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I have a rather basic question about quantum field states. Usually the Heisenberg picture is used in QFT whose effect is that quantum states are time-independent and the operators on the states carry all the time-dependence.

But what's about in a interacting QFT a 2-particle state in the far past: $|E_{\bf p_1}, {\bf p_1}, E_{\bf p_2} {\bf p_2}>$ which undergoes a scattering among the two particles, in the far future it will be $|E_{\bf k_1}, {\bf k_1}, E_{\bf k_2} {\bf k_2}>$, of course respecting energy-momentum conservation, but the quantum state has changed. Therefore, this 2-particle state evolves in time, it changes in time. So how is this compatible with the Heisenberg picture where quantum states don't evolve? I would appreciate to get a conclusive answer as this aspect seems to be rather basic for understanding quantum field theory.

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    $\begingroup$ You are labelling the states with energy and momentum, which are observables and so in the Hiesenberg picture are time dependant.The state will not change but the energy of the particles in that state will evolve from $E_{i1},E_{i2}$ to $E_{f1},E_{f2}$ $\endgroup$ – By Symmetry Aug 31 '16 at 11:58
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Your question appears to have nothing particular to do with quantum field theory, since the same can happen in ordinary quantum mechanics for any state that you labeled by the eigenvalue of a time-dependent observable, and in particular for time-dependent Hamiltonians.

Suppose you have a time-dependent observable $A(t)$ and an eigenstate $\lvert a_1,t_1\rangle$, which has the eigenvalue $a_1$ at time $t_1$. You'Re now wondering how it can be that we have a probability to find this state is the same as $\lvert a_2,t_2\rangle$ for $a_1\neq a_2$, but this is perfectly possible. Just let, for instance, $a_1,a_2$ be $+1/2,-1/2$ and let $A(t_1)$ be $\sigma_z$ and $A(t_2)$ be $-\sigma_z$, i.e. $A$ measures the spin in z-direction at both times, but the sign is flipped between the two states, and we have $\lvert a_1,t_1\rangle = \lvert a_2,t_2\rangle$.

Now, for the case of an interacting QFT that you ask about, you should first think about the picture-free way to write down what you're asking about: With the time-evolution operator "$U(-\infty,\infty)$" and the states $\lvert p_1,p_2\rangle$ and $\lvert k_1,k_2,\rangle$, you want to compute $$ \langle p_1,p_2\vert U(-\infty,\infty)\vert k_1,k_2\rangle$$ and since the time-evolution is computed from a time-dependent Hamiltonian, you have no guarantee that any of the momentary eigenstates $\lvert p_1,p_2\rangle$ stays an eigenstate throughout the evolution. In fact, this cannot happen if there's a non-zero amplitude for $k_1\neq p_1$ or $k_2\neq p_2$, but there's nothing wrong with this in either picture.

Lastly, let me remark that the existence of the pictures in QFT to begin with is a contentious issue, due to Haag's theorem making the rigorous existence of a unitary equivalence between the fields acting on the asymptotically free Hilbert spaces and the interacting Hilbert spaces impossible.

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It actually does not work this way. When you prepare a certain “in” state for a scattering event which contain say n particles, then as a result of scattering you can have a number of different final “out” states containing m particles where m can be different from n. The “in” state should be expressible as a linear combination of all the final states as the Heisenberg state should not change with time. So basically,

$|k_1,….k_n ; in> = \Sigma |p_1,…. p_m ; out> S_{p_1……p_m;k_1….k_n}$

The coefficient $S_{fi}$ are elements of S matrix which basically describe matrix elements from an initial state to final state. For example, two electrons can be scattered into two muons or two electrons or whatever, but the two electrons in the initial state will of course be different from the two electrons state in the final state as the latter would fall in that linear combination with a matrix coefficient.

Moreover you actually work in interaction representation ( Yang- Feldman excluded) in qft , where time dependence is split into two parts: the time dependence of noninteracting systems is carried by the operators, while remaining time dependence is taken by a state vector. So you can avoid simple problems by opting for interaction representation in place of Heisenberg representation.

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