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Normally supersymmetric quantum field theories have Lagrangians which are supersymmetric only on-shell, i.e. with the field equations imposed. In many cases this can be solved by introducing auxilary fields (field which don't carry dynamical degrees of freedom, i.e. which on-shell become a function of the other fields). However, there are cases where no such formulation is known, e.g. N=4 super-Yang-Mills in 4D.

Since the path integral is an integral over all field configurations, most of them off-shell, naively there is no reason for it to preserve the on-shell symmetry. Nevertheless the symmetry is preserved in the quantum theory.

Of course it is possible to avoid the problem by resorting to a "Hamiltonian" approach. That is, the space of on-shell field configurations is the phase space of the theory and it is (at least formally) possible to quantize it. However, one would like to have an understanding of the symmetries survival in a path integral approach. So:

How can we understand the presence of on-shell symmetry after quantization from a path integral point of view?

Dilaton
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    Dear @Squark, surely you may write $N=4$ in the $N=1$ superspace, making the $N=1$ subalgebra manifest even off-shell and even in the path integral, can't you? The path integral for the $N=1$ language is trivialy equivalent to the $N=0$ "in components" formulation – the only slightly nontrivial statement behind this assertion is that the measure flip including the aux. fields doesn't spoil SUSY. So in this sense, I think that SUSY is manifest even in the non-SUSY $N=0$ "in components" formalism of the path integral, off-shell. If you see some problems with this conclusion, tell me details. – Luboš Motl Nov 05 '11 at 19:16
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    The equivalence between N=1 and N=0 is by integrating over the auxiliary fields, as far as I see. Hence it is not quite manifest in the N=0 language. For N=4 you can sure use the N=1 superspace, moreover the GIKOS approach apparently allows making an N=3 sub supergroup manifest. However this doesn't prove the whole symmetry is preserved. –  Nov 05 '11 at 20:19
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    OK, I think I see the answer. Once you can prove the equivalence between N=0 and N=1 you can get N=4 by choosing different N=1 subsupergroups –  Nov 05 '11 at 20:27
  • Well, I need to chew a bit more on the N=4 SYM case, but, in the meanwhile, consider N=1 SYM in 10D. Already we have no off shell formulation and the N is minimal. –  Nov 05 '11 at 20:42
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    I was just wondering about the same question. On physicsforums somebody posted a citation from Sohnius: "This situation could in principle be disastrous for quantum corrections: there the fields must be taken off-shell, away from their classical paths through configuration space. What this does for on-shell supersymmetric theories is quite unclear Obviously, if there exists - even unknown to us - some unique off-shell version, there should be no problem. But what happens if the theory were intrinsically only on-shell supersymmetric (as in N=4 and N=8 supergravity) is unclear to date." Is this –  Jun 08 '12 at 17:47
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    @Moritz: The reason is that the SUSY variations are always reduced by products of equations of motion times a function of another independent field in the path integral, of a different Fermi-Bose character. In this case, the equations of motion are satisfied. This should put Sohnius's concern to rest. – Ron Maimon Jun 13 '12 at 17:33

4 Answers4

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The point of view you give of quantizing phase space is not so good. If you have a classical phase space, there will be more than one quantum analog, differing by commutator terms that vanish classically. I never found the idea that you start with a classical system particularly lucid--- the quantum system is defined by the path integral, and the answer must always be found in the path integral itself.

In 4d, as you said in the comments, the problem is sidestepped by using N=1 superspace plus the appropriate R symmetry, which rotate the different SUSY's into each other, ensuring that the full SUSY is there. In order for the theory to be SUSY, you just have to check that the theory is N=1 and that the R-symmetry is good. This is usually manifest, since the R symmmetry is simple global rotation of fields into each other. But this doesn't work in 10 or 11 dimensions.

