Abt QFT with John

John Rennie

Jan 13, 2019 07:58

@taritgoswami we should probably discuss your question here rather than in the problem solving room.

tarit goswami

@Blue Ooh

@JohnRennie Ok

Are you free now?

John Rennie

@taritgoswami particles are neither particles nor waves, they just sometimes behave like particles and behave like waves. The best description we have of particles is from quantum field theory, and that describes particles as states of quantum fields.

tarit goswami

States of quantum fields means? can you explain it with some more details?

John Rennie

Are you an undergraduate, still at school, or at some other point in the education system?

tarit goswami

I am an Undergraduate

John Rennie

Jan 13, 2019 08:04

OK, do you know what a Fourier transform is?

tarit goswami

Yeah

Don't know very well actually

John Rennie

OK. Suppose you have some function that is a function of position in space, $f(x,y,z)$. We can Fourier transform $f$ and what this does is express $f$ as an infinite sum of plane waves.

That is, we write $f$ as something like:

$$ f(\mathbf x) = \int a(\mathbf k) \sin(\mathbf k\cdot \mathbf x) $$

where $\mathbf k$ is the wave vector.

tarit goswami

Ok

John Rennie

Now, if you remember back to when you first learned about quantum mechanics you probably started with the Schrodinger equation for a free particle, and you discovered its wavefunction was an infinite plane wave, just like the infinite plane waves in the Fourier transform above. Yes?

tarit goswami

Yeah

John Rennie

Jan 13, 2019 08:14

OK. What we do in quantum field theory is to say that the function $f$ represents a quantum field, and the plane waves in the integral are kind of the wavefunctions for free particles. So our quantum field is built up from the states of free particles.

tarit goswami

By function of position in space do you mean the wave function itslef?

John Rennie

These individual states in the integral are called Fock states and they behave like quantised simple harmonic oscillators. The ground state of a Fock state is when no particles are present.

The first excited state is when one particle is present, the second excited state is when 2 particles are present and so.

So quantum field theory describes particles as the excitations of the Fock states from which the quantum field is built. Each Fock state describes particles with a particular momentum $\mathbf p$.

So you can create a particle of momentum $\mathbf p$ by adding energy to the correspnding Fock state to excite it to a higher energy level.

If you're now thinking WTF I sympathise :-)

tarit goswami

Yeah I will need some time to digest it :p

John Rennie

When you get the hang of this it makes a lot of things suddenly a lot clearer.

For example suppose you collide two electrons in a collider, how do extra particles get created in the collision?

tarit goswami

@JohnRennie No idea :p

John Rennie

Jan 13, 2019 08:22

It happens because the Fock states describing the incoming electrons can transfer their energy to the Fock states of other quantum fields, and this creates new particles corresponding to those Fock states.

So the creation and annihilation of particles is just shuffling energy around between different Fock states.

tarit goswami

The individual states in the integral are Fock state, and that's the quanum state of free particle .. means Fock states are quantum states of Free particle.. am I correct? please correct me

user2723984

I find examples illuminating: if you solve the equations for a free real scalar field (Klein Gordon) you get something like $$ \int \frac {d^3p}{(2\pi)^3} a_{\mathbf{p}}e^{-ipx}+a_{\mathbf{p}}^\dagger e^{ipx}$$

whoops I pressed send way before my message was ready lol

didn't mean to intrude myself so rudely

John Rennie

@taritgoswami yes, though this is all rather arm waving and approximate. But you can think of the Fock states as describing free particles.

user2723984

anyway if you then calculate the hamiltonian you get $$\int \frac{d^3 p}{(2\pi)^3}a_\mathbf{p}^\dagger a_\mathbf{p}+\frac{1}{2}[a_\mathbf{p}, a_\mathbf{p}^\dagger]$$if you interpret $a$ and $a^\dagger$ as the usual ladder operators you can see that the field is a continuum of quantum harmonic oscillators, and that the operators can create particles at any momentum

John Rennie

@user2723984 I suspect this might be a bit much for @taritgoswami :-)

tarit goswami

Jan 13, 2019 08:33

@JohnRennie Yeah :(

John Rennie

@taritgoswami I'm giving you a oversimplified and potentially misleading "pop science" description in an attempt to give you that basic ideas. @user2723984 is doing it properly :-)

user2723984

oh sorry

no it's just that having already took quantum mechanics

I thought seeing that the hamiltonian of a certain field is an integral over hamiltonians of harmonic oscillators might make clearer what you meant by what you said before :)

tarit goswami

@JohnRennie I am not getting this line - chat.stackexchange.com/transcript/message/48476681#48476681

user2723984

but if not ignore what I wrote

John Rennie

@taritgoswami if you're just calculating the wavefuction of a free particle you get a plane wave, and you get the amplitude of the plane wave by normalising the wavefunction. Yes?

tarit goswami

Jan 13, 2019 08:39

Yeah

John Rennie

You normalise to one because you know there is one particle and you know the particle must be somewhere, so the integral over the whole wavefunction has to be one.

tarit goswami

Yes

John Rennie

The Fock state is a bit different. It can represent one particle or two particles or $n$ particles for any integer $n$. So when you normalise it the result is one, or two, or $n$ depending on how many particles are present in that state.

tarit goswami

Ok

John Rennie

@taritgoswami anyway, the point is that particles are complicated things in QFT, so you have to be careful about describing them as waves or particles.

They are objects that can behave like waves or behave like particles depending on the system. It is sometimes convenient to use an approximate description of the article as a wave, and sometimes use an approximate description of the particle as a particle.

Polarisation is easy to understand when describing the particle as a wave, so that's the wave we'd normally describe it.

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