gptejms said:
Ok then let me know what you would do--you have QFT and then somebody comes up with the Schrodinger equation.What interpretation would you give to \psi?Let this question be open to all.
I don't understand your question completely. I decided to describe my understanding of wave functions, quantum fields, Schroedinger equation and relativistic wave equations in quantum mechanics and in QFT. Hopefully, this will answer your question. This is going to be quite long. Please bear with me. Those who read S. Weinberg "The quantum theory of fields" vol. 1 may recognize that here I am trying to retell his book in few paragraphs.
First, there is no fundamental difference beween quantum mechanics and QFT. QFT is simply quantum mechanics applied to a class of systems in which the number of particles can change. For this reason, the Hilbert space of QFT cannot be a space with a fixed number of particles. The Hilbert space of QFT is constructed as a direct sum of n-particle spaces (or sectors), where n varies from 0 to infinity. This direct sum is called the Fock space. In the Fock space, there is a 1-dimensional no-particle subspace (n=0), which is called vacuum. There are also 1-particle subspaces (sectors n=1) for each type of particle in the theory: 1-electron sector, 1-photon sector. Then there are 2-particle sectors (n=2) with different combinations of particles: "2 electrons", or "2 photons", or "1 electron plus 1 photon", ... etc. Take a direct sum of all these sectors and you obtain the Fock space which is the Hilbert space of states in QFT.
All standard
Rules of Quantum Mechanics work without change in this Fock space. Any state of the system is described by a state vector. Observables are described by Hermitian operators. You can build orthonormal bases there, define a Hamiltonian, etc. The only difference is that the number of particles is not fixed. If you have an arbitrary state vector |\Psi \rangle in the Fock space, you can take projections of this vector on sectors with different numbers of particles n=0,1,2,3,... In each of these sectors you can have an orthonormal basis (e.g. a basis of position eigenvectors). So, in each of these sectors you can have an n-particle wave function. Then the total wave function corresponding to the state \Psi \rangle is a suporposition of all these wave functions with complex coefficients. To write all this down would require a cumbersome notation. I will give you just a couple of simple examples.
Suppose that the state vector |\Psi \rangle lies entirely in a 1-particle sector corresponding to a massive spinless particle. One can define a basis of position eigenvectors | \mathbf{r} \rangle in this sector and find the wave function of |\Psi \rangle in this representation \psi (\mathbf{r}) = \langle \mathbf{r} | \Psi \rangle.
The time evolution of any state vector in the Fock space is generally described by
-i \hbar \frac{\partial}{\partial t} | \Psi(t) \rangle = H | \Psi(t) \rangle (1)
where H = \sqrt{\mathbf{P}^2c^2 + M^2 c^4} is the Hamiltonian defined in the entire Fock space, \mathbf{P} and M are operators of the total momentum and total mass, respectively. Let us now consider the simple case in which there are no interaction terms in the Hamiltonian H. In particular, each n-particle sector remains invariant with respect to time evolution. In the 1-particle subspace described above, we can multiply eq. (1) by the bra vector \langle \mathbf{r} | from the left, and take into account that M is reduced to multiplication by a number m, which the mass of the particle. Then we get the Schroedinger equation for the wave function of one free particle in QFT
-i \hbar \frac{\partial}{\partial t} \psi(\mathbf{r}, t) = <br />
\sqrt{-\hbar^2 c^2 \nabla^2 + m^2 c^4} \psi(\mathbf{r}, t)
A similar construction can be repeated in the case of a two-particle state. The wavefunction may be written as \psi (\mathbf{r}_1, \mathbf{r}_2), and the Schroedinger equation is
-i \hbar \frac{\partial}{\partial t} \psi (\mathbf{r}_1, \mathbf{r}_2)= <br />
\Bigl( \sqrt{-\hbar^2 c^2 \nabla_1^2 + m^2 c^4} + \sqrt{-\hbar^2 c^2 \nabla_2^2 + m^2 c^4}\Bigr)\psi (\mathbf{r}_1, \mathbf{r}_2)
In the case of multiparticle systems with interactions that can change the number of particles, it is almost impossible to write the Schroedinger equation in this notation, so it is more preferable to use the abstract form (1).
Now, you may ask, where are quantum fields? I didn't need this concept so far. In fact, quantum fields have no relationship to particle wave functions. Their role in QFT is completely different. They are needed in order to write down the interaction part V of the full Hamiltonian H = H_0 +V. This is a non-trivial construction if we want to have a Hamiltonian that satisfies the principle of relativistic invariance. I can tell you the story of quantum fields and their wave equations (Dirac, Klein-Gordon, etc.) if we can agree that what I wrote so far makes sense.
Eugene.
