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?


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

  • $\begingroup$ I understand now, but is there any logical explanation that we start with orthogonal vector to $|w \rangle$? $\endgroup$
    – user237867
    Mar 1 '20 at 17:02
  • 2
    $\begingroup$ 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 $\endgroup$ Mar 1 '20 at 17:45
  • 1
    $\begingroup$ I'll add that to my answer. $\endgroup$ Mar 1 '20 at 17:45
  • $\begingroup$ 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? $\endgroup$
    – user237867
    Mar 10 '20 at 18:15
  • $\begingroup$ I don't really understand your question. Please elaborate. $\endgroup$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy