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Given a space of states $|\rangle$, $|x\rangle$, $|x,y\rangle$, with the creation operators such as $\hat{\phi}(x)|y,z\rangle=|x,y,z\rangle$ for creating a particle at position $x$ and so on.

How does this relate to path integrals?

e.g.

$$\Delta(x,0;y,t) = \int \phi(x,0)\phi(y,t) \exp(i S[\phi] ) D\phi$$

How do we get this using creation and annihilation operators?

I tried setting $|\rangle = \Psi_0[\phi]$ for some ground state but $a^+(x)\Psi_0[\phi] \neq \phi(x)\Psi_0[\phi]$ so then I got stuck. Also because then I'd end up with the ground state inside the path integral which wouldn't right.

Edit

Let me clarify a bit.

Say you have a wave functional $\Psi[\phi,t]$ which satisfies the second quantized Shrodinger equation:

$$ i\frac{d}{dt}\Psi[\phi,t] = \left(-\frac{\delta^2}{\delta\phi(x)^2}-\nabla \phi(x)^2 +m^2\phi(x)^2\right) \Psi[\phi,t]$$

and it has a ground state of the form:

$$ \Psi_0[\phi] = \exp\left( \int \phi(x)s(x-y)\phi(y) \right)$$

This is the vacuum state $|>$

Now I want to find the Feynman propagator. So I want something like:

$$\Delta(x,0;y,t) = \int \Psi[\phi_0]\Phi[\phi_t] \exp(i S[\phi] ) D\phi$$

for some particular wave functionals. But I want to find the one particle states to put in. If I set $\Psi[\phi_t] = \phi_t(x)\Psi_0[\phi_t]$ for example then I get the ground state inside the integral which is not what I want. Is it correct to expand the wave functional as:

$$\Psi[\phi,t] = \left(\psi_t + \int \psi_t(x)\phi(x)dx^3 + \int \psi_t(x,y)\phi(x)\phi(y) dx^3 dy^3+...\right)\Psi_0[\phi]$$

where the terms correspond to the states $|x,y,..;t>$. Or should I be using operators like $a^+(x)$ where the hamiltonian is:

$$H=\int\{a^+(x),a^-(x)\}dx^3$$

and

$$a^\pm(x) = \frac{\delta}{\delta \phi(x)} \pm \int s(x-y) \phi(y) dy^3$$

Whichever way I look at it I always end up with the ground state in the integral instead of just: $$\Delta(x,0;y,t) = \int \phi(x,0)\phi(y,t) \exp(i S[\phi] ) D\phi$$

See here it is just $\phi(x)$ not $\phi(x)\Psi_0[\phi] =\hat{\phi}(x)|>$. What am I doing wrong? How can I get rid of the ground state from the integral?

Edit 2

The only thing I can think of is that the ground state is not the same as the no-particle state. If the no-particle state is $\Psi_{NP}[\phi]=const$ then I think this solves it. Is this true or is the no-particle state equal to the ground state? But in that case what do the operators $a^\pm(x)$ correspond to? Do they mean excitations in the field as opposed to creation of particles?

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    $\begingroup$ First of all $a,a^{\dagger}$ create and destroy particles with momentum $k$, no details on the position are given: $a^{\dagger}|0\rangle=|k\rangle$. However, quantisation in terms of creation and annihilation operators in only possible in some special (linear) cases. The path integral approach is the general one and, by definition, one cannot retrieve $a,a^{\dagger}$ back. Correlation functions, though, are expectation values on the vacuum, and in this respect you may use either ways to compute them, comparing them both. $\endgroup$
    – gented
    Oct 11, 2015 at 16:01
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    $\begingroup$ There are no $\lvert x,y,z; t\rangle$ states because QFT doesn't really have a position operator. The one-particle states are usually momentum states, or superpositions thereof. What is $\Delta(x,0,y,t)$ supposed to be? I also have no idea what $\Psi[\phi_t] = \phi_t(x)\Psi_0[\phi_t]$ is supposed to mean, given that the r.h.s. depends on $x$ (and is thus not a functional of a field configuration) but the l.h.s. does not. $\endgroup$
    – ACuriousMind
    Oct 11, 2015 at 19:32
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    $\begingroup$ $\Delta(x,0,y,t)$ is the Feynman propagator to get from (x,0) to (y,t). I don't see your objection to the equation. The equation $x=3$ is valid even though the RHS doesn't depend on $x$. Also, you can Fourier transform any momentum states to position states. I don't know what you mean there are no states of this form. $|x,y>$ is the amplitude of finding particles both at x and y. It is symmetric for bosons and anti-symmetric for fermions. $\endgroup$
    – user84158
    Oct 11, 2015 at 19:41
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    $\begingroup$ There is no such thing. You can't Fourier transform momentum states into position states because QFT doesn't have a position operator. This is not QM where we have two operators $x,p$ with canonical commutation relations. We have the (four-)momentum operator, but there is no position operator in QFT, see this answer and this answer. What you are trying to do is ill-defined. $\endgroup$
    – ACuriousMind
    Oct 11, 2015 at 20:05
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    $\begingroup$ Those are states in quantum mechanics, not in quantum field theory. $\endgroup$
    – ACuriousMind
    Oct 11, 2015 at 20:17

2 Answers 2

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I have looked at this. Which seems to give more of a clue.

The ground state can be written as:

$$<0|\phi> = \int\limits^{\eta_0=\phi}e^{i S[\eta]} D\eta $$

The transition function can be written as:

$$<\phi|U(t,t')|\psi> = \int\limits^{\eta_t=\phi}_{\eta_{t'}=\psi}e^{i S[\eta]} D\eta $$

So:

$$D(x-y) = <0|\phi_0(x)\phi_t(y)|0> = <0|\phi_t>\phi_t(x)<\phi_t|U(t,t')|\phi_{t'}>\phi(y)<\phi|0> $$

$$= \int \left(\int\limits^{\eta_t=\phi_t}e^{i S[\eta]} D\eta \phi_t(x) \int\limits^{\eta_t=\phi_t}_{\eta_{t'}=\phi_{t'}}e^{i S[\eta]} D\eta \phi_{t'}(y)\int\limits_{\psi_t=\phi_t}e^{i S[\eta]} D\eta \right) D\phi D\psi $$

$$= \int \phi_t(x)\phi_{t'}(y) e^{iS[\phi]} D\phi$$

More or less

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I believe the answer is that the ground state is not the same as the no particle state. Therefor the wave functional should be expanded as:

$$\Psi[\phi,t] = \psi_t + \int \psi_t(x)\phi(x)dx^3 + \int \psi_t(x,y)\phi(x)\phi(y) dx^3 dy^3+...$$

without anything to do with the ground state. Where $\psi(x,y)$ for example is the amplitude for detecting particles at both x and y.

I believe the $a^\pm(x)$ are (de-)excitations in the field at $x$ and not creation/annihilation operators for particles.

I think this is where the confusion lay.

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