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Whenever one needs to calculate correlation functions in QFT using perturbations one encounters the following expression:

$\langle 0| some\ operators \times \exp(iS_{(t)}) |0\rangle$

where, depending on the textbook S is either (up to a sign)

  1. $\int \mathcal{L}dt$ where $\mathcal{L}$ is the interaction Lagrangian


  2. $-\int \mathcal{H}dt$ where $\mathcal{H}$ is the interaction Hamiltonian.

It is straightforward to prove that if you do not have time derivatives in the interaction terms these two expressions are equivalent. However these expressions are derived through different approaches and I can not explain from first principles why (and when) are they giving the same answer.

Result 1 comes from the path-integral approach where we start with a Lagrangian and do perturbation with respect to the action which is the integral of the Lagrangian. Roughly, the exponential is the probability amplitude of the trajectory.

Result 2 comes from the approach taught in QFT 101: Starting from the Schrödinger equation, we guess relativistic generalizations (Dirac and Klein-Gordon) and we guess the commutation relations to be used for second quantization. Then we proceed to the usual perturbation theory in the interaction picture. Roughly, the exponential is the time evolution operator.

Why and when are the results the same? Why and when the probability amplitude from the path integral approach is roughly the same thing as the time evolution operator?

Or restated one more time: Why the point of view where the exponential is a probability amplitude and the point of view where the exponential is the evolution operator give the same results?

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There is a change in sign either in 1 or in 2. – juanrga Dec 4 '12 at 21:17
up vote 2 down vote accepted

Starting from the Hamiltonian formulation of QM one can derive the path-integral formalism (see chapter 9 in Weinberg's QFT volume 1), where the Hamiltonian action is found to be proportional to $\int \mathrm{d}t (pv - H)$.

For a subclass of theories with "a Hamiltonian that is quadratic in the momenta" (see section "9.3 Lagrangian Version of the Path-Integral formula" in above textbook), the term $(pv - H)$ can be transformed into a Lagrangian $L_H = (pv - H)$. Then the Lagrangian action is proportional to $\int \mathrm{d}t L_H$. Both actions give the same results because one is exactly equivalent to (and derived from) the other.

$$ \int \mathrm{d}t (pv - H) = \int \mathrm{d}t L_H$$

Moreover, when working in the interaction representation you do not use the total Hamiltonian but only the interaction. The derivation of the Hamiltonian action is the same, except that now the total Hamiltonian is substituted by the interaction Hamiltonian $V$. Again you have two equivalent forms of write the action either in Hamiltonian or Lagrangian form.

If you consider Hamiltonians whose interaction $V$ does not depend on the momenta, then the $pv$ term vanishes and the above equivalence between the actions reduces to

$$ - \int \mathrm{d}t V = \int \mathrm{d}t L_V$$

where, evidently, the interaction Lagrangian is $L_V = -V$

This is what happens for instance in QED, where the interaction $V$ depends on both position and Dirac $\alpha$ but not on momenta.

Note: There is a sign mistake in your post. I cannot edit because is less than 10 characters and I have noticed the mistake in a comment to you above, but it remains.

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Isn't the exponential in form 1) also an operator? Then it is not really the probability amplitude of a trajectory, as you claim. Both forms seem to be derived from the same origin, namely the Hamiltonian formalism, and both exponentials have the status of the time evolution operator in the interaction picture. In the path-integral approach the expression would be the formally quite different beast: $$\langle 0|\; \text{some operators}\;e^{i S[\hat{\phi}]} |0\rangle = \int D \phi \; \text{some functions (fields)} \; e^{i S[\phi]},$$ where now the exponential on the right hand side is a c-number that indeed can be interpreted as the amplitude of a field configuration $\phi(x)$. The integral is a sum over all such weighted field configurations.

If you want to see quickly why this should be the same as your expression, then remember that the ground state $|0\rangle$ is not an eigenstate of the evolution operator or the field operators, in general. It can, however, be expanded as a sum of field eigenstates (coherent states). You can then see immediately that the expansion will give a sum of c-number contributions, which also represent field configurations weighted by an exponential function. If you want to actually do this properly you will have to do some other stuff like discretise space-time and insert complete sets of field eigenstates at each point. This is covered in most field theory texts, e.g. Peskin & Schroeder, or Altland & Simons.

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thanks for the corrections, however they do not change my question: Why the point of view where the exponential is a probability amplitude and the point of view where the exponential is the evolution operator give the same results? – Krastanov Dec 4 '12 at 20:15
Well, if you really want to understand why they give the same results, you have to just go through the derivation, as with anything else in mathematical physics. What I was trying to say is that both exponentials ultimately represent amplitudes weighting possible field configurations. In order to calculate transition probabilities, expectation values or whatever, you always sum over the amplitudes for indistinguishable paths between initial and final states. – Mark Mitchison Dec 4 '12 at 21:53
(contd.) In the path integral formalism, this sum is explicit, and we generally speak of a dominant contribution coming from the classical stationary path $\delta S/\delta \phi = 0$, with quantum fluctuations around the stationary path suppressed by a factor $e^{i S/\hbar}$. In the Hamiltonian formalism the sum is kind of 'hidden' inside the operators, and appears because non-zero commutation relations between operators mean that the interesting states (e.g. the ground state, particle number eigenstates...) will not be eigenstates of the interaction Hamiltonian/Lagrangian. – Mark Mitchison Dec 4 '12 at 22:01
(contd.) So in the operator formalism, it is instead commutation relations that imply fluctuations. But the physical meaning is the same. I like to think of operators as algebraic book-keeping devices that reproduce results of the path-integral formalism. But this is all horribly vague and imprecise, and will probably irritate some other user of this site who has a different personal take on the meaning of the formalism. If you want to understand for yourself, you really need to go through the derivation from one formalism to the other in, say, one of the references I gave. – Mark Mitchison Dec 4 '12 at 22:06

Well, another reason why this may be the case is that the action is the same thing in both cases. You either start with the action $S = \int L$, or you start with the so-called phase space action of the Hamiltonian $S = \int p{\dot q} - H$. Given the definition of the Hamiltonian, it should be clear that these two expressions are formally identical if you have an invertible mapping from the set $(p, q) \rightarrow (q, {\dot q})$. Since interaction terms typically don't involve the time derivatives of the configuration variables (though there are exceptions), it's hardly surprising that the part of the formalism that involves only interactions will come out nearly identically, formally.

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I do not get the part about starting from the "phase space action". I do not think that I have ever done something like that (I am a novice.). – Krastanov Dec 3 '12 at 20:57
@Krastanov: Start with the "phase space action" I put up above, and vary it with respect to $p$ and $q$. The result will be that you get Hamilton's equations of motion out. It's just a cute way of making the Hamiltonian formalism look more "principle of least action"-y. – Jerry Schirmer Dec 3 '12 at 22:17

From $L=H_0-V$ and $H=H_0+V$ one sees that (for simple enough theories) the Lagrangian interaction and the Hamiltonian interaction differ only by a sign.

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what conditions must hold a theory to be considered simple enough for both formulations be equivalent? – lurscher Dec 3 '12 at 20:25
I think that you restated the naive reason that I already gave for why all this works. However I do not see why it is true from first principles. Sorry for the vagueness of the comment. – Krastanov Dec 3 '12 at 20:55
@Krastanov: there are no firster principles for this. For a general Lagrangian not of the form $L=T-V$, the equivalence is not true. – Arnold Neumaier Dec 4 '12 at 7:57

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