A part of the answer suggested by Qmechanic, the Schwinger Dyson equation (Heisenberg equation of motion) is obeyed by the field operators, so the variation really does vanish as an operator, in a certain sense described precisely below. But this is not a complete answer because it isn't true that everything that vanishes classically when the equations of motion are satisfied also vanishes quantum mechanically, in the path integral. The reason is that the quantities in the path integral can be multiplied at the same point in space and time, and there are singularities of the coinciding operator products which make the naive equations of motion fail in exactly the type of expressions that occur in the action.

For a trivial example, consider the product of two X operators

$$\langle x(t) x(t')\rangle = 0 $$

In the trivial Euclidean free-particle (Brownian motion) 1-d (quantum mechanics, not field theory) path integral $S= \int \dot{x}^2$. The operator X obeys the equation of motion $\ddot{X}=0$, so the second derivative is zero, but of course it isn't at the coinciding point (or else propagator would vanish at all times with the reasonable boundary conditions). The correct second derivative of the correlation function is

$$ \langle \ddot{x}(t) x(t')\rangle = \delta(t-t')$$

The equation of motion holds except at coincident points. The same is true in field theories for the same reason--- the propagator is sourced at the point where the two operators coincide by a delta-function with unit amplitude (if the fields are canonically normalized).

Further, since $\ddot{X}=0$, you might think that the following transformation is a symmetry

$$ \dot x + \epsilon f(t)\ddot{X} $$

Since the variation of S to first order in $\epsilon$ is $\int \dot{x}\ddot{x}f(t)$. So naively, using the equation of motion $\ddot{x}=0$, you would naively find that this is a symmetry. But this is obvious nonsense--- if f is nonconstant (if f is constant, the transformation is a time translation) this is obviously not a symmetry of the action. since it is a time translation by a different amount at different times, and there is a unit of time defined by the diffusion.

The reason is that the $\ddot{x}$ is multiplied by a coinciding $\dot{x}$, and when there are coinciding quantities, the equations of motion are not necessarily satisfied. The actual variation in S is

$$ \int f(t) {d\over dt} {\dot{x}^2\over 2}$$

and you see that it is a perfect derivative if f is constant, assuming Stratonovich convention for time derivatives--- centered differences--- and this is the symmetry of time translation in this case), but it is not a symmetry if f is not constant.

When are equations of motion satisfied in a path integral?

When you have a path integral:

$$ \int e^{iS} D\phi $$

The integral is invariant under a shift of integration variables $\phi(x)+\delta\phi(x)$, where $\delta\phi$ is an arbitrary function of x, since each integral is separately translation invariant. There is no determinant for this transformation--- it's just a shift in the integration variable by a constant at each time.

The change in the integrand when you do the shift is, given by expanding the thing to first order.

$$ \int e^{iS(\phi+\delta\phi)} D\phi = \int e^{iS} D\phi + i\int (\int {\delta S\over \delta \phi(x)}\delta\phi(x) d^dx) e^{iS} D\phi $$

and from this, you learn that the equations of motion are satisfied. This is the correct demonstration of the equations of motion.

$$ {\delta S\over \delta\phi} = 0 $$

This is the equation of motion for the $\phi$ field only, only one field. Now suppose you have some other insertions (composite operators $O_k$):

$$ \int O_1(x_1) O_2(x_2) ... O_n(x_n) e^{iS} D\phi $$

Now doing the same shift, to first order, you find

$$ \int (O_1 +\delta O_1(x_1)) ... (O_n+\delta O_n(x_n)) e^{iS} (1+i\delta S) D\phi = \int e^{iS} D\phi $$

So you find that the equation of motion for the one field $\phi$ is satisfied even with composite operator insertions except at coinciding points:

$$ -i\, {\delta S\over \delta \phi(x)} \, O_{1}(x_1) ... O_{n}(x_n) = {\delta O_1 (x_1)\over \delta \phi(x)} O_2(x_2) ... O_n(x_n) + O_1(x_1){\delta O_2(x_2)\over \delta\phi(x)}...O_n(x_n) + ... + O_1(x_1)O_2(x_2) ... O_{n-1}(x_{n-1}){\delta O_n(x_n)\over \delta \phi(x)}.$$

This is the operator equation for the equation of motion multiplying arbitrary expressions of other fields. The expression is known as the Schwinger-Dyson equation.

