# Srednicki Problem 3.4

1. Dec 15, 2013

### WannabeNewton

1. The problem statement, all variables and given/known data
Recall that $T(a)^{-1}\varphi(x)T(a) = \varphi(x - a)$ where $T(a) = e^{-iP^{\mu}a_{\mu}}$ is the space-time translation operator and $P^{\mu}$ is the 4-momentum operator.

(a) Let $a^{\mu}$ be infinitesimal and derive an expression for $[P^{\mu},\varphi]$.
(b) Show that the time component of your result is equivalent to the Heisenberg equation of motion $\dot{\varphi} = i[H,\varphi]$.
(c) For a free field, use the Heisenberg equation to derive the Klein-Gordon equation.
(d) Define a spatial momentum operator $\mathbf{P} = -\int d^{3}x (\pi \nabla\varphi)$. Use the canonical commutation relations to show that $\mathbf{P}$ obeys the relation you derived in part (a).
(e) Express $\mathbf{P}$ in terms of $a(\mathbf{k})$ and $a^{\dagger}(\mathbf{k})$.

2. Relevant equations

The Hamiltonian density is given by $\mathcal{H} = \frac{1}{2}\pi^2 + \frac{1}{2}(\nabla \varphi)^2 + \frac{1}{2}m^2 \varphi^2$ where $\pi$ is the field conjugate momentum as usual.

The equal time commutation relations for the field and the field conjugate momentum are $[\varphi(x,t),\varphi(x',t)] = 0$, $[\pi(x,t),\pi(x',t)] = 0$, and $[\varphi(x,t),\pi(x',t)] = i\delta^{3}(\mathbf{x} - \mathbf{x'})$.

You probably already noticed but $\varphi$ is a real field. Also I'll use the notation $\varphi' \equiv \varphi(x',t)$ with $\pi'$ and $\nabla'$ defined similarly just to make things cleaner.

3. The attempt at a solution

(a) $\varphi(x^{\mu} - a^{\mu}) = \varphi(x^{\mu}) - a_{\mu}\partial^{\mu}\varphi +O(a^2)$ and $T(a^{\mu}) = I - ia_{\mu}P^{\mu}+O(a^2)$ so $T^{-1}\varphi T = (I + ia_{\mu}P^{\mu})(\varphi - ia_{\mu}\varphi P^{\mu}) = \varphi(x^{\mu}) + ia_{\mu}[P^{\mu},\varphi]+O(a^2)$ hence $i[P^{\mu},\varphi] = -\partial^{\mu}\varphi$ since $a_{\mu}$ was arbitrary.

(b) $\mu = 0$ gives $i[H,\varphi] = -\partial^{0}\varphi = \dot{\varphi}$

(c) We have $2[\varphi,H] = [\varphi,\int d^3{x'}(\pi'^2 + (\nabla'\varphi')^2 + m^2 \varphi'^2)] = \int d^{3}x'([\varphi,\pi'^2] + [\varphi,(\nabla'\varphi')^2] + m^2[\varphi,\varphi'^2])$; the equal time commutator can be pulled into the integral since the commutator is evaluated at the field point $x$ whereas the integral is over $x'$.

Now $[\varphi,\nabla'\varphi'] = \nabla'[\varphi,\varphi'] = 0$ hence $[\varphi,(\nabla'\varphi')^2] = 0$. Similarly $[\varphi,\varphi'^2] = 0$. This leaves us with $2[\varphi,H] = \int d^{3}x'[\varphi,\pi'^2]$.

We have $[\varphi,\pi'^2]= i\delta^{3}(\mathbf{x} - \mathbf{x}')\pi' + (\pi' \varphi)\pi' - (\pi' \varphi) \pi' + i\delta^{3}(\mathbf{x} - \mathbf{x}')\pi' = 2i\delta^{3}(\mathbf{x} - \mathbf{x}')\pi'$ hence $[\varphi,H] = i\int d^{3}x' \delta^{3}(\mathbf{x} - \mathbf{x}')\pi' = i\pi$. The Heisenberg equation of motion thus gives $\dot{\varphi} = \pi$.

Similarly, $2[\pi,H] = \int d^{3}x'([\pi,\pi'^2] + [\pi,(\nabla'\varphi')^2] + m^2[\pi,\varphi'^2])$; $[\pi,\pi'] = 0$ so $[\pi,\pi'^2] = 0$ and $[\pi,\varphi'^2] = - 2i\delta^{3}(\mathbf{x} - \mathbf{x'})\varphi'$.

