Sakurai's proof of the Optical Theorem

[vega12]

Right now, I'm self-studying from J. J. Sakurai's book Modern Quantum Mechanics. In section 7.3, Optical Theorem, there is one step in the proof that he uses that escapes me. His proof involves using the transition operator T defined as:

$V \mid\psi^{(+)} \rangle = T \mid\phi \rangle \\$

where $$\mid\phi \rangle$$ is the free particle state and $$\mid\psi^{(+)} \rangle$$ is the scattered state. In the proof for the optical theorem, he says "Now we use the well-known relation":

$\frac{1}{E - H_0 - i\epsilon} = Pr. \left( \frac{1}{E-H_0}\right) + i \pi \delta(E-H_0)$

But I am at a loss as to what "Pr." means. I can follow the rest of the proof except for this one part, so a bit of clarification would be much appreciated. Thanks!

 

[vanhees71]

That means the principle value. The equation has to be read as an equation about distribution, i.e., applied to a test function it reads (for $$z_0 \in \mathbb{R}$$)

$$\int_{\mathbb{R}} \mathrm{d} z \frac{f(z)}{z-z_0-\mathrm{i} 0^+} = \mathrm{P} \int_{\mathbb{R}} \frac{f(z)}{z-z_0} + \ii \pi f(z_0),$$

Where the principal value of the integral is defined by

$$\mathrm{P} \int_{\mathbb{R}} \mathrm{d} z \frac{f(z)}{z-z_0} = \lim_{\epsilon \rightarrow 0^+} \left [\int_{-\infty}^{z_0-\epsilon} \mathrm{d} z \frac{f(z)}{z-z_0} + \int_{z_0+\epsilon}^{\infty} \mathrm{d} z \frac{f(z)}{z-z_0} \right ].$$

 

[vega12]

Thanks for the quick reply. I'm still unsure how exactly I can arrive at that specific relation, but at least I now know what it means so can understand how to use it. Could you give me an idea how I would go about showing it?

Upon getting your response I decided to see if I can calculate the desired inner product without using that relation and get the same result to at least justify it. It went as follows:

$$\Im \langle\mathbf{k}\mid T\mid\mathbf{k}\rangle = \Im \left[ \left( \langle\psi^{(+)}\mid - \langle\psi^{(+)}\mid V \frac{1}{E-H_0-i\epsilon}\right)V\mid\psi^{(+)}\rangle\right]$$

The first inner product is real so the imaginary part is zero. We only need to calculate the second one.

$$\langle\psi^{(+)}\mid V \frac{1}{E-H_0-i\epsilon}V\mid\psi^{(+)}\rangle$$
$$= \int \mathrm{d}^3 k^\prime \langle\mathbf{k}\mid T^\dag \mid \mathbf{k^\prime} \rangle \langle \mathbf{k^\prime} \mid \frac{1}{\tfrac{\hbar^2}{2m}(k^2 - k^{\prime 2} - i\epsilon)} T \mid \mathbf{k}\rangle$$
$$= - \frac{2m}{\hbar^2} \int \mathrm{d}\Omega^\prime \int_0^\infty \mathrm{d}k^\prime k^{\prime 2} \frac{\left| \langle \mathbf{k^\prime}\mid T \mid \mathbf{k} \rangle \right| ^2}{k^{\prime 2} - k^2 + i\epsilon}$$
$$= - \frac{m}{\hbar^2} \int \mathrm{d}\Omega^\prime \int_{-\infty}^\infty \mathrm{d}k^\prime k^{\prime 2} \frac{\left| \langle \mathbf{k^\prime}\mid T \mid \mathbf{k} \rangle \right| ^2}{(k^\prime - k e^{-\epsilon / 2 k^2})(k^\prime + k e^{-\epsilon / 2 k^2})}$$

Using residue theorem (which is where I believe the delta function comes out in the distribution method but am still shaky on that part)

$$= \frac{m \pi k i}{\hbar^2} \int \mathrm{d}\Omega^\prime \left| \langle \mathbf{k^\prime}\mid T \mid \mathbf{k} \rangle \right|^2$$

All together this gives

$$\Im \langle\mathbf{k}\mid T\mid\mathbf{k}\rangle = \frac{- m \pi k}{\hbar^2} \int \mathrm{d}\Omega^\prime \left| \langle \mathbf{k^\prime}\mid T \mid \mathbf{k} \rangle \right|^2$$

So yeah, at least in the end I was able to complete the steps in the proof without using that relation.

 

[matonski]

Check out this thread.

 

[vega12]

Wow, I spent quite some time searching the net and here and couldn't find that thread. My apologizes as it seems my question was answered there quite well. I was even finally able to locate the theorem on wikipedia: http://en.wikipedia.org/wiki/Sokhatsky%E2%80%93Weierstrass_theorem" . It even gives a simple proof of it.