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Many books have described the path integral for non-relativistic quantum. For example, how to get the Schrödinger equation from the path integral. But no one told us the relativistic version. In fact, the relativistic version is impossible to be perfect, it must be replaced by quantum field theory. But why?

The answer I want is not that the quantum mechanics will give us a negative energy or negative probability. We need a answer to explain why non-relativistic Lagrangian $$L=\frac{p^2}{2m}$$ can lead a correct Schrödinger equation? Why if we replace it by relativistic Lagrangian $$L=-mc^2\sqrt{1-v^2/c^2},$$ we can not obtain any useful information? So how can we think of the correctness of the non-relativistic path integral?

And how a quantum field can give a relativistic quantum theory?

Here give us a simple answer Special relativity version of Feynman's "Space-Time Approach to Non-Relativistic Quantum Mechanics" but it is useless. It doesn't tell us how to develop it step by step in mathematical frame and physical frame...

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I think Kasper Peeters' answer to the linked question is what you are asking for. The relativistic worldline path integral you want is equation 1.7 of the linked article. If I'm misunderstanding the question, how is this inadequate? –  Michael Brown Jul 12 '13 at 3:18
    
Thanks a lot. These articles are especially important and useful. Give me some time to understand them. –  Jianlin Su Jul 12 '13 at 3:41
    
However, I find that eq 1.7 is what I can't understand mostly. How can we use $\tau$ as an independent parameter?It obviously depends on $x$ and $t$ ! As the least action princle said, the particle chose the pathes from which is throught the point $(x_1,t_1)$ and $(x_2,t_2)$,not –  Jianlin Su Jul 12 '13 at 4:52
    
$\tau_1$ and $\tau_2$ !!! –  Jianlin Su Jul 12 '13 at 4:58
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$\tau$ is the parameter on the path: $x(\tau),t(\tau)$. The fact that it is also the proper time just means it obeys $d\tau^2 = \eta_{\mu\nu}dx^\mu dx^\nu$, which is no more a contradiction than using the arc length to parameterise an ordinary curve in space. Try doing the variational derivation of a straight line as the shortest path in flat space! Eq 1.7 is saying: for each possible total proper time $s$ do path integral of paths which start and end at $(x_1,t_1),(x_2,t_2)$ in proper time $s$, then add those up for all $s>0$. This includes all paths which go through the desired endpoints. –  Michael Brown Jul 12 '13 at 6:18

2 Answers 2

It is not quite true that we don't obtain any useful information. The relativistic particle action is indeed $$ S = -m_0c^2 \int dt_{\rm proper} $$ When you substitute your correct formula for $dt_{\rm proper}$ and Taylor expand the Lorentz factor in it, the integral has the factor of $dt_{\rm coordinate}(1-v^2/c^2+\dots)$. The first term proportional to $1$ is constant and the second term gives you the usual $mv^2/2$ part of the non-relativistic action.

At least formally, the relativistic action above may be used to deduce the propagators for a relativistic spinless particle – such as the Higgs boson. The addition of a spin isn't straightforward. However, the "proper time" action above may be easily generalized to the "proper area of a world sheet" action (times $-T$, the negative string tension) for string theory, the so-called Nambu-Goto action, and this action admits spin, interactions, many strings/particles, and is fully consistent. The "proper time" action is therefore the usual starting point to motivate the stringy actions (see e.g. Polchinski's String Theory, initial chapters).

The reason why a single relativistic particle isn't consistent without the whole machinery of quantum fields is a physical one and it may be seen in the operator formalism just like in the path integral formalism. Any valid formalism has to give the right answers to physical questions and the only right answer to the question whether a theory of interacting relativistic particles without particle production may be consistent is No.

In the path integral formalism, we could say that it brings extra subtleties to have a path integral with a square root such as $\sqrt{1-v^2/c^2}$. To know how to integrate such nonlinear functions in an infinite-dimensional functional integral, you have to do some substitutions to convert them to a Gaussian i.e. $\exp(-X^2)$ path integral.

