Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Going to as little details as possible, here is a statement from Wald's text on QFT in curved spacetimes(I am not quoting the book)

He considers two vector spaces ${\cal S}$ and ${\cal H}$.

Note - For more details about ${\cal S}$ and ${\cal H}$, read this box. I believe for most part of the question, the following details are irrelevant, but I will provide them nonetheless. Otherwise, skip below.

He starts by considering the solution space of a classical system ${\cal S}$ with symplectic structure $\Omega$. This has a natural vector space structure. He complexifies it to ${\cal S}^{\mathbb C}$ and extends $\Omega$ to ${\cal S}^{\mathbb C}$ by complex linearity on each variable. He then defines the map $(\cdot, \cdot): {\cal S}^{\mathbb C} \times {\cal S}^{\mathbb C} \to {\mathbb C}$ on ${\cal S}^{\mathbb C}$ as $$ (y_1, y_2) = - i \Omega( \overline{y_1}, y_2) $$ This satisfies all the properties of an inner product except positive-definiteness. He then considers the subspace ${\cal H}$ of ${\cal S}^{\mathbb C}$ on which the inner product above is positive-definite. (There are of course many such choices of ${\cal H}$. Any one of them will do.)

He then shows that there is a one-one onto map $K: {\cal S} \to {\cal H}$. He shows that one can define a real inner product $\mu: {\cal S} \times {\cal S} \to {\mathbb R}$ on ${\cal S}$. He then goes on to show that one can use this to define a complex inner product on $\cal H$ as $$ \left( K y_1, K y_2 \right)_{\cal H} = \mu(y_1, y_2) - \frac{i}{2} \Omega(y_1, y_2)~\forall~y_1, y_2 \in {\cal S} $$ where $\Omega: {\cal S} \times {\cal S} \to {\mathbb R}$ is an antisymmetric function on ${\cal S}$, i.e. $\Omega(y_1, y_2) = - \Omega(y_2, y_1)$. He then uses the Cauchy-Schwarz Inequality for ${\cal H}$. This reads $$ \left( K y_1, K y_1 \right)_{\cal H} \left( K y_2, K y_2 \right)_{\cal H} \geq \left| \left( K y_1, K y_2 \right)_{\cal H} \right|^2 \geq \left| \text{Im} \left( K y_1, K y_2 \right)_{\cal H} \right|^2 $$ Expanding it out, he writes $$ \mu(y_1, y_1) \mu(y_2, y_2) \geq \mu(y_1, y_2)^2 + \frac{1}{4} \Omega(y_1, y_2)^2 \geq \frac{1}{4} \Omega(y_1, y_2)^2 $$ More specifically $$ \boxed{ \mu(y_1, y_1) \mu(y_2, y_2)\geq \frac{1}{4} \Omega(y_1, y_2)^2 } $$

Now, here is the statement that confuses me

Indeed, since $K$ is one-to-one and onto and since the Schwarz inequality on ${\cal H}$ always can be ``saturated", we obtain the following stronger version of the last inequality: For each $y_1 \in {\cal S}$ we have $$ \mu(y_1, y_1) = \frac{1}{4} \max_{y_2 \neq 0} \frac{ \Omega(y_1, y_2)^2}{\mu(y_2, y_2)} $$ Here's my question Q. Where did he get the above expression from?

He seems to be claiming that the boxed inequality is always saturated for some vector $y_2 \in {\cal S}$. Is that true? Why?

PS - I will understand if some people think that this question is more of a math question than a physics one. But, I thought that it might be possible that the answer relies on some of the assumptions we make in physics, so I asked it here. Any comments will be helpful

share|improve this question
    
Wow - much more detailed than the early version :) I think you could get a good answer either here or on math - and I'm personally not against leaving the question here. –  Chris White Jul 16 '13 at 3:48
    
As for addressing the question, this is just a passing thought: I don't have the text, and I'm not entirely familiar with what's going on, but should $\max_{y_2\neq0}$ be replaced with $\sup_{y_2\neq0}$? As in, is it possible that no $y_2$ actually saturates, but that there always exists a sequence that converges as needed? Of course this question is moot if the underlying space is nice and compact or something, but again I also don't really know what I'm talking about... –  Chris White Jul 16 '13 at 4:06
    
The book specifically says $\max$. Also, I think I have figured it out. Given a $y_1 \in {\cal S}$, any choice $y_2 \in {\cal S}$ such that $Ky_2 = \lambda K y_1$ with $\lambda^* = - \lambda$ saturates the boxed inequality. Now, we must only confirm that given the relation for $y_2$ above, one can always find a $y_2$ given any $y_1$. But this is obviously possible since $K$ is an invertible map. –  Prahar Jul 16 '13 at 4:10

1 Answer 1

up vote 1 down vote accepted

Let $y_1, y_2$ $2$ complex vectors and let $<,>$ be a complex inner product defined by $<y_1,y_2> = \vec y_1^*.\vec y_2$.

Let $\vec a$ and $\vec b$ the real and imaginary part of $\vec y$ :
$\vec y = \vec a + i \vec b$

Then :

$$<y_1,y_2> = (\vec a_1 .\vec a_2 + \vec b_1 .\vec b_2) + i (\vec a_1 .\vec b_2 - \vec b_1 .\vec a_2) = u(y_1,y_2) + iv(y_1,y_2)$$

The Cauchy-Schwartz inequality gives :

$$<y_1,y_1><y_2,y_2> ~~\ge ~~|<y_1,y_2>|^2$$

We note that : $<y_1,y_1> = u(y_1, y_1)$, so we have :

$$u(y_1,y_1)~~\ge ~~ \frac{u^2(y_1,y_2) + v^2(y_1,y_2)}{u(y_2,y_2)}$$

Now, fixing a particular $y_1$, we limit the set of $y_2$ to those which respect $u(y_1,y_2) =0$. So, we have now :

$$u(y_1,y_1)~~\ge ~~ \frac{ v^2(y_1,y_2)}{u(y_2,y_2)}~~ ~~ ~~ ~~ ~~ (1)$$

Now, take explicitely $y_2$ defined by $ \vec a_2 = - \vec b_1, \vec b_2 =\vec a_1,$, we see that $\vec a_1 .\vec a_2 + \vec b_1 .\vec b_2 = 0$, that is $u(y_1,y_2) = 0$, so this choice is coherent with our previous hyphothesis.

Morevoer, we have $v(y_1,y_2) = \vec a_1^2 + \vec b_1^2 $, and $u(y_2,y_2) = \vec a_1^2 + \vec b_1^2 $, so we have, for this particular $y_2$.

$$u(y_1,y_1)~~= ~~ \frac{ v^2(y_1,y_2)}{u(y_2,y_2)}~~ ~~ ~~ ~~ ~~ (1)$$

So, we see, that the inequality $(1)$ is effectively saturated by our choice of this particular $y_2$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.