# Dimensions of klein-gordon field

1. Jul 31, 2012

### iangttymn

Does anyone know how to write the classical solution to the Klein-Gordon equation in NON natural units? Where do all the c's and h's go?

2. Jul 31, 2012

### fzero

The KG equation is just the mass-shell relation

$$\left[ E^2 - |\vec{p}c|^2 - (mc^2)^2\right] \phi =0.$$

This can be written in terms of derivatives by setting

$$E = i\hbar \partial_t,~~~\vec{p} = -i\hbar \nabla.$$

3. Jul 31, 2012

### iangttymn

I understand that. I'm looking for the non natural units version of
\begin{eqnarray}
\phi = \int \frac{d^3p}{(2\pi)^3 2E_{\boldsymbol{p}}} \big(a(\boldsymbol{p}) e^{ip_{\mu}x^{\mu}} + a^{\dagger}(\boldsymbol{p}) e^{-ip_{\mu}x^{\mu}}\big)
\end{eqnarray}

4. Jul 31, 2012

### fzero

The KG equation doesn't fix the units of the field, but we can use the corresponding Lagrangian to do so. We want

$$\frac{1}{\hbar} \int d^4x (\partial \phi)^2$$

to be dimensionless.

The argument of the exponents must be dimensionless, so we should replace $px$ by $px/(\hbar c)$. Next, it's conventional to take

$$[a(\mathbf{p}) , a^\dagger (\mathbf{p}')] = \delta^3(\mathbf{p}-\mathbf{p}'),$$

so $a(\mathbf{p})$ has units of $(\mathrm{momentum})^{-3/2}$.

Now, we can convert lengths to (inverse) energies by dividing by $\hbar c$ and momenta to energies by multiplying by $c$. You might want to give it a try yourself and we can check your answer.

5. Jul 31, 2012

### iangttymn

Alright …
So the Lagrangian density
\begin{eqnarray}
\mathcal{L} \propto (\partial \phi)^2
\end{eqnarray}
should be integrated over $d^4x$ to give units of action. The Lagrangian density has units
\begin{eqnarray}
[(\partial\phi)^2] = \frac{1}{l^2} [\phi]^2.
\end{eqnarray}
And
\begin{eqnarray}
[d^4x] = l^4.
\end{eqnarray}
So, this gives
\begin{eqnarray}
[d^4x \mathcal{L}] = l^4 \frac{1}{l^2} [\phi]^2 = l^2 [\phi]^2 \equiv \frac{m l^2}{t},
\end{eqnarray}
or
\begin{eqnarray}
[\phi]^2 = \frac{m}{t}.
\end{eqnarray}
So the field has units of the square root of mass per time ... ?

For $a(\boldsymbol{p})$, I hadn't thought of the commutation relation - I'm actually only interested in in the classical (real) solutions at this point. But,
\begin{eqnarray}
[\delta(\boldsymbol{p})] = \frac{1}{[p]^3} = \bigg(\frac{t}{ml}\bigg)^3
\end{eqnarray}
so as you said,
\begin{eqnarray}
[a] = \bigg( \frac{t}{ml}\bigg)^{\frac{3}{2}}.
\end{eqnarray}

So, from the natural units expression above,
\begin{eqnarray}
\phi = \alpha \int \frac{d^3 p}{(2\pi)^3 2E_p} \big( a(\boldsymbol{p}) e^{\cdots} + \cdots),
\end{eqnarray}
where $\alpha$ captures whatever units are missing. This gives
\begin{eqnarray}
\sqrt{\frac{m}{t}} &=& [\alpha] \bigg( \frac{ml}{t}\bigg)^3 \frac{1}{ \frac{ml^2}{t^2}} \bigg( \frac{t}{ml}\bigg)^{\frac{3}{2}} \\
&=& [\alpha] \sqrt{ \frac{mt}{l} }.
\end{eqnarray}
So
\begin{eqnarray}
[\alpha] = \frac{ \sqrt{l}}{t}.
\end{eqnarray}

But when I try to find the appropriate factor of $\hbar$ and $c$ (assuming $G$ doesn't enter in - that would be weird), I try
\begin{eqnarray}
[\hbar^ac^b] = \frac{m^a l^{2a+b}}{t^{a+b}}.
\end{eqnarray}
Setting this equal to the units for $\alpha$, $a$ must be $0$, but then there is no solution for $b$.

Not sure what I've done wrong.

6. Jul 31, 2012

### fzero

I think it's a bit cleaner to work through it the following way. Start with

$$\phi = \alpha \int \frac{d^3p}{(2\pi)^3 2E_{\boldsymbol{p}}} \big(a(\boldsymbol{p}) e^{ip_{\mu}x^{\mu}/(\hbar c)} + \cdots \big)$$

and compute

$$\frac{1}{\hbar} \int d^4x (\partial \phi)^2 \sim \frac{1}{\hbar} \int d^4x \int d^3p d^3 p' \frac{\alpha^2}{E_p E_{p'}} \frac{p^\mu p'_\mu}{(\hbar c)^2}e^{i (p-p')x/(\hbar c)} \left[a^\dagger a \right],$$

where I've dropped purely numerical factors and haven't tried to keep track of the order of the terms in the $\left[a^\dagger a \right]$ factor. The integral over $x$ is a delta function in momentum space, so we have

$$\frac{\alpha^2}{\hbar (\hbar c)^2} \int d^3p \frac{p^\mu p_\mu}{E_p^2} \left[a^\dagger a \right] \sim \frac{\alpha^2}{\hbar (\hbar c)^2} \frac{1}{c^3} \int d^3(pc) \frac{1}{c^2} \frac{(p c)^2}{E_p^2} c^3 \frac{\left[a^\dagger a \right]}{c^3}$$

$$\sim \frac{\alpha^2}{c (\hbar c)^3} \int d^3(pc) \frac{(p c)^2}{E_p^2} \frac{\left[a^\dagger a \right]}{c^3},$$

where, on the RHS of the first line, I've included factors of $c$ to convert the remaining factors into units of energy. In the last line, it's rewritten so that we can see that the units in the integral expression cancel, so we can just set

$$\frac{\alpha^2}{c (\hbar c)^3}=1$$

to determine $\alpha$ in terms of the dimensionful constants.

7. Jul 31, 2012

### iangttymn

That is cleaner. Thanks!

8. Jul 31, 2012

### Bill_K

First, you better get the formula right!

Way back up at the top, where you had (2π)3(2Ep) in the denominator, it should be (2π)3/2√(2Ep). The (2π)3/2 doesn't matter, but the √(2Ep) does. In particular, if you don't put in the √, it's inconsistent with the usual commutation relations [a(k), a*(k')] = δ3(k - k').

The formula starts off with a straight four-dimensional Fourier transform, with a delta function restriction to the mass shell:

φ(x) = (2π)-3/2∫d4k δ4(k2 - m2) b(k) eik·x

Doing out the k0 integral gives you

φ(x) = (2π)-3/2∫d3k/(2k0) b(k) eik·x

which is still relativistically invariant because d3k/(2k0) is the invariant volume element on the mass hyperboloid. However the states created/destroyed by b(k) are also relativistically invariant, i.e. they're normalized to

<k|k'> = k0 δ3(k - k')

and the b's therefore obey the rather strange looking commutation relations

[b(k), b*(k')] = k0 δ3(k - k')

To get the normal commutators you have to replace b(k) by

b(k) = √(2k0) a(k)

So in the original formula, (2k0)-1 b(k) gets replaced by (2k0)-1/2 a(k). And thus you need a square root!