I am just learning QED and could not understand the role of wave function. Is the basic equation in QED the Schrodinger Equation? Is the difference between Quantum mechanics and QED just they have different Hamiltonians.
Quantum River
Relativistic & non-relativistic quantum description of any physical object (particle/field) can be achieved by at least 2 equivalent methods:
1)Schrodinger: the time-evolution of the system is given by the state vector. operators are time-independent.
So, in the coordinate representation of:
i) point particle;
\hat{X}\rightarrow x
\hat{P}\rightarrow -i \frac{\partial}{\partial x}
we have (depending on the Hamiltonian) the usual Schrodinger/Dirac equation:
i\frac{\partial}{\partial t}\Psi (x,t) = H(x, -i\partial_{x}) \Psi (x,t)
Here, the wavefunction (probability amplitude) is the component of the state |\Psi (t)\rangle in the direction |x> :
|\Psi (t)\rangle = \int dx \Psi (x,t) |x\rangle
ii) continuous object (field);
\hat{\varphi}(\vec{x}) \rightarrow \phi (\vec{x})
\hat{\pi}(\vec{x})\rightarrow -i \frac{\delta}{\delta \phi (\vec{x})}
we have the Schrodinger representation of (relativistic/nonrelativistic) field theory:
i\frac{\partial}{\partial t} \Psi [\phi ,t] = H(\phi (\vec{x}), -i\frac{\delta}{\delta \phi (\vec{x})})\Psi [\phi ,t]
Notice that this equation is a functional differential equation for the wave functional \Psi [\phi ,t] of the quantum field. As in the case of point particle, this number (probability amplitude) reoresents the component of the state |\Psi (t)\rangle in the direction |\phi \rangle :
|\Psi (t)\rangle = \int \cal{D}\phi \Psi [\phi,t] |\phi \rangle
and the square of the modulus of the amplitude is proportional to the probability that the field has the configuration \phi (\vec{x}) at t (i.e., the field is at the point \phi in the function space).
Functional differential can be thought of as an
infinite set of
coupled partial differential equations. So you can imagine the "fun" when trying to solve them. Indeed, very few methods are known to solve these equations. This is why textbooks avoid the Schrodinger representation in field theory. Instead, they work with the following (equivalent) operator representation:
2) Heisenberg: the dynamics is controlled by operators. the state vectors are time-independent.
For QM of particle, you have
\frac{d}{dt}\hat{X}(t) = [iH , \hat{X}(t)]
and similar equation for \hat{P}(t) .
In the theory of fields, the (operator) Heisenberg equations are:
\partial_{t} \hat{\varphi}(\vec{x},t) = [iH , \hat{\varphi}(\vec{x},t)]
\partial_{t} \hat{\pi}(\vec{x},t) = [iH , \hat{\pi}(\vec{x},t)]
Now, consider (for example) a massless scalar field theory;
H = 1/2 \int d^{3}x (\hat{\pi}^{2} + |\nabla \hat{\varphi}|^{2})
In the operator representation, Heisenberg's equations give the operator equation of motion;
\partial_{\mu}\partial^{\mu} \hat{\varphi}(\vec{x},t) = 0
As you might know, this equation is (very) easy to solve (naturally) in terms of crreation & annihilation operators. (I assume, you know what to do next with the solution).
In the Schrodinger representation, we can work with a coordinate base |\phi \rangle for Fock space where the (time-independent) operator \varphi is diagonal;
\hat{\varphi}(\vec{x}, 0) |\phi \rangle = \phi (\vec{x}) |\phi \rangle
(the eigenvalue \phi is an ordinary function of \vec{x} )
This equation together with
\hat{\pi}(\vec{x}, 0) = -i \frac{\delta}{\delta \phi (\vec{x})}
turn the Schrodinger equation
i\partial_{t} |\Psi \rangle = H |\Psi \rangle
into a functional differential equation;
i\partial_{t} \Psi [\phi ,t] =1/2 \int d^{3}x (-\frac{\delta^{2}}{\delta \phi^{2}(\vec{x})} + |\nabla \phi |^{2}) \Psi [\phi ,t]
where
\Psi [\phi ,t] \equiv \langle \phi | \Psi (t) \rangle
If you have the "required skills", then you can solve this equation and
reproduce all the results of the operator representation.In the Schrodinger representation we do not need to speak about (eventhough we can) creation and annihilation operators.
To summarize:
In a given classical system [such system is any 3-word selection from the pairs; (particle,field),(free,interactive),(relativistic,non-relativistic)]:
1)the operator representation is possible if, and only if, the Schrodinger representation is possible.
2) the corresponding quantum theory (QM or QFT) can be solved in any operator representation if, and only if, it can be solved by an appropriate Schrodinger equation.
So, It is not true to say that the functional Schrodinger equation does not solve interacting field theories. Indeed, the Rayleigh-Schrodinger perturbation theory (which is familliar from QM) can (always) be used to calculate the S-matrix. However, it is a lot easier to determined the S-matrix in terms of the Green's functions of the operator representation.
Remember that (apart from overly simplified mathematical models) interacting field theories
do not have exact solutions. This fact (naturally) does not depend on the representation(see 2 above). How the S-matrix is calculated does vary from representation to representation.
regards
sam
