# Dirac notation Schwarz Inequality Proof

1. Jun 20, 2014

### interdinghy

1. The problem statement, all variables and given/known data
This isn't really a problem so much as me not being able to see how a proof has proceeded. I've only just today learned about Dirac notation so I'm not too good at actually working with it. The proof in the book is:

|Z> = |V> - <W|V>/|W|^2|W>

<Z|Z> = <V - ( <W|V>/|W|^2 ) W| V - ( <W|V>/|W|^2 ) W> =

<V|V> - ( <W|V><V|W>/|W|^2 ) - ( <W|V>*<W|V>/|W|^2 ) + ( <W|V>*<W|V><W|W>/|W|^4| ) >= 0

This last step has me completely lost, and if someone could explain it I'd really appreciate it. It looks like he's pulled out the <V| and |V> and combined them to get <V|V>, but I don't get where the minus sign comes from, or how he gets any of the rest of the equation, so I'm assuming I'm wrong about where the <V|V> comes from.

I'm deeply sorry about the notation here, I typed this on a phone or I would've used LaTeX.

2. Relevant equations

<Z|Z> >= 0

3. The attempt at a solution

Mentioned in the first section.

2. Jun 20, 2014

### kreil

So you know that

$$|Z\rangle = |V\rangle - \frac{\langle W|V\rangle}{|W|^2} |W\rangle$$

So what do you think $\langle Z|$ equals?

3. Jun 21, 2014

### interdinghy

$\langle Z|$ should just equal $$\langle V| - \frac{\langle W|V\rangle}{|W|^2} \langle W|$$

right? So if I look at $\langle Z|Z \rangle$, I have to distribute, but I can't see how he's getting the terms he's getting, namely because I don't think I'm too comfortable with Dirac notation yet.

1: $\langle V | V \rangle = \langle V | V \rangle$ obviously,

2: $\langle V | - \frac{\langle W|V\rangle}{|W|^2} W \rangle$ seems like it should equal $- \frac{\langle W|V\rangle \langle V|W \rangle}{|W|^2}$

3: Probably the most confusing term for me:

$\langle - \frac{\langle V|W \rangle}{|W|^2}W|V \rangle$

This seems to me like it should just be $- \frac{\langle V|W \rangle \langle W|V \rangle}{|W|^2}$ but this term doesn't appear anywhere in his final formulation really. I mean I guess $\langle V|W \rangle$ is equivalent to his $\langle W|V \rangle ^*$, but why does he get that conjugate instead of the $\langle V|W \rangle$ that I got?

4: then $\langle - \frac{\langle W|V\rangle}{|W|^2} W|- \frac{\langle W|V\rangle}{|W|^2} W \rangle$ is the first term where I think I'm getting confused. For some reason the term he gets from this has a complex conjugate in it, which I assume means he's moving $| - \frac{\langle W | V \rangle}{|W|^2} \rangle$ to the bra side, which I assume makes the top term a complex conjugate, leaving me with $\langle - \frac{\langle W|V\rangle}{|W|^2}\frac{\langle W|V\rangle ^*}{|W|^2} W| W \rangle = \frac{ \langle W|V \rangle \langle W|V \rangle ^* \langle W|W \rangle}{|W|^4}$

At the end, my final answer would look like: $\langle V | V \rangle - \frac{\langle W|V\rangle \langle V|W \rangle}{|W|^2} - \frac{\langle V|W \rangle \langle W|V \rangle}{|W|^2} + \frac{ \langle W|V \rangle \langle W|V \rangle ^* \langle W|W \rangle}{|W|^4}$

This seems like it may at least be equivalent to his formulation based on what I said in 3, but even this took kind of a mental leap for me in part 4 with moving a term over to the bra side. Is that a correct step? If so, why?

Thanks so much for any help. I'm losing sleep over this.

4. Jun 21, 2014

### kreil

No problem, anyone that gets a science degree loses some sleep over a problem or two, so I know what boat you're in!

Does it help if I say that,

$|Z\rangle = \langle Z|^{\dagger}$
$\langle Z| = |Z\rangle ^{\dagger}$

and that,

$\langle W |V \rangle ^{\dagger} = \langle W | V \rangle ^* = \langle V|W\rangle ...?$

(In other words, the inner product of a bra and a ket is a complex number, and the Hermitian conjugate of a complex number is simply its complex conjugate.)

This all means that,

$$\langle Z| = |Z\rangle^{\dagger} = \left( |V\rangle - \frac{\langle W|V\rangle}{|W|^2} |W\rangle \right) ^{\dagger} =\left( \langle V | - \frac{\langle W|V \rangle ^*}{|W|^2} \langle W| \right)$$

So,

$$\langle Z | Z \rangle = \left( \langle V | - \frac{\langle W|V \rangle ^*}{|W|^2} \langle W| \right) \left( |V\rangle - \frac{\langle W|V\rangle}{|W|^2} |W\rangle \right)$$
http://en.wikipedia.org/wiki/Bra–ket_notation#Inner_products_and_bras

5. Jun 21, 2014

### interdinghy

Yes, that clears it up completely, thank you.

I can now actually see the part in the book where the asterisk denoting that the initial setup involved a complex conjugate has been worn away due to shoddy printing, so I feel a lot better about this whole thing in general now.

Thanks so much!

6. Oct 4, 2015

### fricke

Hmm.. I don't get why the proof starts with that |Z>
Mind to explain it, please? thanks :)

7. Dec 17, 2016

### Shephine Shaji

Towards the end they consider, <W|V><V|W> = |<V|W>|, which I cannot understand. Please explain someone.