Trying to make the connection with QM as a 1 parameter QFT

Lanza52
Messages
63
Reaction score
0
I'm seeing lots of underlying connections between the canonical formulism of QFT and QM. But I'm getting a bit confused by their differences. I'll just write down my thought process:

QM is a one parameter system (t) in a space with three quantized operators (x,y,z)
QFT is a four parameter system (t,x,y,z) in a space with one/four/four quantized operators (scalar/spinor/vector).

In QM we develop the hamiltonian formalism and treat our one parameter as "special."
In QFT we develop the hamiltonian formlism and treat one of our four "equal" parameters as special. And I lose sleep at night because of it. Anyways...

In QM we have the S/H pictures where we apply the development of our parameter t to the states/operators.
In QFT we treat our special parameter with privilege once again and apply the development of t to the states/operators.
In QFT, I think, but I'm not sure, that the rest of the Schrodingafication/Heisenbergation is assumed throughout and we apply all the rest of the parameter dependence upon the field operators. IE ∂ψ/∂x^i = 0 for all states ψ and i=1,2,3.

I actually think typing this up has cleared up my thoughts well enough, but I'm still inquiring further as I read my textbook. But I can't find anybody that develops these concepts AT ALL. Maybe an obscure mention, but not completely.
Also, I posted this earlier as I wasn't quite as far along in my logical development. I'll just repost it here if anybody has any desire to respond to it.

I'm a little confused in the allocation of t,x,y,z dependence between the states and fields in all the different pictures.

In QM, we had a state ψ and an operator A. In the SP, ψ is a function of t and A isn't. In the HP, A is a function of t and ψ isn't.

In QFT, we have a state ψ and an operator A. But we also give ourselves three new dependent variables for our system to do whatever it wants with; x,y,z. I don't see a reason why we don't have xyzHeisenberg pictures and xyzSchrodinger pictures. Not necessarily saying they would be useful, but their complete lack of mention in P&S, Zee, Srednicki, Zuber etc have me puzzled. Can't we pick between an operator relation of any of these?

A(0)ψ(t,x,y,z) --- A(x,y,z)ψ(t) --- A(t)ψ(x,y,z) --- A(t,x,y,z)ψ(0)

Also, if anybody has a resource that talks about this, I'd like to read it.
 
Physics news on Phys.org
The difference between QM/QFT is not Newtonian mechanics/special relativity but particles/fields. In relativistic QFT, we don't treat t more special than special relativity itself does. In relativistic QFT, you have a unitary representation of the Poincare group on your Hilbert space, so that all time translations/space translations/boosts/rotations are treated on equal footing. Because this representation is assumed to be strongly continuous, you can write down an infinitesimal version of it. The infinitesimal version of time-translation gives you the Schrödinger equation and you can also switch to the Heisenberg picture, just as in QM.
 
Lanza52 said:
In QM we develop the hamiltonian formalism and treat our one parameter as "special."
In QFT we develop the hamiltonian formlism and treat one of our four "equal" parameters as special. And I lose sleep at night because of it. Anyways...

The point is that you quantize only fields on the surface (mass shell) k^2=m^2. What you really single out is the k vector normal to that hypersurface, see
http://www.thphys.uni-heidelberg.de/~duo/skripten/schmidt_QFT1.pdf
I think you can get the same thing using derivatives with respect to proper time (which is a lorentz scalar) but this will again lead you to that constraint and you have to use special methods (Dirac brackets) to take care of it.
 
  • Like
Likes 1 person
Lanza52 said:
Also, if anybody has a resource that talks about this, I'd like to read it.

If I remember right, Klauber (http://quantumfieldtheory.info/) spends a lot of time trying to draw parallels between NRQM and QFT as he goes through all the basic derivations. You might give that a shot.
 
DrDu said:
The point is that you quantize only fields on the surface (mass shell) k^2=m^2. What you really single out is the k vector normal to that hypersurface, see
http://www.thphys.uni-heidelberg.de/~duo/skripten/schmidt_QFT1.pdf
I think you can get the same thing using derivatives with respect to proper time (which is a lorentz scalar) but this will again lead you to that constraint and you have to use special methods (Dirac brackets) to take care of it.

Thanks, that source has a lot of great alternative derivations.
 
NRQM is formally the same as 0+1 dimensional QFT, i.e. QFT in 0 spatial dimensions and 1 time dimension. The "fields" of NRQM are the quantities x, y, z.

In QFT you have three approaches:
(a) Work in the Heisenberg picture of the canonical formulation, in which case fields are operator functions of the spatial coordinates and that evolve with time according to a Hamiltonian. The Hamiltonian is built from field operators and their canonically conjugate momenta.

(b) Work in the Schrodinger picture of the canonical formulation, in which a wave functional evolves in time. The wave functional is a state vector which we can think of as a linear combination of all eigenstates of the field operators. The field operators are time-independent.

(c) Do a path integral over the fields, which are functions of 4 coordinates, of exp(iS), where S is a functional of the fields.

Now replace fields with x,y,z and remove all spatial dimension to get the three approaches to NRQM:

(a) Work in the Heisenberg picture of the canonical formulation, in which case x, y, z are operators that evolve with time according to a Hamiltonian. The Hamiltonian is built from x, y, z and their canonically conjugate momenta.

(b) Work in the Schrodinger picture of the canonical formulation, in which a wave function evolves in time. The wave function is a state vector which we can think of as a linear combination of all eigenstates of the x, y, and z operators. The x, y, and z operators are time-independent.

(c) Do a path integral over particle paths ##(x(t), y(t), z(t))##, which are functions of the time coordinate, of exp(iS), where S is a functional of the path.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
According to recent podcast between Jacob Barandes and Sean Carroll, Barandes claims that putting a sensitive qubit near one of the slits of a double slit interference experiment is sufficient to break the interference pattern. Here are his words from the official transcript: Is that true? Caveats I see: The qubit is a quantum object, so if the particle was in a superposition of up and down, the qubit can be in a superposition too. Measuring the qubit in an orthogonal direction might...

Similar threads

Replies
23
Views
1K
Replies
13
Views
3K
Replies
33
Views
3K
Replies
7
Views
2K
Replies
7
Views
2K
Replies
13
Views
2K
Back
Top