Total Momentum Operator for Klein Gordon Field

[Samama Fahim]

Homework Statement

Prove that $$P^i=\int d^3x T^{0i} = -\int d^3 \pi \partial_i \phi$$

Relevant Equations

$$T_{\nu}^{\mu}=\frac{\partial\mathcal{L}}{\partial(\partial_{\mu} \phi)}\partial_{\nu}\phi-\delta_{\nu}^{\mu}\mathcal{L}$$

As
$$\hat{P_i} = \int d^3x T^0_i,$$

and

$$T_i^0=\frac{\partial\mathcal{L}}{\partial(\partial_0 \phi)}\partial_i\phi-\delta_i^0\mathcal{L}=\frac{\partial\mathcal{L}}{\partial(\partial_0 \phi)}\partial_i\phi=\pi\partial_i\phi.$$

Therefore,

$$\hat{P_i} = \int d^3x \pi\partial_i\phi.$$

However, in Peskin & Schroeder it's given with a negative sign:

$$\hat{P^i} = -\int d^3x \pi\partial_i\phi.$$

How do I go from $$T^{0}_i$$ to $$T^{0i}$$?

 

[ergospherical]

Does Peskin & Schroeder use a (+---) metric signature? In which case $\partial^i \equiv - \partial_i$ and\begin{align*}
P^i &= \int d^3 x T^{0i} = \int d^3 x \frac{\partial\mathcal{L}}{\partial(\partial_{0} \phi)}\partial^{i}\phi = \int d^3 x \pi \partial^i \phi = -\int d^3 x \pi \partial_i \phi
\end{align*}

 

[Samama Fahim]

Does Peskin & Schroeder use a (+---) metric signature? In which case $\partial^i \equiv - \partial_i$ and\begin{align*}
P^i &= \int d^3 x T^{0i} = \int d^3 x \frac{\partial\mathcal{L}}{\partial(\partial_{0} \phi)}\partial^{i}\phi = \int d^3 x \pi \partial^i \phi = -\int d^3 x \pi \partial_i \phi
\end{align*}

Yes they do use the signature (+---). Why would $$\partial^i$$ become $$-\partial_i$$ in that case?

 

[ergospherical]

Recall that with (+---) then $\eta = \eta^{-1} = \mathrm{diag}(1,-1,-1,-1)$ and\begin{align*}
v^0 = \eta^{0j} v_j = \eta^{00} v_0 = v_0
\end{align*}whilst, for any fixed $i = 1,2,3$,\begin{align*}
v^i = \eta^{ij} v_j = \eta^{ii} v_i = -v_i
\end{align*}

 

[Samama Fahim]

Does Peskin & Schroeder use a (+---) metric signature? In which case $\partial^i \equiv - \partial_i$ and\begin{align*}
P^i &= \int d^3 x T^{0i} = \int d^3 x \frac{\partial\mathcal{L}}{\partial(\partial_{0} \phi)}\partial^{i}\phi = \int d^3 x \pi \partial^i \phi = -\int d^3 x \pi \partial_i \phi
\end{align*}

Could you please confirm one more thing:

If

$$\phi = \int d^3k \frac{1}{(2 \pi)^3 \sqrt{2\omega_k}} \left(a(\vec{k})e^{-i(\omega_kx_o-\vec{k}\cdot \vec{x})}+a^{\dagger}(\vec{k})e^{i(\omega_kx_0-\vec{k}\cdot \vec{x})}\right).$$

then we should have

$$\partial_i\phi=\int d^3k \frac{1}{(2 \pi)^3 \sqrt{2\omega_k}} \left(a(\vec{k})(ik_i)e^{-i(\omega_kx_o-\vec{k}\cdot \vec{x})}+a^{\dagger}(\vec{k})(-ik_i)e^{i(\omega_kx_0-\vec{k}\cdot \vec{x})}\right).$$

Is that correct?

 

[ergospherical]

That looks fine to me.

 

[Samama Fahim]

That looks fine to me.

But Jakob Schwichtenberg in No-Nonsense Quantum Field Theory, p. 600 writes:

$$\partial_i\phi=\int d^3k \frac{1}{(2 \pi)^3 \sqrt{2\omega_k}} \left(a(\vec{k})(-ik_i)e^{-ikx}+a^{\dagger}(\vec{k})(ik_i)e^{ikx}\right).$$

Here $$kx \equiv k_{\mu}x^{\mu}$$

 

[ergospherical]

Apologies, Schwichtenberg's is correct (it's just due to the index placement on $k$, which I'd overlooked). Remember that $kx = k_{\mu} x^{\mu} = k_0 x^0 + k_i x^i = k^0 x^0 - k^i x^i$, so that $\partial_i(kx) = k_i = - k^i$.

 

[Samama Fahim]

Apologies, Schwichtenberg's is correct (it's just due to the index placement on $k$, which I'd overlooked). Remember that $kx = k_{\mu} x^{\mu} = k_0 x^0 + k_i x^i = k^0 x^0 - k^i x^i$, so that $\partial_i(kx) = k_i = - k^i$.

Actually on p. 58, Schwichtenberg defines $$p_{\mu}p^{\mu}$$ to be $$p_0p_0-\vec{p} \cdot \vec{p}$$. And he also uses the metric signature $$(1,-1,-1,-1)$$ and not $$p_0p_0+\vec{p} \cdot \vec{p}$$

 

[ergospherical]

That's correct, yes. By $\mathbf{k} \cdot \mathbf{x}$ one means $k^i x^i$.

 

[Samama Fahim]

That's correct, yes. By $\mathbf{k} \cdot \mathbf{x}$ one means $k^i x^i$.

$\vec{k} \cdot \vec{x} = k_ix_i$, is that true also?

 

[ergospherical]

$\vec{k} \cdot \vec{x} = k_ix_i$, is that true also?

It would be, yes, simply because $k^i x^i = (-k_i)(-x_i) = k_i x_i$.
But be careful to note that the symbol $\partial_i \equiv \partial / \partial x^i$ refers to differentiation w.r.t. $x^i$, not $x_i$.

It all works out, but you have to be careful not to trip up on signs anywhere!

 

[Samama Fahim]

Thank you so much for all the responses.

 

[Samama Fahim]

But on p. 311, Schwichtenberg writes $$kx \equiv k_0x_0-k_ix_i$$ and goes on to take the derivative $$\partial_0 e^{ik_0x_0} = ik_0 e^{ik_0x_0} $$

 

[ergospherical]

It's still fine, because $k_0 = k^0$ and $x_0 = x^0$. When using (+---), i.e. when $\eta_{00} = 1$, you can raise/lower a $0$ index without having to change the sign.

 

[Samama Fahim]

Four-divergence is

$$\partial_{\mu}K^{\mu}=\partial_0K^0+\partial_iK^i.$$

However, on p. 160 Schwichtenberg writes

$$0=\partial_{\mu}T^{\mu}_0$$
$$=\partial_0T^0_0-\partial_iT^i_0.$$

Why the minus sign? Here T denotes momentum-energy tensor.

 

[ergospherical]

That does look like a blunder, though it doesn't change the argument much.\begin{align*}
\partial_{\mu} T^{\mu}_0 = \partial_0 T^0_0 + \partial_i T^i_0 &= 0 \\ \\

\implies \int_V d^3 x \partial_0 T^0_0 &= -\int_V d^3 x\partial_i T^i_0 \\

\implies \int_{V} d^3 x \partial_0 T^0_0 &= -\int_{\partial V} d^2 S_i T^i_0 = 0

\end{align*}so that $\dfrac{d}{dx_0} \displaystyle{\int_{V} d^3 x T^0_0} = 0$.