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Consider the Lagrangian density of a complex scalar quantum field:

$$ \mathcal{L} = (\partial_\mu\varphi^\dagger)(\partial^\mu\varphi) - m^2\varphi^\dagger\varphi $$

With the conjugate momenta $\pi^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\varphi)} = \partial^\mu\varphi^\dagger$ and $(\pi^\dagger)^\mu = \partial^\mu\varphi$. From this we immediately obtain the stress-energy tensor:

$$ T^{\mu\nu} = \pi^\mu\partial^\nu\varphi - g^{\mu\nu}\mathcal{L} = \pi^\mu(\pi^\dagger)^\nu - g^{\mu\nu}((\partial_\mu\varphi^\dagger)(\partial^\mu\varphi) - m^2\varphi^\dagger\varphi) $$ where $g^{\mu\nu}$ is the flat Minkowski metric, with signature $(+---)$. We take the $(0,0)$ component of this tensor, which equals the Hamiltonian density:

$$ \mathcal{H} = \pi^0(\pi^\dagger)^0 - \partial_0\varphi^\dagger\partial^0\varphi + \nabla\varphi^\dagger\cdot\nabla\varphi + m^2\varphi^\dagger\varphi $$

But now, according to the definitions of the conjugate momenta above, this is equal to

$$ \mathcal{H} = \pi^0(\pi^\dagger)^0 - \pi^0(\pi^\dagger)^0 + \nabla\varphi^\dagger\cdot\nabla\varphi + m^2\varphi^\dagger\varphi = \nabla\varphi^\dagger\cdot\nabla\varphi + m^2\varphi^\dagger\varphi $$

Whereas on the exercise sheet, the Hamiltonian density is given as:

$$ \mathcal{H}=\pi^0(\pi^\dagger)^0 + \nabla\varphi^\dagger\cdot\nabla\varphi + m^2\varphi^\dagger\varphi $$ Where did I go wrong? Is it possible that the exercise sheet is wrong? I would email the professor but I doubt he would answer over the easter holidays. So far, I have only seen this derivation with the Lagrangian density multiplied by a factor of $1/2$, in which case the time derivatives of the field do not drop out when calculating the Hamiltonian. Also, as a personal side note, what is it with particle physicists and their abuse of index notation? Vectors and covectors are not the same, you cannot just arbitrarily shove indices around like that.

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    $\begingroup$ It is odd that you write the conjugate momenta as vectors, they are not, they are operator-valued distributions like unto the fields to which they are conjugate. The momentum conjugate to $\varphi$ is defined as:$$\pi={\partial\mathcal L\over\partial(\partial_0\varphi)}.$$ $\endgroup$ Apr 1 at 1:05
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    $\begingroup$ The stress energy tensor is defined:$$T^{\mu\nu}={\partial\mathcal L\over\partial(\partial_\mu\varphi_\rho)}\partial^\nu\varphi_\rho-\mathcal L\delta^{\mu\nu}.$$ Note how there is a sum of terms for each of the fields $\varphi_\rho$. See for example, Peskin and Schroeder, pg. 18. $\endgroup$ Apr 1 at 1:15
  • $\begingroup$ i wish this was an answer so i could mark it as such $\endgroup$
    – paulina
    Apr 1 at 10:04
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    $\begingroup$ Ok. I made an answer out of the comment. $\endgroup$ Apr 1 at 13:25
  • $\begingroup$ Minor note: there is nothing specifically "quantum" about the field you are using here: these computations and definitions are agnostic regarding whether the field ($\phi$) and canonical momentum ($\pi^0$, by your definition, but conventionally just $\pi$) commute. $\endgroup$
    – Rishi
    Apr 1 at 13:47

2 Answers 2

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Where did I go wrong?

Seems like probably there are a number of different places you could have gone wrong. Since this is a homework-like question, I will answer in terms of classical field theory, and let you worry about any potential difference due to transitioning to the quantum field theory.

One potential place you went wrong is that you are seemingly not using the fact that there are two independent fields $\phi$ and $\phi^*$ (or, equivalently, the real and imaginary parts of the complex $\phi$ field). Or, you are using this fact in one of your equations and not the other.

