# Proving the Schwarz Inequality

1. Sep 1, 2009

### IHateMayonnaise

1. The problem statement, all variables and given/known data

Prove the schwarz inequality:

$$|<\alpha|\beta>|^2\leq<\alpha|\alpha><\beta|\beta>$$

2. Relevant equations

$$<\alpha|\alpha>\geq 0$$

$$|\alpha>=|\beta>-\left(\frac{<\alpha|\beta>}{<\alpha|\alpha>}\right )|\alpha>$$

3. The attempt at a solution

The first step would obviously be to evaluate the first equation using the second:

$$<\alpha|\alpha>=\left <|\beta>-\left(\frac{<\alpha|\beta>}{<\alpha|\alpha>}\right )|\alpha>\middle | |\beta>-\left(\frac{<\alpha|\beta>}{<\alpha|\alpha>}\right )|\alpha>\right>$$

..And from here I am kind of stumped. I am familiar with the identity $$<a+b|c>=<a|c>+<b|c>$$, however what would the identity be for $$<a+b|a+b>$$? Am I even going in the right direction here?

In Shanker's Principles of Quantum Mechanics 2nd ed. Pg 17, it says that the next step is:

$$=<\beta|\beta>-\frac{<\alpha|\beta><\beta|\alpha>}{<\alpha|\alpha>}-\frac{<\alpha|\beta>^*<\alpha|\beta>}{<\alpha|\alpha>}$$

I am not understanding this logic. I know that $$<\alpha|\beta>$$ represents the inner product of $$\alpha$$ and $$\beta$$, respectfully, but I do not understand how he gets to that step.

2. Sep 1, 2009

### fantispug

For a start I don't understand your second relevant equation; you've got alpha on both sides. Are you choosing a beta (resp. alpha) such that that equation holds for each given alpha (resp. beta)?

I think you want to define a new quantity $$|\gamma \rangle$$:

$$|\gamma \rangle=|\beta \rangle-\left(\frac{\langle \alpha|\beta \rangle}{\langle \alpha|\alpha \rangle}\right )|\alpha \rangle$$.

Now we want to take the adjoint; that is essentially replace all bras with kets and all complex numbers with their complex conjugates: some examples
If
$$|\gamma \rangle = i |\alpha \rangle$$
then
$$\langle \gamma | = -i \langle \alpha |$$

Or If
$$|\gamma \rangle = \langle \alpha | \beta \rangle |\alpha \rangle$$
then
$$\langle \gamma | = \langle \beta | \alpha \rangle \langle \alpha |$$

(Why? Remember changing the order of the inner product gives the complex conjugate).

Your linearity equations are right. You should be able to follow Shankar if you rewrite the equation for $$\langle \gamma | \gamma \rangle$$ correctly.

3. Sep 1, 2009

### IHateMayonnaise

Thanks for the reply fantispug!

Yes; I apologize I meant for that to be gamma instead of alpha. So are you staying I should take the adjoint of the new quantity, or the adjoint of the inner product of the new quantity with itself?

Also: did you mean:

$$|\gamma \rangle =i|\gamma \rangle$$

$$|\gamma \rangle = i |\alpha \rangle$$

?

So if I rewrite what I had in my original post with gamma instead of alpha, we have:

$$\langle \gamma|\gamma \rangle=\left < |\beta \rangle-\left(\frac{\langle\alpha|\beta\rangle}{\langle\alpha|\alpha\rangle}\right )|\alpha\rangle\middle | |\beta\rangle-\left(\frac{\langle\alpha|\beta\rangle}{\langle\alpha|\alpha\rangle}\right )|\alpha\rangle\right>$$

And taking the adjoint (I think Im doing this right):

$$\langle \gamma|\gamma \rangle=\left < \langle\beta|^*-\left(\frac{\langle\beta|\alpha\rangle}{\langle\alpha|\alpha\rangle}\right )\langle\alpha|\middle | \langle\beta|^*-\left(\frac{\langle\beta|\alpha\rangle}{\langle\alpha|\alpha\rangle}\right )\langle\alpha|\right>$$

?

Last edited: Sep 1, 2009
4. Sep 1, 2009

### gabbagabbahey

I'm pretty sure he meant take the adjoint of the Ket $\vert\gamma\rangle$, in order to obtain the Bra $\langle\gamma\vert$, since (as you should have learned)

$$\vert\gamma\rangle^\dagger=\langle\gamma\vert$$

for any Ket $\vert\gamma\rangle$

Again, I'm pretty sure he meant exactly what he posted....he is simply providing an example for the ket defined by the equation $|\gamma \rangle \equiv i |\alpha \rangle$ (this is a different gamma than the one you will use in your problem, it is just an example to demonstrate how to turn kets into bras)

For this example,

$$\langle\gamma\vert=\vert\gamma\rangle^\dagger=\left(i\vert\alpha\rangle\right)^\dagger=-i\langle\alpha\vert$$

No, if $|\gamma \rangle=|\beta \rangle-\left(\frac{\langle \alpha|\beta \rangle}{\langle \alpha|\alpha \rangle}\right )|\alpha \rangle$. then

$$\langle\gamma\vert=|\gamma \rangle^\dagger=\left(|\beta \rangle-\left(\frac{\langle \alpha|\beta \rangle}{\langle \alpha|\alpha \rangle}\right )|\alpha \rangle \right)^\dagger$$

5. Sep 2, 2009

### IHateMayonnaise

Got it, thanks yall!