QFT - Commutator relations between P,X and the Field operator

[Drew Carey]

Hi all,
I haven't been able to find an answer online but this seems like a pretty basic question to me. What are the commutator relations between the position/momentum operators and the field operator?
I'm not even certain what the commutation relations between X/P and a single ladder operator are in the QFT context. With the harmonic oscillator this is trivial, but it QFT we don't really define the ladder operators in terms of P/X, but rather as Fourier components of a classical field (and then proceed with canonical quantization (at least that's how I learned it)).

A simple motivation for this can be to show that a 2nd quantization Hamiltonian commutes with P for example (thus deducing that momentum is conserved in the system), however this would require knowledge of the commutator relations with the field operator.

Thanks for the help!

 

[Drew Carey]

Perhaps this isn't well defined?
I can treat this by writing the 2nd quantized momentum or position operators in terms of the field ladder operators and then express my desired commutator as a pretty trivial commutator between ladder operators.

Is this the only way to go about this?
Doesn't this work only if the functions making up the field are momentum eigen functions?

 

[WannabeNewton]

There is no position operator in second quantization. Position is a coordinate label for the fields $\varphi_i$ and their conjugate momenta $\pi_i$. However you can regard e.g. a single scalar field $\varphi(x)$ as creating a particle at the event $x$.

 

[atyy]

There is no position operator in second quantization.

There is no position operator in relativistic second quantization, but how about non-relativistic second quantization? Non-relativistic second quantization is just a rewriting of the non-relativistic Schroedinger equation for many identical particles, so I naively expect there to be a position operator.

 

[Drew Carey]

There is no position operator in second quantization. Position is a coordinate label for the fields $\varphi_i$ and their conjugate momenta $\pi_i$. However you can regard e.g. a single scalar field $\varphi(x)$ as creating a particle at the event $x$.

Interesting, thanks, I haven't realized this; I'll look into this. However it does digress a bit from my question which I am willing to restrict to the momentum operator which definitely can be 2nd quantized...

 

[WannabeNewton]

There is no position operator in relativistic second quantization, but how about non-relativistic second quantization? Non-relativistic second quantization is just a rewriting of the non-relativistic Schroedinger equation for many identical particles, so I naively expect there to be a position operator.

Check out the bottom of page 23 onwards of Tong's notes: http://www.damtp.cam.ac.uk/user/tong/qft/two.pdf

Interesting, thanks, I haven't realized this; I'll look into this. However it does digress a bit from my question which I am willing to restrict to the momentum operator which definitely can be 2nd quantized...

I'm not sure as to why it's a digression. Your original question explicitly asks for commutation relations between the position and momentum operators in (presumably relativistic) QFT in the second quantization framework. While a 3-momentum operator can be easily defined in this framework, and a 4-momentum operator as well in the usual way, a position operator obtained directly from a simple second quantization procedure is at odds with Lorentz covariance if one keeps time as a label. There are alternatives of course, such as http://en.wikipedia.org/wiki/Newton–Wigner_localization, but there is no "the position operator" in relativistic QFT unlike the case with the 4-momentum operator and your original post does hinge on this after all.

 

[atyy]

Check out the bottom of page 23 onwards of Tong's notes: http://www.damtp.cam.ac.uk/user/tong/qft/two.pdf

Looking at it quickly, I didn't see where it addressed the question. Could you point out more specifically?

Also, how about something like equation 48 of http://www.pma.caltech.edu/~mcc/Ph127/c/Lecture1.pdf . Couldn't the position operator be converted to second quantized form in a similar way?

 

[atyy]

So here is an interesting comment "Does it make sense to talk of the momentum p_i of a single particle i? The answer is NO, since the particles are indistinguishable." http://www.phys.lsu.edu/~jarrell/CO...s/Sandeep_Pathak/second_quantization_orig.pdf

Also "This is due to the fact that the observables of an assembly of N identical particles are permutation invariant, symmetric, functions of the single-particle dynamical variables and hence have vanishing matrix elements between states of the assembly belonging to different permutation symmetry classes of tensors. These classes distinguish the different super-selection sectors of the Hilbert space." http://www.scholarpedia.org/article/Second_quantization

 

[WannabeNewton]

Looking at it quickly, I didn't see where it addressed the question. Could you point out more specifically?

