# Derivation of momentum for the complex scalar field

1. Nov 18, 2015

### CAF123

The conserved 4-momentum operator for the complex scalar field $\psi = \frac{1}{\sqrt{2}}(\psi_1 + i\psi_2)$ is given in terms of the mode operators in $\psi$ and $\psi^{\dagger}$ as $$P^{\nu} = \int \frac{d^3 p}{(2\pi)^3 }\frac{1}{2 \omega(p)} p^{\nu} (a^{\dagger}(p) a(p) + b^{\dagger}(p) b(p))$$
This is just stated in my notes but I would like to see how to get to it using the mode operators. The lagrangian for the complex scalar field is $$\mathcal L = \partial_{\mu} \psi^{\dagger} \partial^{\mu} \psi - m^2 \psi^{\dagger} \psi.$$ The the stress energy tensor associated with this theory is $$T^{\mu \nu} = \frac{\partial \mathcal L}{\partial (\partial_{\mu}\psi)} \partial^{\nu} \psi + \partial^{\nu} \psi^{\dagger} \frac{\partial \mathcal L}{\partial (\partial_{\mu} \psi^{\dagger})} - \mathcal L,$$ which using the lagrangian gives $$T^{\mu \nu} = \partial^{\mu} \psi^{\dagger} \partial^{\nu} \psi + \partial^{\nu} \psi^{\dagger}\partial^{\mu} \psi - \mathcal L$$
Then $$P^{\nu} = \int T^{0 \nu} d^3 x = \int (\partial^{0} \psi^{\dagger} \partial^{\nu} \psi + \partial^{\nu} \psi^{\dagger}\partial^{0}\psi - \mathcal L) d^3 x$$$$= \int (\partial^{0} \psi^{\dagger} \partial^{\nu} \psi + \partial^{\nu} \psi^{\dagger}\partial^{0}\psi - \partial_0 \psi^{\dagger} \partial^0 \psi - \partial_i \psi^{\dagger} \partial^i \psi + m^2 \psi^{\dagger}\psi) d^3 x$$
Would someone be able to agree/disagree with me in what I have written above? I proceeded to put in the standard mode expansions for $\psi$ and $\psi^{\dagger}$ but I don't see, from the outset, how the m^2 term in the above integral comes to not appear in the expression for the momenta.

Thanks!

2. Nov 18, 2015

### Dr.AbeNikIanEdL

Shouldn't the term proportional to $\mathcal L$ be $-g^{\mu\nu}\mathcal L$ instead of just $-\mathcal L$ (at least there should be some tensor involved as $\mathcal L$ is scalar)? This would remove the $m^2$ term from the spatial components $P^i$, and on first sight it seems correct to appear in the $P^0$ component, though for me it is too late today to write out concrete formulas .

3. Nov 18, 2015

### CAF123

Hi Dr.AbeNikIanEdL,
Ah yes, I think it is so, thanks. Still it looks like from the form of the equation for $P^{\nu}$ that the zeroth component of this should not be dependent on m either. Do you think so too? I've put the mode expansions for the fields into $T^{0\nu}$ and done the integrations but haven't obtained the correct result yet for either $P^0$ or $P^i$. I'll show more steps if I don't have it tomorrow. Thanks.

4. Nov 19, 2015

### CAF123

$$\psi = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2 w(p)} (a(p) e^{-ipx} + b^{\dagger}(p)e^{ipx})$$ and so $$\psi^{\dagger} = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2 w(p)} (a^{\dagger}(p) e^{ipx} + b(p)e^{-ipx})$$
Then I proceed to calculate all the pieces before substituting in. So i'll need a $\partial_0 \psi^{\dagger}, \partial_0 \psi, \partial^i \psi, \partial^i \psi^{\dagger}$ which I can compute using $\pm i \partial_0 p.x = \pm i \partial_0 (p_ot - \mathbf p \cdot \mathbf x) = \pm i p_o$ and $\pm i \partial_i p.x = \pm i \partial_i (p_ot - p_jx_j) = \mp i \partial_i p_jx_j = \mp i p_i$ Do you agree with this?
Then $$P^0 = \int T^{00} d^3 x = \int (2\partial^0 \psi^{\dagger} \partial^0 \psi - \mathcal L) d^3 x\,\,(1)$$ The integral of the first term there gives me $$\int \frac{d^3p}{2 (2\pi)^3 }(a^{\dagger}(p) a(p) + b(p) b^{\dagger}(p) - a^{\dagger}(p)b^{\dagger}(p)e^{2iw(p)t} - b(p) a(p) e^{-2iw(p)t})$$ I think the last two terms there are zero because they are oscillating sines and cosine and over the whole spectrum of p, will vanish. Do you agree? This means, after normal ordering on the second term to interchange the order of the operators, I am left with the result I wanted except that I still have a piece due to L remaining from eqn. (1). So somehow, unless my argument above is wrong, this L term must go away otherwise I'll have m's around too which are not present in the result. I thought maybe the equations of motion would make L vanish but it is not the case as far as I can see.
Thanks!
Edit: I correct a sign above which in the end gives me $$\int \frac{d^3 p}{2 w(p)^2 (2\pi)^3} m^2 (a^{\dagger}a + bb^{\dagger})$$ The factors with a trailing complex exponential cancelled in a term in the lagrangian. Now my thinking now is to write $m^2/w(p)^2 = m^2/E^2 = m^2/(m^2 c^4) = c^{-4}$. In my units c=1 so this seems to give the right answer.
Edit2: No, that can't be right. E^2 is not m^2c^4 in general.