It is not just true as a vacuum expectation value, because it is satisfied independent of the path integral boundary conditions, which don't have to be vacuum, and you don't need to do a path integral from time minus infinity to time infinity, the identity holds in an arbitrary state.

This Schwinger-Dyson argument for insertions of any type with lots of fields running around implies the following rules for using the equation of motion:

  1. The equations of motion derived from varying $\phi$ are satisfied as operator equations away from insertions.
  2. the equation of motion derived from varying $\phi$ are satisfied even at cross-insertions of other fundamental fields (independent path-integrations) other than $\phi$.
  3. the equations of motion fail only at insertions of composite or elementary operators which vary when you shift the field whose equation of motion you are talking about, and the failure is of the ward-identity type.

So it isn't true that anything that vanishes classically has no effect. But when you have a product of the equation of motion multiplying functions of fields which are different from the one you vary to get the equation of motion, these are still identically zero.

In the SUSY case, your variation involves products of fields with their SUSY partners, which are independent integration variables. The equations of motion are satisfied in the products that you reduce, so using the equations of motion in the SUSY closure is justified, but you check it theory by theory by hand, by looking to see which equation of motion you use, and what it is multiplying.

The interpretation is that when you do the SUSY transformation, you also have to slide the fields around a little bit in a measure preserving way to finish the transformation.

