Normalization of free scalar field states

[soviet1100]

Hi,

if we adopt the convention, $a^{\dagger}_\textbf{p} |0\rangle = |\textbf{p}\rangle$

then we get a normalization that is not Lorentz invariant, i.e. $\langle \textbf{p} | \textbf{q} \rangle = (2\pi)^3 \delta^{(3)}(\textbf{p} - \textbf{q})$.

How do I explicitly show that this delta function (of 3-vectors) is not lorentz invariant, say for a boost in the p^3 direction ? What is $\langle \textbf{p'} | \textbf{q'} \rangle$ where the primed frame is the boosted frame?

I thought since $\int d^{3}\textbf{p} \hspace{2mm} \delta^{(3)}(\textbf{p} - \textbf{q}) = 1$ is lorentz invariant (integral over all space, so includes q)

and as $\hspace{3mm} d^3 \textbf{p} = \gamma^{-1} d^3 \textbf{p'}$, (primed frame is boosted frame)

so $\hspace{5mm} \delta^{(3)}(\textbf{p} - \textbf{q}) = \gamma\,\delta^{(3)}(\textbf{p'} - \textbf{q'})$.

so that $\int \gamma\,\delta^{(3)}(\textbf{p'} - \textbf{q'}) \gamma^{-1} d^3 \textbf{p'} = 1$ stays Lorentz invariant.

However, Peskin gets for the same calculation (p.22-23, eqn.2.34) the result

$\delta^{(3)}(\textbf{p'} - \textbf{q'}) = \delta^{(3)}(\textbf{p} - \textbf{q}) (\frac{E}{E'})$ using a method that I don't really understand well.

Have I made a mistake somewhere?

 

[George Jones]

I thought since $\int d^{3}\textbf{p} \hspace{2mm} \delta^{(3)}(\textbf{p} - \textbf{q}) = 1$ is lorentz invariant

Why?

so $\hspace{5mm} \delta^{(3)}(\textbf{p} - \textbf{q}) = \gamma\,\delta^{(3)}(\textbf{p'} - \textbf{q'})$.

so that $\int \gamma\,\delta^{(3)}(\textbf{p'} - \textbf{q'}) \gamma^{-1} d^3 \textbf{p'} = 1$ stays Lorentz invariant.

You can't just impose this because you want it to be true. The mathematical relationship between $\delta^{(3)}(\textbf{p} - \textbf{q})$ and $\delta^{(3)}(\textbf{p'} - \textbf{q'})$ has to be investigated mathematically. Peskin and Schroeder does this at the top of page 23.

However, Peskin gets for the same calculation (p.22-23, eqn.2.34) the result

$\delta^{(3)}(\textbf{p'} - \textbf{q'}) = \delta^{(3)}(\textbf{p} - \textbf{q}) (\frac{E}{E'})$ using a method that I don't really understand well.

What don't you understand about the math at the top of page 23.

 

[soviet1100]

Sorry, I don't know what I was thinking. Lack of sleep. Of course, $\int d^{3}\textbf{p} \hspace{2mm} \delta^{(3)}(\textbf{p} - \textbf{q})$ does not have to be Lorentz invariant. I deserve a slap for that, please disregard that trail of thought.

Here's what P&S do: They consider a boost in the 3-direction, with the unprimed frame as the boosted frame as:

$p'_3 = \gamma\,(p_3 + \beta E), \hspace{4mm} E' = \gamma\,(E + \beta\,p_3 )$

and then they use the identity $\delta(f(x) - f(x_0)) = \frac{1}{|f'(x_0)|}\,\delta(x - x_0) \hspace{4mm}$ to get:

$\delta^{(3)}(\textbf{p} - \textbf{q}) = \delta^{(3)}(\textbf{p'} - \textbf{q'}) \frac{dp'_3}{dp_3}$

The subsequent steps, I follow. It's just the one above that I don't get. Here is my attempt:

First, isn't the delta function identity $\delta[g(x)] = \sum_i \frac{\delta(x-x_i)}{|f'(x_i)|}$, where the x_i are roots of the function g(x)?

To get a relation between the boosted $\delta^{(3)}(\textbf{p} - \textbf{q})$ and the unboosted $\delta^{(3)}(\textbf{p'} - \textbf{q'})$, I suspect P&S have used the identity on the former to bring out p'₃.

I take $g(\textbf{p'}) = \textbf{p} - \textbf{q}, \hspace{4mm} \textbf{p} = (p_1, p_2, p_3) = (p'_1, p'_2, \gamma(p'_3 - \beta E_p')), \hspace{4mm} \textbf{q} = (q_1, q_2, q_3) = (q'_1, q'_2, \gamma(q'_3 - \beta E_q'))$. We want the values of p' that will give a zero for g(p') as defined above.

