# Cauchy-Schwarz inequality in Shankhar's Quantum Mechanics

I'm trying to understand proof of this inequality. But I have some problems.

So, Shankar starts a proof with definition a new vector $$|z \rangle$$:

$$|z \rangle = |v\rangle - \frac{\langle w|v \rangle}{|w|^2} |w \rangle.$$

And here, I would like to know, why we suppose vector in this form? I know that it's projection of one vector on the second one.

My second question is, that we multiply this equation by $$\langle z|$$. $$\langle z|z\rangle = \left\langle v - \frac{\langle w|v \rangle}{|w|^2} w \right|\left.v-\frac{\langle w|v \rangle}{|w|^2} |w \right\rangle\\ = \langle v|v \rangle - \frac{\langle w|v \rangle \langle v|w \rangle}{|w|^2} - \frac{\langle w|v \rangle ^*\langle w|v \rangle}{|w|^2} + \frac{\langle w|v \rangle ^*\langle w|v \rangle \langle w|w \rangle}{|w|^4}$$

What I don't understand is a complex conjugate in the last two expression. Why then, we don't conjugate the first and the second one?

EDIT:

So - we're multiplying ket-z times bra-z. So, if the ket-z is: $$|z \rangle = |v\rangle - \frac{\langle w|v \rangle}{|w|^2} |w \rangle.$$ and bra-z is: $$\langle z |= \langle v|- \frac{\langle w|v \rangle}{|w|^2} \langle w|.$$

And what concernes me is this term above: $$\langle w|v \rangle,$$ because if $$| z \rangle = [(\langle z |^*)^T].$$ So shouldn't bra-z be $$\langle z |= \langle v|- \frac{\langle v|w \rangle}{|w|^2} \langle w|.$$

With regards to your first question, the vector $$| z\rangle$$ is chosen so it is orthogonal to $$| w\rangle$$. To see this, note that $$\langle a|(|b\rangle+|c\rangle)=\langle a|b\rangle+\langle a|c\rangle$$ implies $$\langle w|z\rangle=\left\langle w\left|v-\frac{\langle w|v\rangle}{|w|^2}|w\rangle\right.\right\rangle =\langle w|v\rangle+\left\langle w\left|-\frac{\langle w|v\rangle}{|w|^2}|w\rangle\right.\right\rangle.$$ Let us be careful now about the treatment of the second term, which is related to your second question. We have that $$\langle a|(\lambda |b\rangle)\rangle=\lambda\langle a|b\rangle$$. Therefore, taking $$|a\rangle=|w\rangle=|b\rangle$$ and $$\lambda=-\frac{\langle w|v\rangle}{|w|^2}$$, we see that $$\left\langle w\left|-\frac{\langle w|v\rangle}{|w|^2}|w\rangle\right.\right\rangle=-\frac{\langle w|v\rangle}{|w|^2}\left\langle w\left|w\right.\right\rangle$$ Finally, recalling that $$|w|^2:=\langle w|w\rangle$$, we conclude that $$\left\langle w\left|-\frac{\langle w|v\rangle}{|w|^2}|w\rangle\right.\right\rangle=-\frac{\langle w|v\rangle}{|w|^2}|w|^2=-\langle w|v\rangle,$$ and $$\langle w|z\rangle=\langle w|v\rangle-\langle w|v\rangle=0.$$ As a consequence, you get that $$|v\rangle$$, $$|z\rangle$$ and $$\frac{\langle w|v\rangle}{|w|^2}|w\rangle$$ form a right triangle with hypotenuse $$|v\rangle=|z\rangle +\frac{\langle w|v\rangle}{|w|^2}|w\rangle.$$ Pythagoras' theorem then guarantees that $$|v|^2=|z|^2+\frac{|\langle w|v\rangle|^2}{|w|^4}|w|^2=|z|^2+\frac{|\langle w|v\rangle|^2}{|w|^2}\geq\frac{|\langle w|v\rangle|^2}{|w|^2}.$$ We conclude that $$|v|^2|w|^2\geq|\langle w|v\rangle|^2$$. This is the proof given in this Wikipedia article.

If you understood this computation, the answer to your your second question should be clear. Remember that $$\langle a|b\rangle=\langle b|a\rangle^*$$ and so $$\langle (\lambda a)|b\rangle=\lambda^*\langle a|b\rangle$$.

Response to edit:

Your first formula for $$\langle z|$$ is wrong. The second one is the correct one $$\langle z|=\langle v|-\frac{\langle v|w\rangle}{|w|^2}\langle w|$$

• I understand now, but is there any logical explanation that we start with orthogonal vector to $|w \rangle$?
– user237867
Mar 1 '20 at 17:02
• The idea is that $|z\rangle$, $|v\rangle$ and $\frac{\langle w|v\rangle}{|w|^2}|w\rangle$ form a right triangle whose hypotenuse is $|v\rangle$. The Cauchy-Scwartz inequality then comes as a direct consequence of Pythagora's theorem Mar 1 '20 at 17:45
• I'll add that to my answer. Mar 1 '20 at 17:45
• I have one more question about $\langle z |$. So, if the $[(|z \rangle)^*]^T= \langle z|$, why this term $\langle w|v \rangle$ isn't conjugate?
– user237867
Mar 10 '20 at 18:15
• I don't really understand your question. Please elaborate. Mar 11 '20 at 15:25

If you would take a look at Shankar chapter 1.2 Inner product spaces, he explains that the complex inner product is linear in the ket part and skew-symmetric. Meaning $$\langle v|w\rangle=\langle w|v\rangle^*$$ and $$\langle v|\alpha w + \beta z\rangle=\alpha \langle v|w\rangle + \beta \langle v | z \rangle$$. Now try and work out the expression and you should be fine.

As for your first question, it is just a definition. I don't know the motivation but I guess it might be a clever one such that the proof works out nicely.