# Schwarz inequality with bra-ket notation

1. Mar 13, 2015

### Maylis

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

2. Relevant equations

3. The attempt at a solution
Hello,

I just want to make sure I am doing this right
$$<a|b> = a_{x}^{*}b_{x} + a_{y}^{*}b_{y} + a_{z}^{*}b_{z}$$
$$= [(1-i)|x>][-i|x>] + (2 |y>)(-3 |y>) + (0|z>)(|z>)$$
$$=(-i + i^{2})|x> - 6 |y> + 0|z>$$
$$=(-1-i)|x> - 6 |y>$$

Then to find $\mid<a|b> \mid^{2}$, I need to take the complex conjugate since this is a complex vector
$$\mid<a|b> \mid^{2} = [(-1-i)|x> - 6|y> + 0|z>][(-1+i)|x> - 6|y> + 0|z>]$$
$$= 2|x> + 36 |y> + 0|z>$$

Now I want to make sure that this is even right before I spend time evaluating the other inner products to determine if Schwarz inequality is true.

Last edited: Mar 13, 2015
2. Mar 13, 2015

### Dick

The inner product is just a number not a vector. Try that again. In your formula $a_x$ is just the coefficient of $(1+i)|x>$, i.e. $(1+i)$.

3. Mar 13, 2015

### Maylis

ok, so then
$$<a|b> = 1 - i^{2} - 6 = -4$$
$$<a|a> = 1 + 1 + 4 = 6$$
$$<b|b> = 1 + 9 + 1 = 11$$

So clearly $16 \le 6(11)$

Thanks

4. Mar 13, 2015

### Dick

Better slow down a bit. I don't think <a|b> comes out to -4. Want to show the steps that lead to that?

5. Mar 13, 2015

### Maylis

Sure,

$$(1-i)(-i) + 2(-3) + 0(1)$$
$$=-i+i^{2} - 6$$
$$=-i-7$$
So then the absolute value squared is $(-i - 7)(i - 7) = -i^{2} + 49 = 50$

So $50 \le 66$

6. Mar 13, 2015

### Dick

That's better.