# Importance of local conservation of probability

In almost every textbook of quantum mechanics we can find the derivation of the local conservation of probability. $$\nabla\cdot\vec{J}+\partial_t (\psi^*\psi)=0$$ where $\vec{J}$ is probabilty current. It seems like an important aspect of quantum physics.

Furthermore, this topic is a must in the discussion of finding a relativisic wave equation. We can see how this is done for the Klein-Gordon equation and Dirac equation.

But then, suddenly, nobody talks about that anymore. I dont see it in quantum electrodynamics or in string theory or any other quantum theory. Is local conservation of probability perhaps not as important as it seems? Do you know any reference(book or paper) deriving the local conservation of probability in the context of QFT, QED or String Theory?

• Note that similar things can be said about pretty much all aspects of the Schrödinger formalism: The reality is that people typically just don't work in terms of wavefunctions in QFT. It is possible, though. – Danu May 31 '15 at 6:22

One may think it is maybe possible to reproduce the reasoning that leads to this equation from the Schrödinger one in any quantum system where the Schrödinger equation holds (i.e. surely QFT and QED, I don't know about string theory). However, this is actually rather difficult, and it is not possible to obtain the nice form of the continuity equation (in my opinion) apart from some special situation. In addition, I would like to point out that the local conservation equation is a direct consequence of the Schrödinger equation (and of the particular form of "standard" Hamiltonians in non-relativistic QM), and does not add any further insight than the Schrödinger equation.

One problem in QFT is that we often do not know the precise form of the interacting Hamiltonian, however we may suppose that we know it and it is of the form $H=H_0(\Pi)+V(\Phi)$; where $\Phi$ is the quantum field operator and $\Pi$ its conjugate momentum.

Another problem is that $\Pi$ does not necessarily behave like a derivation operator on the wavefunction $\Psi$, and $V(\Phi)$ as a multiplicative operator. However, this can be made possible using a "trick" at least in a special case, i.e. when the Hilbert space of the QFT is a Fock space $\Gamma(\mathscr{H})$, where $\mathscr{H}$ is the one-particle (separable) Hibert space. In fact there is a construction, called Q-space, that unitarily identifies the Fock space with an $L^2(\Omega,d\mu)$ space of functionals $\Psi(\phi)$ on $\Omega$ with Gaussian measure $d\mu(\phi)$. In this space, the Fock space field $\Phi(x)$ acts as the multiplication by the function $\phi(x)$, and the momentum $\Pi(x)$ as the functional derivative $-i\partial_{\phi(x)}$.

Now in the Q-space form the Hamiltonian becomes $H=H_0(\partial_\phi)+V(\phi)$, and this is analogous to the usual $L^2(\mathbb{R}^d)$ form of QM, and therefore the conservation equation $$\partial_{\phi(t)} J\Bigl(\Psi(t,\phi),\partial_{\phi(t)}\Psi(t,\phi(t))\Bigr)+\partial_t\Bigl(\Psi(t,\phi)^*\Psi(t,\phi)\Bigr)=0\; ;$$ may be recovered (with a suitable current $J$, that in the case of $H_0(\Pi)\simeq \Pi^2$ has the usual form).

I'd say the answer to your question "Is local conservation of probability perhaps not as important as it seems?" should be positive, at least for relativistic quantum theory. Indeed, is probability really conserved? I'd say not necessarily. For example, particle-antiparticle pairs can be created in some processes. Is probability conserved in such processes? I doubt it, unless you say that a positron can be regarded as an electron with "negative probability distribution". What is important in relativistic quantum theory is current conservation, not probability conservation, whereas it is pretty much the same for non-relativistic quantum theory, as the energy is too low there for pair creation.

• 'For example, particle-antiparticle pairs can be created in some processes. Is probability conserved in such processes? I doubt it, unless you say that a positron can be regarded as an electron with "negative probability distribution"' What probability do you mean? If two particles appear at certain point as an event, that could have some probability density that is conserved. That conservation need not be violated when the event actually happens - there is a difference between probability of event and the event itself. – Ján Lalinský May 31 '15 at 9:04
• @JánLalinský: I mean probability of finding a particle at some place. I guess this is the probability typically related to the current in nonrelativistic quantum theory. – akhmeteli May 31 '15 at 9:39
• So you are saying that concept of probability density meant for stable number of particles cannot be applied in a theory where the particles can be created, right? – Ján Lalinský May 31 '15 at 9:47
• @JánLalinský: No, that's not what I'm saying. You can use the concept of probability density, however, the idea of probability conservation (and that is what OP asks) at least needs some modifications in this case. – akhmeteli May 31 '15 at 9:53
• I am not sure how you can separate the probability density from its conservation. If function $\rho(x)$ is probability density for configuration $x$ then that already implies the integral $\int \rho(x)\,dx$ equals 1 and is conserved. – Ján Lalinský May 31 '15 at 13:37