Ron Maimon
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  • I liked your answer, but after a second reading I realized that the notation didn't make sense or was confusing. I fixed it. I hope you don't mind. – Diego Mazón Jun 26 '13 at 01:35
  • @drake: Reading the corrections--- I want to make sure the main point of this is clear: the equation of motion obeys a ward identity with other insertions, so that the product of the phi equation of motion times an operator is equal to zero away from the collision, and is equal to the variation of that operator under a shift in phi at the point of the collision (times a delta function). So the product stays identically zero when the fields are independent variables, or when you have a composite operator that doesn't vary when you vary phi, which happens to be the case for the SUSY closures. – Ron Maimon Jun 26 '13 at 15:17
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    I edited your answer again because the last equation was still wrong, even dimensionally wrong, and there is not enough space in these comments to insert that equation. What you say is totally right, but not the expression. You should also say, if you don't like expectation values, that your fields are in the Heisenberg picture. – Diego Mazón Jun 26 '13 at 21:21
  • And the equation I've just written is the same equation as I wrote times $\delta \phi_i$ and summed over $i$ (and without kets). – Diego Mazón Jun 26 '13 at 21:34
  • I think you cannot see my last edit (6 hours ago) since it hasn't been peer reviewed (?). In your last version of the answer (about 9 hours ago), the right hand side of the last equation was wrong because there were fields missing in each term. I did understand your notation in the LHS $\delta S$ and in my last edit I defined it. Thanks. – Diego Mazón Jun 27 '13 at 03:54
  • In my first edit, which was totally correct, I essentially improved and cleaned up the notation, wrote expectation values to be consistent with your quantum mechanical example (although their absence is not a mistake at all), and I did appropriately correct your last equation. I made a second edit to change again your last equation that was WRONG and after your edit is still WRONG, even dimensionally WRONG, and you aren't able or don't want to see it. – Diego Mazón Jun 27 '13 at 19:34
  • Right now, I'm seeing my 2nd edit in which on the top says: "Thanks for your edit! This edit will be visible only to you until it is peer reviewed." However, you must be so conceited that you cannot accept you are WRONG and that I did make a 2nd edit even though you cannot see it. – Diego Mazón Jun 27 '13 at 19:37
  • I wrote an answer in order for you to see it. – Diego Mazón Jun 27 '13 at 19:48
  • But I understood where the confusion came from--- you thought the $\phi_k$'s were other fundamental fields in the Lagrangian--- not necessarily (and the relation would be trivial then). The $\phi_k$'s are just arbitrary composite local operators, and the right hand side has the variational derivative. Since it is not clear to you yet (and you know your stuff) I changed the notation, and fixed the "dimensional inconsistency" (which just means you didn't like it that I factored out the stupid delta function without changing the notation for $\delta \phi$, ok fixed) – Ron Maimon Jun 28 '13 at 00:49
  • But even in that case, isn't your last eq. wrong? Aren't you missing in, for example, the first term of the RHS $O_2(x_2)...O(x_n)$ and the same thing in the other terms? Really?! – Diego Mazón Jun 28 '13 at 01:01
  • @drake: Yes! You are right, of course, stupid me. I just noticed this too, I am sorry. – Ron Maimon Jun 28 '13 at 01:06
  • Hallelujah!! (No worries) You didn't check the dimensions of your eq. in the whole discussion, did you? – Diego Mazón Jun 28 '13 at 01:15
  • @drake: I just noticed when I did the example that I was missing the other O's, it was dimensionally wrong too. I give up on edits, the editor is impossibly slow (every keystroke takes 10 seconds to register, and the time gets longer and longer). The point is: equation of motion for the $\phi$ field, times the other operators, is only nonzero at the collisions, where the contribution is the variation of these operators under the variation of $\phi$, I did it from memory, I made a botchy notation mess, but I really understand this, and it was annoying that the fixes were changing the meaning. – Ron Maimon Jun 28 '13 at 01:15
  • @drake: I understand what you are saying, I agree with you on the physics, and the content you gave was fine, although not written the way or with the meaning I intended at all. I am sorry for sniping at you, I got annoyed. Thanks for making the answer nice, and good luck. – Ron Maimon Jun 28 '13 at 02:50
  • A triviality: one can always exponentiate the determinant $\text{det}, a=\exp{(\text{tr}\ln{a})}$ so that it can be considered part of $\delta \mathcal {L}$ so the equations work as they're written. Now I'm wondering if a quantum theory can have a symmetry that its classical limit hasn't. I mean, if the Jacobian in the path integral meassure can compensate the variation in the classical action. – Diego Mazón Jun 29 '13 at 03:56
  • It's not possible since the effective $\delta \mathcal L$ that comes from the Jacobian doesn't have real part. – Diego Mazón Jun 29 '13 at 04:05
  • @drake: Yes, I know about the reexponentiation, it doesn't matter for the case in the question where you are only shifting fields by other independent fields, since then the determinant is one. The problem with compensation is that the classical variation is order 1, while the determinant variation is order $\hbar$, so this only can happen if there is no classical limit. Maybe 2d Gross-Neveu type models (four Fermi), where Bosonization can reveal new symmetry, but in this case it's not going to be a determinant compensation either, but a symmetry which is only evident in one formulation. – Ron Maimon Jun 29 '13 at 22:13
  • @drake: Oh, I see what you mean now about real part--- this is not necessarily fatal, you can transform the fields with an i too so that the determinant is non-real, but then perhaps there is a field redefinition in real and imaginary parts so that it becomes fields shifted by independent fields. Yes, it's an issue. The Lagrangian becomes imaginary. But perhaps one can see it as a smooth deformation to get to PT quantum mechanics? I don't know. – Ron Maimon Jun 29 '13 at 22:19
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How can we understand the presence of on-shell symmetry after quantization from a path integral point of view?

One can derive a Schwinger-Dyson equation associated with the current conservation, also known as a Ward identity; see e.g. Peskin and Schroeder, An Introduction to Quantum Field Theory, Section 9.6; or Srednicki, Quantum Field Theory, Chapter 22.

