I'm working through my old notes on QFT (cf. Ref 1) and I'm not quite sure how to approach the derivation of the Hamiltonian density for a free scalar field (question 2.3 on page 19) and the subsequent derivation of the components of the energy-momentum four-vector in terms of the new field operators. Could you point me in the right direction?



  1. M Dasgupta, AN INTRODUCTION TO QUANTUM FIELD THEORY, Lecture presented at the School for Experimental High Energy Physics Students Somerville College, Oxford, September 2009.
  • $\begingroup$ Do you need full derivation? $\endgroup$ – user8817 Sep 4 '13 at 14:56

First question.

Let's have Lagrangian $$ L = \frac{1}{2}\left( (\partial \varphi )^{2} - m^{2}\varphi^{2}\right) . $$ Stress-energy tensor is equal to $$ T^{\alpha \beta} = \frac{\partial L}{\partial_{\alpha }\varphi }\partial^{\beta}\varphi - g^{\alpha \beta}L = \partial^{\alpha}\varphi \partial^{\beta}\varphi - \frac{g^{\alpha \beta }}{2}((\partial \varphi )^{2} - m^{2}\varphi^{2}). $$ Then $$ H = \int T^{00}d^{3}\mathbf r = ..., \quad \mathbf P^{i} = \int T^{0i} d^{3}\mathbf r = ..., \qquad (.1) $$ and you have an answer.

Second question.

Let's have the solution for scalar real field: $$ \varphi (x) = \int \left( b_{\mathbf k}^{*}e^{ikx} + b_{\mathbf k}e^{-ikx}\right)\frac{d^{3}\mathbf k}{\sqrt{(2 \pi )^{3}2\epsilon_{\mathbf k}}}, \quad kx = \varepsilon_{\mathbf k}x^{0} - (\mathbf k \cdot \mathbf x). \qquad (.2) $$ So you can rewrite $(.1)$ as $$ \mathbf P = \int \mathbf k b_{\mathbf k}b_{\mathbf k}^{*}d^{3}\mathbf k , \quad H = \int \epsilon_{\mathbf k} b_{\mathbf k}b_{\mathbf k}^{*}d^{3}\mathbf k \qquad (.3) $$ (where I ignored the infinite constant summand as we can do for the free field).

Then you want to rewrite operators of $(.3)$ expressions in terms of

$$ \hat {\varphi} , \quad \hat {\frac{\partial L}{\partial \dot {\varphi }}} = \hat {\dot {\varphi }} = \hat {\pi} . $$ You can do next steps.

  1. To multiply $(.2)$ by $e^{-i(\mathbf q \cdot \mathbf x )}$ and then take an integration over $\frac{d^{3}\mathbf r}{\sqrt{(2 \pi )^{3}}}$. You'll get $$ \int \varphi (\mathbf r , t) e^{-i (\mathbf q \cdot \mathbf r )}\frac{d^{3}\mathbf r }{\sqrt{(2 \pi )^{3}}} = ... = \frac{1}{\sqrt{2 \epsilon_{\mathbf q}}}\left( b_{-\mathbf q}^{*}e^{i \epsilon_{\mathbf q} t} + b_{\mathbf q}e^{-i \epsilon_{\mathbf q} t}\right) . \qquad (.4) $$
  2. To take t-derivation of $(.4)$. You'll get $$ \int \pi (\mathbf r , t) e^{-i (\mathbf q \cdot \mathbf r )}\frac{d^{3}\mathbf r }{\sqrt{(2 \pi )^{3}}} = i\sqrt{\frac{\epsilon_{\mathbf q} }{2}}\left( -b_{\mathbf q}e^{- i\epsilon_{\mathbf q} t} + b_{-\mathbf q}^{*}e^{i \epsilon_{\mathbf q} t} \right) \qquad (.5) $$
  3. Now you have from $(.4), (.5)$ a system of linear equations. You can express $b_{\mathbf q}$ from these equations as function of $\varphi , \pi$.
  4. You can use the hermitian nature of $\hat {b}_{\mathbf q}, \hat {b}^{+}_{\mathbf q}$. By assuming the $\hat {b}_{\mathbf q}$ expression you can get $\hat {b}^{+}_{\mathbf q}$ expression. Finally, you'll get $$ \hat {b}_{\mathbf q } = \int \left( \epsilon_{\mathbf q} \hat {\varphi} + i \hat {\pi} \right)e^{i q x}\frac{d^{3}\mathbf r }{\sqrt{2 \epsilon_{\mathbf q} (2 \pi )^{3}}} , \quad \hat {b}_{\mathbf q }^{+} = \int \left( \epsilon_{\mathbf q} \hat {\varphi} - i \hat {\pi} \right)e^{-i q x}\frac{d^{3}\mathbf r }{\sqrt{2 \epsilon_{\mathbf q} (2 \pi )^{3}}}. $$
  5. Use these expressions for the transformation of $(.3)$ expression.

If you want some details of derivations which were used in steps above, I'll help you.

  • 1
    $\begingroup$ Ok. I follow the approach in the first part and I'm working through it now. The approach in the second part makes sense to me (bringing it into ${\pi}$ etc.) up to about step 4, but this is a lack of knowledge on my part. I'm going to read more... Thank you very much for your assistance! $\endgroup$ – d3pd Sep 5 '13 at 7:09

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