2
$\begingroup$

My lecture notes define the conjugate momentum of a scalar field via:

$$\pi = \dot{\psi}$$

Where

$$\psi = \int \frac{d^3p}{(2\pi)^{3}}\frac{1}{\sqrt{2E_p}}\left(a_p e^{i\vec{p}\cdot \vec x} + a_p^\dagger e^{-i\vec p \cdot \vec x}\right) $$

and claim that this gives

$$\pi = \int \frac{d^3p}{(2\pi)^{3}}\sqrt{\frac{E_p}{2}}\left(a_p e^{i\vec p \cdot \vec x} + a_p^\dagger e^{-i\vec p \cdot \vec x}\right)$$

whilst working in Schodinger picture. But clearly $\psi$ doesn't even depend on time. Am I right thinking that what is stated in my lecture notes is wrong and the definition

$$\pi = \dot{\psi}$$

is valid in the Heisenberg picture only? And in order to obtain the above expressions, which are in Schrodinger picture, one needs to take the Heisenberg picture expressions:

$$\psi = \int \frac{d^3p}{(2\pi)^{3}}\frac{1}{\sqrt{2E_p}}\left(a_p e^{-ip\cdot x} + a_p^\dagger e^{ip\cdot x}\right) $$

$$\pi = \int \frac{d^3p}{(2\pi)^{3}}\sqrt{\frac{E_p}{2}}\left(a_p e^{-ip\cdot x} + a_p^\dagger e^{ip\cdot x}\right)$$

(where I now used the 4-vector notation) and then turn them into Schrodinger picture?

$\endgroup$
  • $\begingroup$ Could you please clarify your question, I am not sure what you are asking about? I am also not sure why are you bringing Schrödinger and Heisenberg up in a QFT topic. In QFT (and most of QM for that matter) the two pictures are practically used together. $\endgroup$ – Y2H May 28 '17 at 13:39
  • $\begingroup$ See the edit, is that clearer now? Well, I want to get my definitions right and for the above expressions, since $\pi$ is defined as a time derivative of $\psi$, their time dependence seems quite relevant. $\endgroup$ – Piotr May 28 '17 at 13:47
  • $\begingroup$ The thing is, the definition must be valid in both pictures since deriving a function with respect to time is the same in the matrix representation as deriving the operator. $\endgroup$ – Y2H May 28 '17 at 14:00
2
$\begingroup$

First forget QFT for a while and think about Classical Field Theory. Consider the Klein-Gordon field more precisely. Its Lagrangian is

$$\mathcal{L}(\phi,\partial_\mu\phi)=\dfrac{1}{2}\partial^\mu \partial_\mu \phi-\dfrac{1}{2}m^2\phi^2$$

In this Lagrangian the variable is $\phi$. Now since $\phi$ is a function defined on spacetime, $\phi$ depends on time in particular and you can compute in a reference frame $\dot{\phi} = \partial_0 \phi$.

One then defines the conjugate momentum, as in Classical Mechanics, to be

$$\pi = \dfrac{\partial \mathcal{L}}{\partial(\partial_0\phi)}.$$

For this field, what do we get? If you work this out you'll find $\pi = \dot{\phi}$.

This is all classical. Then with this in hands you can quantize.

After all, quantizing the field means that you want to turn $\phi,\pi$ into operators obeying

$$[\phi(x),\phi(y)]=[\pi(x),\pi(y)]=0$$

$$[\phi(x),\pi(y)]=i\delta(x-y).$$

Thus you already need $\pi$ to talk about quantization. As in Quantum Mechanics, you need both position and momentum to impose the canonical commutation relations.

By the way, there's a small detail. The commutation relations are taken at equal times. In that case they hold among the Schrodinger picture operators $\phi(\mathbf{x}),\pi(\mathbf{y})$, since they are defined at the same initial time.

So if you want to compute $\pi$ from $\phi$, you can do it classicaly and then impose the canonical commutation relations, or you can do it in the Heisenberg picture and you'll get the same results.

