# QED Feynman

1. Nov 1, 2006

### eb227

On page 50 he states " The probability of an event is equal to the square of the length of an arrow called the "probility amplitude".
What, if any, is the relationship between Feynman's "arrow" and Shroedinger's wave function "psi".

2. Nov 1, 2006

### vanesch

Staff Emeritus

It's the same. The Schroedinger wave function "psi" is a specific "coordinate representation" of the "arrow" Feynman talks about (in the basis of functions of position).

3. Nov 1, 2006

Staff Emeritus
Psi is a complex number. In the Argand or polar representation of complex numbers they are vectors ("little arrows") in the geometric plane spanned by the vectors 1 = (1,0) and i = (0,1).

A complex number a + ib has the polar representation of a vector of magnitude or length m making an angle of theta with the positive real axis, with m = sqrt(a^2+b^2) and theta = arctan (b/a).

Then the probabllity generated by psi = (a + ib) is (a+ib)(a - iB) = (a^2+b^2), which is the square of the magnitude, or length, of the vector in the polar represntation.

4. Nov 1, 2006

### eb227

Does that imply that Shroedinger's wave equation is the central mathematical equation of QED as well as Quantum Mechanics?

Last edited: Nov 1, 2006
5. Nov 1, 2006

### Parlyne

No. But, the fundamental ideas in Schroedinger's equation do get carried over.

However, Schroedinger's equation can't be used in QED because it is not valid for relativistic particles. Instead, you have to use the Dirac equation, which makes the mathematics much more complicated. (Dirac wavefunctions must be spinors, rather that scalars.)

6. Nov 1, 2006

Staff Emeritus
Plus Feynman's method becomes nonconvergent in Minkowski spacetime. A problem that is fixed by Wick rotation, at the cost of losing the pretty motivation. I don't know if Feynman mentions that in QED.

7. Nov 2, 2006

### vanesch

Staff Emeritus
Well, you can still use Schroedinger's equation in the relativistic version, in the sense:

hbar/i d/dt |psi> = H |psi>

if you place yourself in a specific reference frame.
Only, |psi> stands for the state vector, say, in Fock space.
But usually, in QFT, we write the Schroedinger equation in time-integrated form, which leads us to the Dyson recursion formula (from which the Feynman diagrams are derived in the canonical quantisation).

The Dirac equation is now the equation of the field operator (and not of the quantum state, or "wave field" or whatever).

It is in fact a very funny interplay, that originally, the Schroedinger equation was (for a single particle), seen as a kind of field equation for a field psi over space/time in a non-relativistic version, that Dirac tried to generalise it to a relativistic field equation, and that it then turned out that Dirac's "field equation" needed a "second quantisation". That is, we had to look upon the Dirac equation as a *classical* field equation now, and that we had to apply the formalism of quantum mechanics to this classical field, and that in this quantum theory of the quantization of a classical field, the Schroedinger equation turned up, again, as the time evolution equation.
So, after all, the Dirac equation is NOT the relativistic variant of the Schroedinger equation, though it was initially derived that way.

The Schroedinger equation is the time evolution equation in Hilbert space ; the Dirac equation is a classical field equation.

However, things are even more confusing ! It can now be seen that a quantized classical field is equivalent to the quantum description of a set of particles with variable number, obeying Bose or Fermi statistics. It is then a matter of taste to see a QFT as the quantum version of a classical field, with "emerging particles" ; or to see a QFT as the quantum version of a set of particles with a given statistics and variable number, and see the "field" emerging. They are equivalent.

8. Nov 2, 2006

### masudr

That is an important point that many introductory sources omit. It took me a while to work that out for myself.