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

1. Aug 2, 2013

### Lanza52

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)

2. Aug 2, 2013

### rubi

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.

3. Aug 2, 2013

### DrDu

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.

4. Aug 2, 2013

### Chopin

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.

5. Aug 2, 2013

### Lanza52

Thanks, that source has a lot of great alternative derivations.

6. Aug 3, 2013

### The_Duck

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.