Anyone any thoughts?

Last edited: Nov 19, 2015
5. Nov 20, 2015

### vanhees71

I'm a bit short of time these days, but I want to answer since the OP has addressed me via a conversation. It's a bit much to type. So I attach the calculation for $\vec{P}$ as a scan. I hope you can read my bad hand-writing. The calculation for $H$ is analogous.

#### Attached Files:

• ###### momentum-kg-field.pdf
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1.6 MB
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6. Nov 20, 2015

### CAF123

Hi vanhees71,
Many thanks for the document. I understood much of what you did but I have a few questions. At the beginning, you wrote that $$\vec \pi = -:\dot{\psi}^{\dagger} \nabla \psi: + \text{h.c}.$$ Where is this minus at the front coming from? It doesn't come from the definition of $P^{\nu}$ given in my opening post.

Also, at the very bottom of the first page, you have the terms go as $$a_1^{\dagger}a_2 - b_1a_2 - a_1^{\dagger}b_2^{\dagger} + b_2^{\dagger}b_1$$ but in the last line they go as $$a_1^{\dagger}a_1 + b_1a_1 + a_1^{\dagger}b_1^{\dagger} + b_1^{\dagger}b_1$$ Why the change in sign of the middle terms?

Thanks!

Last edited: Nov 20, 2015
7. Nov 21, 2015

### vanhees71

In my notes you find for the four-momentum density
$$\pi^{\nu}=T^{0\nu}=:(\partial^0 \psi^{\dagger})(\partial^0 \psi{\dagger})-\mathcal{L},(\partial^0 \psi^{\dagger})\partial^{\mu} \psi+\text{h.c.}):$$
Now $\partial^{0}=\frac{\partial}{\partial t}$ and $\partial^{j}=\frac{\partial}{\partial x_{j}}=-\frac{\partial}{\partial x^j}$ ($j \in \{1,2,3\}$), and $\vec{\nabla} = (\partial/\partial x^1,\partial/\partial x^2,\partial/\partial x^3)$.
That's where the minus comes from in $\vec{\pi}$.

8. Nov 21, 2015

### CAF123

Is there some typos in what you wrote for $T^{0\nu}$ above? I have that $$T^{0\nu} =: \partial^{0} \psi^{\dagger} \partial^{\nu} \psi + \partial^{\nu} \psi^{\dagger}\partial^{0} \psi - \mathcal Lg^{0\nu}:$$$$\Rightarrow T^{0i} = :\partial^{0} \psi^{\dagger} \partial^{i} \psi + \partial^{i} \psi^{\dagger}\partial^{0} \psi:.$$ I could then write $\partial^i = -\partial_i$ but that would give me in the end $$P^i = \int d^3 x T^{0i} \propto \int d^3 p p_i (a^{\dagger}a + b^{\dagger}b) = -\int d^3 p p^i (a^{\dagger}a + b^{\dagger}b)$$ so i still have the minus error.

9. Nov 21, 2015

### vanhees71

Again: You want to calculate $P^j$, because these are the spatial of contravariant components of the four-momentum, and this is what's denoted by $\vec{P}$ in the 1+3-formalism.

In the 1+3-Formalism you have by definition
$$\vec{\nabla}=(\partial_1,\partial_2,\partial_3).$$
Thus you have
$$\pi^j=:\dot{\psi}^{\dagger} \partial^j \psi:+\text{h.c.}=-\dot{\psi}^{\dagger} \partial_j \psi:+\text{h.c.}$$
In 1+3 notation thus you have
$$\vec{\pi}=-:\dot{\psi}^{\dagger} \vec{\nabla} \psi:+\text{h.c.}$$
Now look at my notes. When you take the derivatives of $\psi$ you get down a factor $\mathrm{i}^2=-1$ from the exponential function, which cancels the $-1$ in the expression for $\vec{\pi}$. The calculation in my notes is complete and yields the right result. Just follow it step by step.