Anyways, we have the momentum density (I will just call it $\pi$, whereas you are using the notation $\pi^0$): $$ \pi = \frac{\partial \mathcal{L}}{\partial \dot \phi}=\dot \phi^* $$ and a second momentum density (because there are two independent fields): $$ \pi^* = \frac{\partial \mathcal{L}}{\partial \dot \phi^*} = \dot \phi\;. $$

Thus, the Hamiltonian density is: $$ \mathcal{H}=\pi \dot \phi + \pi^* \dot \phi^* - \mathcal{L} $$ $$ =2|\dot \phi|^2 - \mathcal{L} $$ $$ =2|\dot \phi|^2 - |\dot \phi|^2 + |\nabla \phi|^2 + m^2|\phi|^2 $$ $$ =|\dot \phi|^2 + |\nabla \phi|^2 + m^2|\phi|^2 $$ $$ =|\pi|^2 + |\nabla \phi|^2 + m^2|\phi|^2 $$


For two independent fields ($\phi$ and $\phi^*$), the classical expression for the stress tensor is: $$ T^{\mu\nu} = \frac{\partial \mathcal{L}}{\partial \partial_\mu\phi}\partial^{\nu}\phi +\frac{\partial \mathcal{L}}{\partial \partial_\mu\phi^*}\partial^{\nu}\phi^* -g^{\mu\nu}\mathcal{L}\;. $$

For example: $$ T^{00} = \pi^*\dot\phi + \pi \dot\phi^* - L = \mathcal{H} $$

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  • $\begingroup$ yes, thats how we derived it in the lectures as well. my issue is with deriving it from the stress tensor $\endgroup$
    – paulina
    Mar 31 at 22:12
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    $\begingroup$ You are probably using the wrong equation for the stress tensor. You need to include both the conjugate momenta in the stress tensor definition, just like I included both in my expression for the Hamiltonian. So, just to be very explicit, you need to include a term for each "individual" field like: $T^{\mu\nu} = \pi^\mu\partial^\nu\phi + {\pi^*}^\mu\partial ^{\nu}\phi^* - g^{\mu\nu}\mathcal{L}$. $\endgroup$
    – hft
    Apr 1 at 1:02
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    $\begingroup$ @paulina I updated this answer to lift my comment above into the answer (and expand on the comment a bit). $\endgroup$
    – hft
    Apr 1 at 17:10
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In field theory texts, e.g. Peskin and Schroeder, one will often see the stress energy tensor written simply as: $$T^{\mu\nu}={\partial\mathcal L\over\partial(\partial_\mu\phi)}\partial^\nu\phi-\delta^{\mu\nu}\mathcal L,$$ where $T^{\mu\nu}$ is a conserved Noether current for any given $\nu=0,1,2,3$. This notation is accurate for the case of a theory with one field of concern, viz. $\phi$. However, one can apply Noether's theorem to the symmetries arising from space-time translation for theories with more than one field just as well. For example, given a theory with a set of fields $\phi_\rho$, where $\rho$ is an index that serves to label the distinct fields. Thus, in more complete notation, such as one finds in the relevant section of Goldstein's chapter 13 of Classical Mechanics, we have that: $$T^{\mu\nu}={\partial\mathcal L\over\partial(\partial_\mu\phi_\rho)}\partial^\nu\phi_\rho-\delta^{\mu\nu}\mathcal L.$$ So given a theory with two fields $\phi$ and $\phi^\dagger$, and their associated conjugate momenta: $$\pi_\rho={\partial\mathcal L\over\partial(\partial_0\phi_\rho)};\;\;\text{and}\;\; \pi_{\rho}^\dagger={\partial\mathcal L\over\partial(\partial_0\phi_\rho)},$$ we can write: $$T^{\mu\nu}={\partial\mathcal L\over\partial(\partial_\mu\phi)}\partial^\nu\phi+{\partial\mathcal L\over\partial(\partial_\mu\phi^\dagger)}\partial^\nu\phi^\dagger-\delta^{\mu\nu}\mathcal L.$$ From which point it is easy to find the Hamiltonian density from $T^{00}=\mathcal H$.

To understand why there is a sum of terms for each field in the stress energy tensor, see any good reference that shows a derivation of the Noether current $j^\mu$ from the transformation $\phi\rightarrow\phi^\prime$ and $\mathcal L\rightarrow\mathcal L^\prime$.

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