Furthermore $[\pi,(\nabla'\varphi')^2] = (\pi \nabla'\varphi')\cdot\nabla'\varphi' - \nabla'\varphi'\cdot((\nabla'\varphi')\pi) \\= (\nabla'\varphi')\cdot\pi (\nabla'\varphi') - i(\nabla'\delta^{3}(\mathbf{x} - \mathbf{x'}))\cdot\nabla'\varphi' - (\nabla'\varphi')\cdot\pi(\nabla'\varphi') - i\nabla'\varphi'\cdot(\nabla'\delta^{3}(\mathbf{x} - \mathbf{x'})) \\= -2i(\nabla'\delta^{3}(\mathbf{x} - \mathbf{x'}))\cdot\nabla'\varphi'$

This leaves us with $[\pi,H] = -i\int d^{3}x'\delta^{3}(\mathbf{x} - \mathbf{x'})\varphi' - i\int d^{3}x'(\nabla'\delta^{3}(\mathbf{x} - \mathbf{x'}))\cdot\nabla'\varphi' = i(\nabla^2\varphi - m^2\varphi)$.

So now here's the part that I'm not sure about. The Heisenberg equation of motion derived above is valid for any scalar quantum field $\varphi$ right? In the derivation itself I think all I used was the fact that $\varphi$ is an arbitrary scalar field, making no reference to a specific scalar field. So it works for the field conjugate momentum $\pi$ as well? If so then we get $i[\pi,H] = -\dot{\pi} = -\ddot{\varphi}$ and at the same time $i[\pi,H] = -(\nabla^2\varphi - m^2\varphi)$ so that $\ddot{\varphi} - \nabla^2\varphi + m^2\varphi = -\partial^{\mu}\partial_{\mu}\varphi+m^2\varphi = 0$ as desired.

Does this all check out i.e. are all my calculations sound? I just want to make sure the calculations are sound so that I can move on to parts (d) and (e). Also if you know of a faster calculation to get the KG equation from the Heisenberg equation of motion then I'd appreciate it if you could show it. Thanks in advance!

2. Dec 15, 2013

### WannabeNewton

Actually part (d) was really short so I may as well post that too; hopefully it checks out :)

We have by definition $P^{i} = -\int d^{3}x'(\pi' \partial'^{i}\varphi')$ and from part (a) above we also have $i[P^{\mu},\varphi] = -\partial^{\mu}\varphi$.

Now $[\pi' \partial'^{i}\varphi',\varphi] \\= \pi'((\partial'^{i}\varphi') \varphi) - (\varphi\pi')\partial'^{i}\varphi' \\= \pi'\partial'^{i}(\varphi'\varphi)-\pi'(\varphi (\partial'^{i}\varphi')) - i\delta^{3}(\mathbf{x} - \mathbf{x'})\partial'^{i}\varphi' \\= \pi'\partial'^{i}[\varphi',\varphi] - i\delta^{3}(\mathbf{x} - \mathbf{x'})\partial'^{i}\varphi' = - i\delta^{3}(\mathbf{x} - \mathbf{x'})\partial'^{i}\varphi'$

hence $i[P^{i},\varphi] = -i\int d^{3}x'[(\pi' \partial'^{i}\varphi'),\varphi] = -\int d^{3}x' \delta^{3}(\mathbf{x} - \mathbf{x'})\partial'^{i}\varphi' = -\partial^{i}\varphi$ as desired.

3. Dec 15, 2013

### WannabeNewton

Also I'm stuck on part (e) so if someone could help me out with part (e) that would be swell :)

The free scalar field solution is $\varphi(x) =\int d\tilde{k} [a(\mathbf{k})e^{i(\mathbf{k}\cdot \mathbf{x} - \omega t)} + a^{\dagger}(\mathbf{k})e^{i(-\mathbf{k}\cdot \mathbf{x} + \omega t)}]$ where $d\tilde{k} = \frac{d^3 k}{(2\pi)^3 2\omega}$ is the Lorentz invariant measure.