I think the answer to your question is because it is difficult to work in the Schrodinger picture in QFT whilst maintaining a reasonable standard of logic. Let $|0\rangle_{H}$ be the vacuum state (ground state of the interacting Hamiltonian) in the Heisenberg picture. In general, the Schrodinger picture does not exist, so there is no Schrodinger picture time evolution, \begin{equation} |0(t)\rangle_{S} = e^{-itH_{S}}|0\rangle_{H} \end{equation} where $H_{S}$ is the Hamiltonian of the interacting theory. However, the Heisenberg picture is good so that operators make sense in the Heisenberg picture. \begin{equation} K(t)_{H}=e^{itH_{S}}K_{S}e^{-itH_{S}} \end{equation} At any time $t$, there is, presumably, a state $|0\rangle_{t}$ (which we can call the vacuum at $t$) which is annihilated by the Heisenberg picture annihilation operator evaluated at $t$, \begin{equation} a(t)_{H}|0\rangle_{t}=0 \end{equation} but, since the Schrodinger picture does not exist, the vacuums $|0\rangle_{t}$ and $|0\rangle_{t'}$ for two different times $t,t'$ are not related by a unitary operator. So, at time $t$, the states that make sense are of the form $K(t)_{H}|0\rangle_{t}$ and all physical questions that can be asked must be answered by these states. In order to answer physical questions, we need to get c-numbers and so the operator $K(t)_{H}$ has to be put into normally ordered form with the annihilation operators to the right. Now, any terms in the normally ordered $K(t)_{H}$ that contain annihilation operators will not contribute in the physical state $K(t)_{H}|0\rangle_{t}$; Dirac, Lectures on Quantum Field Theory, Belfer Graduate School of Science, 1966, page 148 calls them latent terms. Now, these latent terms are still in $K(t)_{H}$, so they may evolve to other terms at $t'$ which are no longer latent in $K(t')_{H}$ and so contribute to the physical state $K(t')_{H}|0\rangle_{t'}$. This means that we can normalize a physical state $K(t)_{H}|0\rangle_{t}$ at one time $t$ but $K(t')_{H}|0\rangle_{t'}$ won't necessarily stay normalized and so it's not possible to logically get a notion of conserved probability.

I would say that the answer to your question relies in the construction of the $S$-matrix. When constructing the $S$-matrix, one requires that it is a unitary operator, hence one that conserves probability. Now, $S$-matrix is a typical object of the Interaction Picture, in which QFT and QED are mainly developed by physicists, so it is in some way "folklore". Indeed, on a formal level Haag's theorem states that, in four-spacetime dimensions, the transformation carrying from the Schroedinger to the interaction picture is not unitary, hence the two pictures are not equivalent. Obviously, one can't keep Haag's theorem on the some foot of a "true" mathematical theorem, since its hypotesis are based on our approximated description of the physical world and are subjected to our current knowledge. As far as I know, people working in QFT are inclined to consider it as the consequence of an over-formalization of the objects describing the theory.

I think that a formal presentation of such topics could be found in . An informal outilne can be found in  and .

 Bogoliubov et alii, "Axiomatic Quantum Field Theory"

 Mandl, Shaw, "Quantum Field Theory"

 Landau, Lifschitz, "Course of theoretical physics IV"

• Haag's theorem is a perfectly true mathematical theorem, and has to be taken into account. Long time ago, and in part by Haag himself, a scattering theory called Haag-Ruelle theory has been developed to take into account Haag's theorem. In addition, in Haag-Ruelle scattering the LSZ reduction formulas have been proven to hold (I think by Hepp). Therefore there is no problem to reconcile Haag's theorem with scattering theory; there are many unsolved problems, however, to construct a rigorous interacting quantum field theory in $3+1$ dimensions. – yuggib May 31 '15 at 14:49
• Sure, I absolutely agree with you about the fact it is rigorous, once the hypotesis has been accepted. What I meant is that a no-go theorem in a "perfectly" working (from experimental viewpoint) theory is usually viewed by physicits (at least, by those I've talked about this) as the consequence of a too strong formalization. They try to avoid "unnecessary" complications. In fact, e.g., LSZ formulas are usually taught in QFT courses and Haag-Ruelle theory is not. – user91126 May 31 '15 at 15:02
• Well the hypotheses on it are quite natural. And here is not a matter of introducing a complication, simply trying to understand what happens if we have a reasonable interacting QFT, and try to do perturbative scattering. It seems to me (at least in my experience) that physicists tend to overreact when they talk about Haag's theorem, as if the latter disrupts perturbative QFT. It simply tells that some care has to be taken, and that the naive picture is not probably the correct one (even if it leads to the correct results anyway). – yuggib May 31 '15 at 15:15
• If you, for example, take a quick look at Haag's theorem on this site (for what it is worth) you will see that many people react over it as if it is a very bad piece of mathematics in disagreement with experience, and therefore physically untrue. I think this is absolutely not the case...and that Haag's theorem is in some sense quite misunderstood. – yuggib May 31 '15 at 15:19
• Yeah, I really agree with your opinion. Indeed, e.g, if this was a post on math.se, I would never made that specification. However, this is physics.se, and it seems to me usefeul to spend some word, within my capabilities, about the relation between the physicists viewpoint and the mathematicians one. I think this is useful for a physics student, since he/she usually studies from books concerned only with "effective" theory and never from those, such as Bogoliubov, dealing with the rigorous theory, and he/she could be confused by the change of paradigm. – user91126 May 31 '15 at 15:39