Edit: the mode decomposition can be achieved in Classical Field Theory, the only thing is that the coefficients will be numbers. The equation of motion is

$$(\Box + m^2)\phi=0$$

Take the Fourier transform in the spatial variable so that denoting the Fourier transform by $\hat{\phi}$ you have

$$\partial^2_t \hat{\phi} +(|\mathbf{p}|^2+m^2)\hat{\phi}=0$$

define $\omega_{p}^2=|\mathbf{p}|^2+m^2$ and $p = (\omega_p, \mathbf{p})$. The equation is parametrized by $\mathbf{p}$ and can easily be solved to give

$$\hat{\phi}(\mathbf{p},t)=a_p e^{-i\omega_p t}+b_p e^{i\omega_p t}$$

now apply the reality condition of the Fourier transform

$$\hat{\phi}(-\mathbf{p},t)=\hat{\phi}(\mathbf{p},t)^\ast.$$

You arrive at the condition

$$a_{-\mathbf{p}}e^{-i\omega_p t}+b_{-\mathbf{p}}e^{i\omega_p t}=a_{\mathbf{p}}^\ast e^{i\omega_p t}+b_{\mathbf{p}}^\ast e^{-i\omega_p t}$$

the linear independence of the exponentials then gives $a_{-\mathbf{p}}=b_{\mathbf{p}}^\ast$ and $b_{-\mathbf{p}}=a_{\mathbf{p}}^\ast$. Now you have

$$\hat{\phi}(\mathbf{p},t)=a_pe^{-i\omega_p t}+a_{-p}^\ast e^{i\omega_p t}$$

now apply the Fourier inversion to get

$$\phi(x)=\int \dfrac{d^3 p}{(2\pi)^3}(a_p e^{-i\omega_p t}+a_{-p}^\ast e^{i\omega_p t}) e^{i \mathbf{p}\cdot \mathbf{x}}$$

if you change variables on the second term making $p \to -p$ since $\omega_{-p}=\omega_p$ you get the formula

$$\phi(x)=\int \dfrac{d^3 p}{(2\pi)^3}(a_p e^{-ipx}+a_p^\ast e^{ipx}).$$

The $\sqrt{2\omega_p}$ is then included for convenience to get a Lorentz invariant result (it amounts for a redefinition of $a_p$). The final answer is

$$\phi(x)=\int \dfrac{d^3 p}{(2\pi)^3}\dfrac{1}{\sqrt{2\omega_p}}(a_p e^{-ipx}+a_p^\ast e^{ipx}).$$

Please, understand that this doesn't derive the Fock space representation. This is just a classical calculation that in turns motivates the mode decomposition in terms of the Fock space ladder operators.

By the way, there's a cleaner and more elegant approach with the spacetime Fourier transform that can be found in the question A question on using Fourier decomposition to solve the Klein Gordon equation.

$\endgroup$
  • $\begingroup$ I think I almost get it. The one little thing is, you say I can compute $\pi$ from $\phi$ classically. But what are the expressions for the fields in classically? Because the expressions in terms of mode operators I wrote above are clearly a result of quantization and don't mean anything in the light of the classical theory. I'm quite happy with the alternative approach of quantizing in Heisenberg picture, obtaining $\pi$ by differentiating and then moving back to Schrodinger picture though, thanks!. $\endgroup$ – Piotr May 28 '17 at 14:40
  • 1
    $\begingroup$ @Piotr, I added one edit, clarifying that the same decomposition can be obtained with numerical coefficients in Classical Field Theory, and in turn can be seen as some sort of motivation for the definition made in Quantum Field Theory. By the way, if you want to derive the conjugate momentum field in terms of creation and annihilation operators, I prefer doing directly in the Heisenberg picture. $\endgroup$ – user1620696 May 28 '17 at 15:01
  • $\begingroup$ Great explanation, I would up-vote this more if I could. Thank you so much! $\endgroup$ – Piotr May 28 '17 at 15:33
-2
$\begingroup$

I think I know what your problem is. You are forgetting that the time dependency can be implicit and doesn’t have to only be explicit. For instance, $\psi$ might depend on time because $x$ and/or $p$ depend on time. In this case the derivative will not be zero.

Also, the definition must be valid in both pictures since deriving a function with respect to time is the same in the matrix representation as deriving the operator with respect to time.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.