Then $\pi = \dot{\varphi} = \int d\tilde{k} [ -a(\mathbf{k})e^{i(\mathbf{k}\cdot \mathbf{x} - \omega t)} + a^{\dagger}(\mathbf{k})e^{i(-\mathbf{k}\cdot \mathbf{x} + \omega t)}]i\omega$

and $\nabla \varphi = \int d\tilde{k} [a(\mathbf{k})e^{i(\mathbf{k}\cdot \mathbf{x} - \omega t)} - a^{\dagger}(\mathbf{k})e^{i(-\mathbf{k}\cdot \mathbf{x} + \omega t)}]i\mathbf{k}$

so we have $\mathbf{P} = \int d\tilde{k}d\tilde{k}' d^{3}x[ -a(\mathbf{k})e^{i(\mathbf{k}\cdot \mathbf{x} - \omega t)} + a^{\dagger}(\mathbf{k})e^{i(-\mathbf{k}\cdot \mathbf{x} + \omega t)}] [a(\mathbf{k'})e^{i(\mathbf{k'}\cdot \mathbf{x} - \omega' t)} - a^{\dagger}(\mathbf{k'})e^{i(-\mathbf{k'}\cdot \mathbf{x} + \omega' t)}]\omega\mathbf{k'}\\ = \int d\tilde{k}d\tilde{k}' d^{3}x[-a(\mathbf{k})a(\mathbf{k'})e^{i(\mathbf{k}+\mathbf{k'})\cdot \mathbf{x}} e^{-i(\omega + \omega') t} + a^{\dagger}(\mathbf{k})a(\mathbf{k'})e^{i(\mathbf{k'}-\mathbf{k})\cdot \mathbf{x}} e^{-i(\omega - \omega') t} \\+a(\mathbf{k})a^{\dagger}(\mathbf{k'})e^{i(\mathbf{k}-\mathbf{k'})\cdot \mathbf{x}} e^{-i(\omega' - \omega) t} - a^{\dagger}(\mathbf{k})a^{\dagger}(\mathbf{k'})e^{-i(\mathbf{k}+\mathbf{k'})\cdot \mathbf{x}} e^{i(\omega + \omega') t}]\omega\mathbf{k'}\\ = -(2\pi)^3 \int d\tilde{k}d\tilde{k}'\delta^{3}(\mathbf{k} +\mathbf{k'})[a(\mathbf{k})a(\mathbf{k'})e^{-i(\omega + \omega') t}+ a^{\dagger}(\mathbf{k})a^{\dagger}(\mathbf{k'})e^{i(\omega + \omega') t}]\omega\mathbf{k'}\\+(2\pi)^3 \int d\tilde{k}d\tilde{k}'\delta^{3}(\mathbf{k} - \mathbf{k'})[a^{\dagger}(\mathbf{k})a(\mathbf{k'})e^{-i(\omega - \omega') t} +a(\mathbf{k})a^{\dagger}(\mathbf{k'})e^{-i(\omega' - \omega) t}]\omega\mathbf{k'}\\ = -\frac{1}{2} \int d\tilde{k}[a(\mathbf{k})a(\mathbf{-k})e^{-2i\omega t}+ a^{\dagger}(\mathbf{k})a^{\dagger}(\mathbf{-k})e^{2i\omega t}]\mathbf{k}\\+\frac{1}{2} \int d\tilde{k}[a^{\dagger}(\mathbf{k})a(\mathbf{k}) +a(\mathbf{k})a^{\dagger}(\mathbf{k})]\mathbf{k}$

I don't know how to simplify this any further. The second term in the final equality looks like it should stay intact since the Hamiltonian for the free scalar field is given by $H = \frac{1}{2} \int d\tilde{k}[a^{\dagger}(\mathbf{k})a(\mathbf{k}) +a(\mathbf{k})a^{\dagger}(\mathbf{k})]\omega$.

I'm guessing it's the first term in the final equality that has to vanish somehow. However I can't seem to make it vanish. I tried using a parity inversion $\mathbf{k}\rightarrow -\mathbf{k}$ and the fact that $[a(\mathbf{k}),a(\mathbf{-k})] = [a^{\dagger}(\mathbf{k}),a^{\dagger}(\mathbf{-k})] = 0$ but the lone $\mathbf{k}$ in the integrand picks up an overall sign that cancels out with the overall sign picked up by the Lorentz invariant measure $d\tilde{k}$, giving me back the original integral even though what I need is an overall negative of the original integral in order for it to vanish. Thanks in advance for any help.

EDIT: Oops nevermind! I forgot that the bounds of the integral also flip under the parity inversion so I have to flip them back and that gives me the overall negative sign that I need to make the first term vanish.

Last edited: Dec 15, 2013
4. Dec 15, 2013

Thanks wbn!