Did you read section 2.8.1. "Recovering Quantum Mechanics"? In the non-relativistic limit of QFT Tong writes down a position operator in terms of the scalar field $\varphi$ as $\vec{X} = \int d^3x \vec{x}\varphi^{\dagger}\varphi$ so that $\vec{X}|x\rangle = \vec{x}|x\rangle$. Or did I misunderstand your question?

Also, how about something like equation 48 of http://www.pma.caltech.edu/~mcc/Ph127/c/Lecture1.pdf . Couldn't the position operator be converted to second quantized form in a similar way?

Assuming I have understood the notation in that paper correctly, for a one body operator and a single scalar field it doesn't look to me like that definition would give something much, if at all, different from Tong's definition.

EDIT: By the way when I said p.23 of Tong's notes I meant p.23 of the pdf and not p.23 of the notes themselves. Sorry about that.

 

[Drew Carey]

Check out the bottom of page 23 onwards of Tong's notes: http://www.damtp.cam.ac.uk/user/tong/qft/two.pdf



I'm not sure as to why it's a digression. Your original question explicitly asks for commutation relations between the position and momentum operators in (presumably relativistic) QFT in the second quantization framework. While a 3-momentum operator can be easily defined in this framework, and a 4-momentum operator as well in the usual way, a position operator obtained directly from a simple second quantization procedure is at odds with Lorentz covariance if one keeps time as a label. There are alternatives of course, such as http://en.wikipedia.org/wiki/Newton–Wigner_localization, but there is no "the position operator" in relativistic QFT unlike the case with the 4-momentum operator and your original post does hinge on this after all.

I'm asking about the commutation relation between P and the field operator, not between P and X (which I realize now doesn't have an obvious 2nd Q parallel), but we can focus on the question in regards to P and the field operator.

 

[WannabeNewton]

I'm asking about the commutation relation between P and the field operator, not between P and X (which I realize now doesn't have an obvious 2nd Q parallel), but we can focus on the question in regards to P and the field operator.

See part (a): https://www.physicsforums.com/showpost.php?p=4603311&postcount=1

 

[atyy]

Did you read section 2.8.1. "Recovering Quantum Mechanics"? In the non-relativistic limit of QFT Tong writes down a position operator in terms of the scalar field $\varphi$ as $\vec{X} = \int d^3x \vec{x}\varphi^{\dagger}\varphi$ so that $\vec{X}|x\rangle = \vec{x}|x\rangle$. Or did I misunderstand your question?



Assuming I have understood the notation in that paper correctly, for a one body operator and a single scalar field it doesn't look to me like that definition would give something much, if at all, different from Tong's definition.

EDIT: By the way when I said p.23 of Tong's notes I meant p.23 of the pdf and not p.23 of the notes themselves. Sorry about that.

Ah yes, I missed that because I was using the wrong page numbers. Yes, that's seems to be the position operator in second quantization, similar to what's in Michael Cross's notes.

 

[Drew Carey]

See part (a): https://www.physicsforums.com/showpost.php?p=4603311&postcount=1

Thanks!
So basically it operates as it would on any function of space?
What is the commutation relation between P and the field ladder operators?

 

[WannabeNewton]

So basically it operates as it would on any function of space?

Basically, yes.

What is the commutation relation between P and the field ladder operators?

Well for a fixed mode, say $\vec{k}_0$, we have $[\vec{P}, a_{k_0}] = \frac{1}{2} \int d\tilde{k}[a^{\dagger}_k a_ka_{k_0} + a_k a^{\dagger}_ka_{k_0}]\vec{k} - \frac{1}{2} \int d\tilde{k}[a_{k_0}a^{\dagger}_k a_k + a_{k_0} a_k a^{\dagger}_k]\vec{k} \\= - \int d\tilde{k}(2\pi)^3 2\omega_0 \delta^3(\vec{k}_0 - \vec{k}) a_k \vec{k} = -\vec{k}_0a_{k_0}$
and similarly for the creation operator of a fixed mode.

P.S. Drew Carey is awesome :)