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From non-relativistic quantum mechanics, a $\frac{1}{2}$- spin system can be represented by a ket vector like:

$$|\psi\rangle = a|+\rangle_{z}+b|-\rangle_{z}. \tag{1}$$

The object on $(1)$, is a ket vector with a fancy name: a (Pauli) spinor. This spinor object will satisfy the Pauli equation, the low-energy dynamical equation for the Dirac equation.

Now, still in non-relativistic quantum mechanics context, the Schröedinger equation can be dealt with in both ways:

$$i\hbar\frac{\partial }{\partial t}|\psi\rangle = H |\psi\rangle, \tag{2}$$

and

$$i\hbar\frac{\partial }{\partial t}\psi(\vec{r}) = H \psi(\vec{r}). \tag{3}$$

We can use Pauli spinors, in both $(1)$ fashion and $(3)$ fashion, since $\psi(\vec{r}) := \langle \vec{r}| \psi\rangle$.

But, in quantum field theory, we never use ket vectors! So my question is:

Why I can't study quantum field theory with a Dirac equation acting in a "Dirac four-spinor" $|\Psi\rangle$ as:

$$(i\gamma^{\mu}\partial_{\mu}-m)|\Psi\rangle = 0? \tag{4}$$

Qmechanic
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BasicMathGuy
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  • Ok, but could you elaborate an answer with math? – BasicMathGuy Nov 27 '22 at 16:27
  • You can do this, it would basically mean you have moved to the schrodinger picture, where the states are again the objects which change in time. – doublefelix Nov 27 '22 at 16:28
  • @doublefelix first: do you have a reference? Second: it means that every field in QFT is in position representation? Because, it seems a bit odd to sat that a vector field $|A\rangle>$ is written as $\langle x^{\mu} | A\rangle := A_{\mu}$ – BasicMathGuy Nov 27 '22 at 16:31
  • @TobiasFünke actually not, because I know what a spinor field means, and I know how to operate with Dirac equation. But I simply didn't grasp why we abandoned ket vectors in QFT. At least in master equations; you see, no one uses klein-gordon equation as $(\square - m^2) |\phi\rangle = 0$, or Dirac as $(4)$. Every time it seems that we are in position representation. – BasicMathGuy Nov 27 '22 at 16:35
  • Okay. Does this help then? I still don't understand your question, tbh, but there are sure experts here that do. I just want to point out that, I think, you're referring to field equations. And not to equations governing the time evolution of states. – Tobias Fünke Nov 27 '22 at 16:38
  • For a reference on solving the Dirac equation for states rather than fields, this is done often in relativistic quantum mechanics. Technically you could move the fields to another representation just like you could with any operator, but I haven't seen it done. – doublefelix Nov 27 '22 at 16:42
  • @TobiasFünke The same equation does govern the states in the Schrodinger picture. It just is most commonly not done in the Schrodinger picture. But that does not stop it from being possible to do. – doublefelix Nov 27 '22 at 16:43
  • Also, in response to OP's question of whether we abandoned ket vectors in QFT: We did not by any means. They are still necessary to get results from experiment and work by the same "postulates" as in Quantum Mechanics in regards to probability. The LSZ formula is the fundamental formula in QFT relating green's functions to measurement results and uses exactly those ket vectors. – doublefelix Nov 27 '22 at 16:44
  • @doublefelix so what is a spinor field? A section of a fiber bundle or a element of a hilbert space? Because the spinor field that occurs in Dirac equation the $\Psi$ is a section of a fiber bundle. – BasicMathGuy Nov 27 '22 at 16:47
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    @doublefelix There is no position basis in QFT. $\langle 0|\phi (x^{\mu}) |v\rangle$ gives you an approximate position basis solution. The real Schrodinger equation of QFT is the one with the integral of Hamiltonian density as its Hamiltonian. This Schrodinger equation is nothing like Klein Gordon or Dirac equation – Ryder Rude Nov 27 '22 at 16:49
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    There's a lot to say here. It is true that there is no position basis with satisfactory properties in QFT. Nonetheless there are kets, in the momentum basis, which not only exist but are necessary to relate the theory to experiment. You could in principle work in the schrodinger picture, in which case it is only the state (a ket) which changes in time. But the dirac equation is almost never used in that way- in normal QFT courses we work in the heisenberg or interaction picture. A spinor field is not an element of the hilbert space, nor is any field. They are operators/operator-valued distribs – doublefelix Nov 27 '22 at 17:04
  • @doublefelix I agree that kets are not abandoned. But I don't think the Dirac Hamiltonian will be appropriate even in the momentum basis. In the momentum basis, we'd be using the Hamiltonian $\int \sum _s (a^{\dagger}a + b^{\dagger}b ) d^3p$. This Hamiltonian has only positive energy eigenkets. The Dirac equation has negative energy states too. – Ryder Rude Nov 27 '22 at 17:18
  • I see now that I was assuming that the Dirac equation would maintain the same form for states as it does for fields, but this is by no means necessarily true and seems to be false based on your argument. In that case the equation itself would truly not apply to states regardless of which picture you are using, and instead there would be a different corresponding equation in the schrodinger picture. – doublefelix Nov 27 '22 at 17:32

1 Answers1

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With our modern understanding of quantum field theory, the Dirac equation is not a generalization of the Schrodinger equation from non-relativistic 1-particle quantum mechanics to relativistic 1-particle quantum mechanics, despite the fact that Dirac discovered his equation by looking for an equation to describe relativistic 1-particle quantum mechanics. In fact, we now understand that it there no so consistent theory of (interacting) 1-particle relativistic quantum mechanics; by combining relativity and quantum mechanics, we are necessarily led to theories with an indefinite number of particles.

Instead, the Dirac equation is best understood as an equation satisfied by a quantum field $\psi(x, t)$. It best understood as a generalization of the Heisenberg equation for the time evolution of operators. In quantum field theory, the spinor field $\psi(x, t)$ is a field (an operator-valued distribution), not a state.

We can and do talk about states in quantum field theory. Usually we work in the Heisenberg or interaction pictures, in which case the time evolution is carried by the operators (fields), or a mix of the operators and states. However we can work in the Schrodinger picture, in which case the state $|\Psi\rangle$ (which, in the field basis, is now a functional, mapping each possible field configuration to a probability amplitude) obeys the functional Schrodinger equation \begin{equation} i \frac{\partial}{\partial t}|\Psi\rangle = H|\Psi\rangle \end{equation} where $H$ is a functional of the fields. For example, for a free, massive spinor field $\psi$ (don't confuse $\psi$ the field with $\Psi$ the state!), $H$ would be (see, eg, Eq 5.8 here) \begin{equation} H = \int d^3 x \bar{\psi} \left(-i \gamma^i \partial_i + m\right) \psi \end{equation}

Andrew
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  • It is important not to conflate the Schrodinger picture with wavefunctional evolution. The Schrodinger picture is a basis-independent idea. The wavefunctional evolution is the Schrodinger picture in the wavefunctional basis. You can, in principle, even have interaction picture in the wavefunctional basis. Schrodinger picture is the idea that all the time evolution is carried by the state vector. This idea works in any basis. – Ryder Rude Nov 27 '22 at 17:22
  • @RyderRude Sure, I agree. – Andrew Nov 27 '22 at 17:33