This may be done by the introduction of an auxiliary time-like parameter along the world line, $\tau$, agreeing with the variable in the aforementioned paper. With a condition relating $\tau$ and $t_{\rm coordinate}$, it may be guaranteed that the new action in the $\tau$ language looks like $$ - m\int d\tau \,e(\tau) \left( \frac{d X^\mu}{d\tau}\cdot \frac{dX_\mu}{d\tau}\right) $$ which is nicely bilinear and the square root disappear. However, this clever substitution or any similar substitution has the effect of allowing the negative-energy solutions, too.

While the relativistic $p^2/2m$ is positive semidefinite, $m/\sqrt{1-v^2/c^2}$ can really have both signs. We may manually try to "forbid" the negative sign of the square root but this solution will always reappear whenever we try to define the path integral (or another piece of formalism) rigorously.

This implies that we have states with energy unbounded from below, an instability of the theory because the particle may roll down to minus infinity in energy. Alternatively, the squared norms of these negative-energy states may be (and, in fact, should be) taken to be negative, traded for the negative energy, which brings an even worse inconsistency: negative probabilities.

The only consistent way to deal with these negative-norm solutions is to "occupy" all the states with negative energies so that any change in the states with negative energy means to add a hole – an antiparticle such as the positron – whose energy is positive (above the physical vacuum) again. At least, this description (the "Dirac sea") is valid for fermions. For bosons, we use an approach that is the direct mathematical counterpart of the Dirac sea but only in some other variables.

It's important to realize that any attempt to ban the negative-energy solutions by hand will lead to an inconsistent theory. The consistent theory has to allow the antiparticles (which may be identical to the particles in some "totally real field" cases, however), and it must allow the particle-antiparticle pairs to be created and destroyed. It's really an inevitable consequence of the combination of assumptions "special relativity" plus "quantum mechanics". Quantum field theory is the class of minimal theories that obey both sets of principles; string theory is a bit more general one (and the only other known, aside from QFT, that does solve the constraints consistently).

Why quantum field theory predicts a theory equivalent to multibody relativistic particles (which are indistinguishable) is the #1 most basic derivation in each quantum field theory course. A quantum field is an infinite-dimensional harmonic oscillator and each raising operator $a^\dagger(\vec k)$ increases the energy (eigenvalue of the free Hamiltonian $H$) by $\hbar\omega$ which is calculable and when you calculate it, you simply get $+\sqrt{m_0^2+|\vec k|^2}$ in the $c=1$ units. So $a^\dagger(\vec k_1)\cdots a^\dagger(\vec k_n)|0\rangle$ may be identified with the basis vector $|\vec k_1,\dots,\vec k_n\rangle$ (anti)symmetrized over the momenta (with the right normalization factor added) in the usual multiparticle quantum mechanics. This has various aspects etc. that are taught in basic quantum field theory courses. If you don't understand something about those things, you should probably ask a more specific question about some step you don't understand. Quantum field theory courses often occupy several semesters so it's unproductive to try to preemptively answer every question you could have.

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+1 good answer; though it inspired me to ask another question: physics.stackexchange.com/questions/70836/… –  Michael Brown Jul 12 '13 at 7:28

The problem still exist: how could we convert them to a Gaussian path integral?

Let us just consider the classical situation, not the quantum. The action is $S=\int -mds$,$ds=\sqrt{dt^2 -dx^2},c=1$. It is no problem to rewrite it to

$S=\int -m[(\frac{dt}{ds})^2-(\frac{dx}{ds})^2]ds$

What I can't accept is how could we use $L=-m[(\frac{dt}{ds})^2-(\frac{dx}{ds})^2]$ and parameter $s$ to calculus variation ? Namely how can we do the following calculation?

$\frac{d}{ds}(\frac{\partial L}{\partial \frac{dx}{ds}})=\frac{\partial L}{\partial x},$ $\frac{d}{ds}(\frac{\partial L}{\partial \frac{dt}{ds}})=\frac{\partial L}{\partial t}$

It is obviously wrong, thought it may gives us correct answer. Because $ds$ depends on $dx$ and $dt$. while we get $\delta dx$ and $ \delta dt$, we also get a $\delta ds \neq 0$,$ds$ can not be constant.

For that, I can't understand how can we use the path integral with the parameter $s$?

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You need to add these questions to your main question as an edit, rather than putting them in the answers section. –  Larry Harson Jul 13 '13 at 13:13

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