How Do You Prove the Schwarz Inequality?

[IHateMayonnaise]

Homework Statement

Prove the schwarz inequality:

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

Homework Equations

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

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

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.

 

[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.

 

[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:

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

instead of

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

?

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

?

 

[gabbagabbahey]

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?

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 $\vert\gamma\rangle$

Also: did you mean:

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

instead of

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

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,

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

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

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

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

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

 

[IHateMayonnaise]

Got it, thanks yall!