Qmechanic
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  • I don't have access to the book. How do you derive it using the path integral if the symmetry only exists on shell? –  Nov 05 '11 at 20:43
  • @Squark: Srednicki's book (up to some small changes) is available on his webpage. – Simon Nov 05 '11 at 22:01
  • OK, and where is the answer to my question in Srednicki's book? In fact it seems to me he doesn't go beyond N=1 4D supersymmetry –  Nov 05 '11 at 23:20
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    I interpreted the question (v1) as asking about on-shell symmetry on general grounds, cf. the title (v1). The references do not have any specific mentioning of $N=4$ 4D supersymmetry. – Qmechanic Nov 06 '11 at 00:26
  • OK. So Peskin and Schroeder have a treatment of on-shell symmetry on general grounds? Which example they consider if not supersymmetry? Also, I still have no access to the book. –  Nov 11 '11 at 20:46
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Quantum field theoretic analog of the classical equations of motion (EOM):

$$ -i\, \delta S (x)\, \phi_{1}(x_1) ... \phi_{n}(x_n) = \sum _{j=1}^n \phi_{1}(x_1)... \delta \phi_j (x)\delta(x-x_j)... \phi_{n}(x_n)= \delta\phi_{1}(x_1)\delta(x-x_1)\phi_2(x_2)... \phi_{n}(x_n)+\phi_1(x_1)\delta\phi_2(x_2)\delta(x-x_2)\phi_3(x_3)...\phi_n(x_n)+...+\phi_1(x_1)...\phi_{n-1}(x_{n-1})\delta\phi_n(x_n)\delta(x-x_n)$$ where $$ \delta S\equiv \sum_{i=1}^n{\delta S\over \delta \phi_{i} (x)}\delta \phi_i(x) $$ that is, $${\delta S\over \delta \phi_{i} (x)}$$ is the Euler-Lagrange EOM for the field $\phi_{i} (x)$. Answer to comment: In case you have local operators $O_i (x_i)$, this is the equation:

$$ -i\, {\delta S (x)\over \delta\phi (x)} O_{1}(x_1) ... O_{n}(x_n) = {\delta O_{1}(x_1)\over \delta \phi(x)}O_2(x_2)... O_{n}(x_n)+...+O_1(x_1)...O_{n-1}(x_{n-1}){\delta O_n(x_n) \over \delta \phi (x)}$$

