# Abstract Schrödinger's equation in QFT

1. Jun 19, 2008

### ledamage

Hi folks!

A QFT question: you start from the lagrangian, compute the hamiltonian via Legendre transform and promote the the fields to operators with canonical equal-time commutation relations. Now you can compute the relation

$$[H,F(x)]=-\mathrm{i}\partial_0 F(x) \ ,$$

where $$H$$ is the hamiltonian and $$F(x)$$ is a polynomial of the fields and/or their conjugate momenta. This implies that the hamiltonian is the generator of time translations. Does this mean that the QFT formalism automatically implies the validity of abstract Schrödinger's equation which states the same?

2. Jun 19, 2008

### lbrits

Quantum field theory is, first and foremost, a quantum mechanical theory. So you shouldn't be surprised to learn that the Schrodinger equation is obeyed. It's not like QFT implies the SE. The SE was there when you started. Incidently, the SE arises in ordinary quantum mechanics in the same way. It's just that people usually derive it heuristically by appealing to de Broglie waves.

3. Jun 19, 2008

### Fredrik

Staff Emeritus
If you describe a system as being in state $|\psi\rangle$, another observer, who is moving with the same velocity as you, and is facing the same direction as you, but is translated in time relative to you (i.e. he sets his clock to zero at a different time), will describe the system as being in a different state $|\psi'\rangle$. The operator that takes your state vector to his (i.e. the map $|\psi\rangle\mapsto|\psi'\rangle$) takes the form $e^{-iHt}$ in both relativistic and non-relativistic quantum mechanics. (This is implied by the properties of whatever space-time we're using together with the basic postulates of quantum mechanics, such as "states are represented by the rays of a Hilbert space").

As you probably know, this operator satisfies the Schrödinger equation. (I'm using units such that $\hbar=1$).

4. Jun 19, 2008

### pmb_phy

I don't know QFT myself. But I do know that it is a quantum theory of fields. Schrodinger's equation, however, is the equation to describe the quantum dynamics of particles, not fields.

Pete

5. Jun 19, 2008

### Fredrik

Staff Emeritus
More generally, there are operators corresponding to all the Lorentz transformations, not just the translations in time. The map $U$ that takes a Lorentz transformation $\Lambda$ to the corresponding operator $U(\Lambda)$ is (almost) a representation of the Poincaré group, with the Hilbert space as the representation space. (I'm too lazy to explain the "almost" right now).

If a subspace is invariant under the action of every $U(\Lambda)$, we can use that subspace as the representation space instead of the larger Hilbert space. (So U becomes a function into the set of operators on that space instead). If a representation space doesn't have any invariant subspaces, the representation is said to be irreducible. It's possible to show that all the states in the representation space of an irreducible representation of the Poincaré group are eigenstates with the same eigenvalue of both $\vec J^2$ (3-vector square of the generators of rotations) and $P^2$ (4-vector square of momentum). The eigenvalue of the former is written as j(j+1) and j is called the spin. The eigenvalue of the latter is written as -m2 and m is called the mass.

The rays of an invariant subspace are therefore interpreted as one-particle states. This is far from obvious when you learn about canonical quantization of classical field theories, but what you're really doing is an explicit construction of an irreducible representation of the Poincaré group. (I think you can take the solutions of the field equation to be the representation space, but this is hardly ever mentioned anywhere, so I'm not 100% sure).

6. Jun 19, 2008

### ledamage

Yes, I think you're right. You can use this relation to group theory to "derive" relativistic wave equations. I'm wondering why this is never told in elementary text books on QFT; in my opinion, this is a crucial fact if you want to know what's really going on.

Okay, but where do I use the SE in QFT, explicitly or implicitly? Apparently, I use methods of classical field theory up to the point where I promote the fields to operators and everything else happens as if by magic...

Last edited: Jun 19, 2008
7. Jun 19, 2008

### Fredrik

Staff Emeritus
Every time you use the time evolution operator.

8. Jun 19, 2008

### Haelfix

"(I think you can take the solutions of the field equation to be the representation space, but this is hardly ever mentioned anywhere, so I'm not 100% sure). "

Modulo quibbles about superselection sectors, I think this is right.

9. Jun 20, 2008

### Demystifier

Yes it does. Just as quantum mechanics of particles, QFT can also be formulated in 3 equivalent ways:
1. Heisenberg picture - the most frequent formulation of QFT
2. Path integral formulation - also very common in practice, especially for non-Abelian gauge theories
3. Schrodinger picture - relatively rare in use, but also can be formulated, and corresponds exactly to your suggestion above.
I recommend you to see the QFT textbook:
B. Hatfield, Quantum Field Theory of Point Particles and Strings (1991)
in which all standard results of QFT are derived in detail in all 3 representations.

10. Jun 20, 2008

### ledamage

Thanks! This could be exactly what I'm looking for. I find it quite helpful to look at the same thing from different perspectives to gain a deeper understanding...

11. Jun 20, 2008

### ledamage

I think, spin is an eigenvalue of the square $$W^\mu W_\mu$$ of the Pauli-Lubanski vector

$$W^\mu \equiv \frac{1}{2}\epsilon^{\mu\nu\varrho\sigma}P_\nu M_{\varrho\sigma}$$

rather than $$\mathbf{J}^2$$ which is not a Casimir of the Poincaré group as far as I know.