟩|2?

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 $|a+bi|=\sqrt{a^2+b^2}$.

 

[Vitani11]

Awesome. That helps a lot. Can I write ∑⟨e_m|f⟩⟨f_n|e_m⟩ = ∑⟨e_m|f⟩⟨f|e_m⟩ = ∑|⟨em|f⟩|²? At least if I can make an argument for f_n = 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.

 

[Vitani11]

⟨f_n| = Σa^m_n|e_m⟩ this I know. So therefore is its conjugate |f_n⟩ equal to Σa^m_n⟨e_m|?

 

[andrewkirk]

⟨f_n| = Σa^m_n|e_m⟩ this I know. So therefore is its conjugate |f_n⟩ equal to Σa^m_n⟨e_m|?

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 $\sum_m a^m{}_n\langle e_m|$?'

To answer that, let's think about this
$$\sum_m a^m{}_n|e_m\rangle
=\sum_m |a^m{}_n\,e_m\rangle$$
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?

 

[Vitani11]

If you take the a^m_n 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 a^m_n 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 a^m_n again. Sorry, this might have sounded ridiculous.

 

[andrewkirk]

but if it is a coefficient it doesn't change

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

 

[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

 

[andrewkirk]

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?

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