Diego Mazón
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  • While what you write is the Schwinger Dyson equation as it appears on Wikipedia, namely $\langle {\delta S\over\delta\phi}O\rangle = \langle {\delta O \over \delta \phi}\rangle$, trivially equivalent to my answer, I read the original articles of Schwinger and Dyson, and a lot of related literature, and please, for my sanity, could you tell me who writes this down first? I had to come up with it myself, and I didn't even know it was called Schwinger Dyson equation, SD was the simpler $\langle {\delta S\over \delta \phi} \rangle=0$ in all sources I ever read. I had to rederive it myself. – Ron Maimon Jun 28 '13 at 17:53
  • I am not claiming originality, but I want to know who writes this, because it just doesn't commonly appear, but it appears on Wikipedia. It's not the Ward identity (although the derivation is nearly identical), it's something else, involving the question of equation of motion times local operators, and this is something that confused the heck out of me for years until I figured it out. Is it in a textbook? I skimmed a few just now and didn't find it. Is it in some review paper? – Ron Maimon Jun 28 '13 at 17:55
  • Once one realizes ${\delta S\over \delta \phi_i}\delta\phi i=\delta \mathcal{L}+\partial{\mu}J^{\mu} $, Ward identity follows trivially from Schwinger-Dyson, and reciprocally. I don't know who was the first, I'd say it was Schwinger, but I am not certain. ${\delta S\over \delta \phi}=0$ (in Heisenberg picture) is Ehrenfest theorem. I think Schwinger-Dyson is in most books under the name of Schwinger-Dyson, Dyson (in condensed matter oriented books) or Ward. I'd be surprised if it is not in Itzykson's, Weinberg's or Bogoliubov's, for example. @RonMaimon – Diego Mazón Jun 28 '13 at 22:23
  • The Ward identity is doing an infinitesimal symmetry transformation, that's what the $\delta \phi_i$ are in your comment, while the identity I derived is doing a shift of the field $\phi$ (not a symmetry). In papers and books, they always derive the Schwinger dyson equation as the operator equation ${\delta S\over \delta \phi} =0$, where $S = \int L(\phi) + J \phi$, including a source term J, and this is sufficient to derive perturbation theory (like Schwinger's/Dyson's papers), but it's not enough to say exactly when the equation of motion works with a coinciding insertion. – Ron Maimon Jun 28 '13 at 22:38
  • Why I am mystified: the Ward identity is a special sort of Schwinger Dyson equation, for shifting the fields by a symmetry, while the equation of motion identity is for just shifting the fields. The identity for equation of motion times a local operator is what I never saw in books, so I had to do myself. I see the transformations involved, they are similar but not the same. I am taking your word for it that this is called "Schwinger Dyson equation", but I am wondering why I never saw the identity for ${\delta S\over \delta\phi}O$ anywhere (I read the books you gave, no). Sohnius didn't too. – Ron Maimon Jun 28 '13 at 22:43
  • I have seen a million people ask this question about SUSY closure, why it is justified to use eqns of motion. I have seen people give glib or wrong answers, or claim it's a Ward identity. But I never saw the actual answer you have above, not this identity. The derivation of the Ward identity and the Schwinger Dyson equation with coinciding insertions are exactly the same, but the field transformation is of a different kind. That's why I am asking for a source, because I checked to see if it appears in Schwinger and Dyson's papers, no, and 6 reviews, no. There SD is just Eherenfest/Heisenberg. – Ron Maimon Jun 28 '13 at 22:49
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    @RonMaimon. In the first identity I wrote in my previous comment, $\delta \phi$ is a general transformation and $J^{\mu}$ is the "formal" Noether current (please, see the identity). In the particular case of a symmetry $\delta \mathcal{L}$ is zero or a total derivative that may be included in the current. With this, Ward is the same as Schwinger-Dyson. And I'm (almost) sure that those books talk about contact or Schwinger's terms more or less explicitly. – Diego Mazón Jun 28 '13 at 22:52
  • Excellent! Thanks. I got it. The formal current thing is the idea that I was missing. – Ron Maimon Jun 28 '13 at 22:54
  • Oh! All we are writing in this question obviously assumes that the measure is invariant. Otherwise one would have an additional term coming from the determinant. I think we agree. – Diego Mazón Jun 28 '13 at 23:14
  • Your Ward identity formula is not really general. what should the Ward identity be for the formal transformation on $S = \int |\partial \phi|^2 + |\partial \eta|^2 + \phi^2 \eta^2 d^4x$ under the formal non-symmetry transformation $\delta \phi = \epsilon \phi \eta$, $\delta \eta = - \epsilon \eta $? The reason I ask is because the general formula is false for this case, as the transformation is not a field shift, but a field-value dependent field shift. so there is a determinant in the measure. – Ron Maimon Jun 28 '13 at 23:16
  • @RonMaimon I wrote the answer before you ask the question... just kidding – Diego Mazón Jun 28 '13 at 23:17
  • Yes, we agree... (I edited the comment too late, hence time travel!) – Ron Maimon Jun 28 '13 at 23:23
  • If you wish to continue this discussion, please do so in a new [chat] room. Thank you. – Manishearth Jun 29 '13 at 02:39
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First of all there is an error in the Schwinger-Dyson equation that was derived in the previous post. There will be n-1 fields on the right-hand side. In any event, the SD equations don't save you because the supercurrent has fields sitting at the same point, so these contact terms come in and spoil the equations of motion. I.e., you get the equation of motion when you differentiate the supercurrent, but it is multiplied by fields at the same points, so all these delta functions come in. Try it at home, you'll see what I mean.

user20602
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