Now, $\textbf{p} - \textbf{q} = (p'_1 - q'_1, p'_2 - q'_2, \gamma[(p'_3 - q'_3) - \beta (E'_p - E'_q)])$

So $\textbf{p'} = \textbf{q'}$ isn't a sufficient condition for a zero of g(p'), is it? One must also have E'_p = E'_q, isn't it so?

Are there any flaws in my working? I suspect there obviously are, for I don't know how to get the factor $\frac{dp'_3}{dp_3}$ out.

 

[George Jones]

First, isn't the delta function identity $\delta[g(x)] = \sum_i \frac{\delta(x-x_i)}{|f'(x_i)|}$, where the x_i are roots of the function g(x)?

Yes. (Typo in the denominator.) What happens if you set $g\left(x\right) = f\left(x\right) - f\left(x_0 \right)$?

So $\textbf{p'} = \textbf{q'}$ isn't a sufficient condition for a zero of g(p'), is it? One must also have E'_p = E'_q, isn't it so?

But the fact that $\left( \Delta E \right)^2 - \left( \Delta p \right)^2$ is frame-invariant gives you ... ?

 

[soviet1100]

Yes. (Typo in the denominator.) What happens if you set $g\left(x\right) = f\left(x\right) - f\left(x_0 \right)$?

Sorry about the typo. $\hspace{2mm} \delta[g(x)] = \sum_i \frac{\delta(x-x_i)}{|g'(x_i)|}$. I get it now:

Setting $g(x) = f(x) - f(x_i)$ gives the zeros of g(x) as x_i and $g'(x) = f'(x)$ so one obtains the form of the identity used by P&S:

$\delta[f(x) - f(x_i)] = \sum_i \frac{\delta(x-x_i)}{|f'(x_i)|}$.

But the fact that $\left( \Delta E \right)^2 - \left( \Delta p \right)^2$ is frame-invariant gives you ... ?

Ah, of course. $\Delta E^2_p - \Delta |\textbf{p}|^2 = \Delta E^2_q - \Delta |\textbf{q}|^2$, and $\textbf{p'} = \textbf{q'}$ implies $E'_p = E'_q$. So, p' = q' is indeed a sufficient condition for a zero of g(x).

As for the obtaining the factor $\frac{dp'_3}{dp_3}$, I've managed to come up with it using intuition and some mathematics that I'm unfamiliar with. Perhaps you would care to comment if it is incorrect, or if there is a better way?

Computing the factor $|g'(\textbf{p'}=\textbf{q'})|$ :

Now, $\textbf{g}(\textbf{p'}) = \textbf{p} - \textbf{q} = (p'_1 - q'_1, p'_2 - q'_2, p_3)$.

According to the definition in the 'addendum' of the first answer (by user 'joshphysics') on this page (linked), one has:

$\left| \frac{d\textbf{g}}{d\textbf{p'}} \right| = \left| \begin{array}{ccc} \frac{dg_1}{dp'_1} & \frac{dg_1}{dp'_2} & \frac{dg_1}{dp'_3} \\ \frac{dg_2}{dp'_1} & \frac{dg_2}{dp'_2} & \frac{dg_2}{dp'_3} \\ \frac{dg_3}{dp'_1} & \frac{dg_3}{dp'_2} & \frac{dg_3}{dp'_3} \end{array} \right| = \left| \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{dp_3}{dp'_3} \end{array} \right| = \frac{dp_3}{dp'_3}$.

(Actually, the link above defines the total derivative of a vector valued function w.r.t a vector as a matrix of various partial derivatives. I just assumed that the bars in the denominator when applied to a square matrix meant the determinant.)

Did I get everything right?

Even if I did, I confess that there are two pieces of math here that I haven't studied before (I'm a 2nd year undergrad). One is the identity involving the delta function (of a function), and the other is the definition of the derivative of a vector valued function w.r.t to another vector. Could you perhaps point me to resources (preferably a book for the latter) that discuss these concepts?

Thanks a ton for the guidance!

 

[soviet1100]

Sorry, disregard my questions in the last post. I proved the identity myself and found a good book on the latter subject - Hubbard's Vector Calculus, Linear Algebra, & Differential Forms (A Unified Approach).

Thanks again.

 

[soviet1100]

Computing the factor $|g'(\textbf{p'}=\textbf{q'})|$ :

Now, $\textbf{g}(\textbf{p'}) = \textbf{p} - \textbf{q} = (p'_1 - q'_1, p'_2 - q'_2, p_3)$.

Corrected: $\textbf{g}(\textbf{p'}) = \textbf{p} - \textbf{q} = (p'_1 - q'_1, p'_2 - q'_2, p_3 - q_3)$