Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This must be the first question everyone asks when starting to study field theory at finite density and zero temperature.

To introduce a finite density one adds a Lagrange multiplier which fixes the number of particles in a non-relativistic field theory or the number of particles minus the number of antiparticles in a relativistic field theories. To be concrete, in a non-relativistic field theory one has the Lagrangian density: $$\mathcal L=i\psi ^*\partial _t \psi -{1\over 2m}\nabla\psi ^*\nabla\psi - V[\psi ^*\psi]$$ and then the same theory at finite density is described by: $$\mathcal L=i\psi ^*\partial _t \psi -{1\over 2m}\nabla\psi ^*\nabla\psi - V[\psi ^*\psi]+\mu \psi ^*\psi \,\,\,\,\,\,\,\, [1]$$ where $\mu$ is the chemical potential and the number of particles is $N=\int \psi ^*\psi$.

What disturbs me is that one may redefine the field: $$\chi\equiv e^{-i\mu t}\psi$$

with $t$ the time. And after replacing in [1]:

$$\mathcal L=i\chi ^*\partial _t \chi -{1\over 2m}\nabla\chi ^*\nabla\chi - V[\chi ^*\chi]$$ The "finite density term" disappears. Moreover, the path integral measure is invariant under this trivial field redefinition so that the finite density QFT seems equivalent to the zero density one. (Those who are familiarized with the non-relativistic limit of relativistic equation —such as Klein-Gordon or Dirac— will recognize this term and field redefinition.)

Something similar happens in the relativistic case.

What am I overlooking?

(Note that the theory is at zero temperature so I am in real time formalism and I don't have to impose periodic boundary conditions (I think). I am interested in time evolution, scattering, etc.)

Reply to Ron's answer:

(I here will overlook Thomas' comments, I will answer him tomorrow.)

Ron makes an interesting point, but I think his answer is not correct. The Hamiltonian density of the theory in the original fields is:$$\mathcal H={1\over 2m}\nabla\psi ^*\nabla\psi + V[\psi ^*\psi]-\mu \psi ^*\psi $$ What he correctly pointed out is that the Hamiltonian in the new fields is not just the previous Hamiltonian written in the new fields. This is right and in classical mechanics it is well-know that one has to add a piece to the Hamiltonian when one makes an explicit time dependent transformation. He pointed out the case of quantum mechanics, and the field theoretic version is: $$i\partial _t \, \chi =i\partial _t \, (e^{-i\mu t}\psi )=i(-i) \mu \, e^{-i\mu t}\psi + e^{-i\mu t}[\psi , H]=[e^{-i\mu t}\psi ,\, \mu N + H] $$ And therefore the Hamiltonian for the field $\chi$ is: $$H'=\mu N + H=\int {1\over 2m}\nabla\chi ^*\nabla\chi + V[\chi ^*\chi]$$ This is of course the Hamiltonian corresponding to the Lagrangian written in terms of $\chi$. In this way, one sees how the "standard" theory (without the $\mu$ term) is recovered in the Hamiltonian formalism as it was in path integral formalism. And this makes a lot of sense because the theory with $\mu = -m$ has to be equivalent to the theory with $\mu=0$ because we are just subtracting the rest energy.

Of course, this is true as long as one does not impose periodic boundary conditions.

share|cite|improve this question

The transformation you made is well known in quantum mechanics--- it is the time-dependent phase transformation that corresponds to the freedom to add a constant to the Hamiltonian. The freedom to make this redefinition is the statement that the zero point of energy is arbitrary.

In order to do it correctly, you have to shift all energies. The chemical potential term is giving you an energy cost per extra particle, and you need to shift it by a constant in order to make this transformation give the same physics. Once you do this, the system looks exactly the same.

Your change of phase only sort-of works relativistically, you have to pick a frame to do the phase change.

share|cite|improve this answer
So setting aside the relativistic issue, your answer is "you are not overlooking anything, one can throw away the density effect trough a field redefinition". Really? It does not make much sense physically, right? Or am I misunderstanding you? – Diego Mazón Aug 21 '12 at 5:24
@drake: No--- the consistent transformation both does the phase rotation and adds a constant to the chemical potential. The transformed Hamiltonian is exactly the same, it's a null transformation. – Ron Maimon Aug 21 '12 at 5:34
What's the relationship between your answer and Thomas'? – Ryan Thorngren Aug 21 '12 at 5:53
@RonMaimon I see (+1)... it is the quantum counterpart of the time derivative of the generating function (because the transformation is time dependent), right? And I guest that Thomas and you are saying the same thing, right? – Diego Mazón Aug 21 '12 at 6:14
@drake: I couldn't figure out what Thomas is saying exactly, although each of the statements taken individually are probably right, I don't think he resolved the issue fully. He never says explicitly that when you make a phase change on the two ends an up-in-time link of the lattice, you need to add the same constant phase shift to the chemical potential to keep the path integral the same. I just know this from the fact that the chemical potential is an energy per unit particle, and all energies are shifted when you do a time dependent phase rotation. – Ron Maimon Aug 21 '12 at 6:28

This is indeed a little more subtle (You can also see the problem by realizing that the free fermion propagator only depends on $p_0-\mu$, so that integrals over $p_0$ are $\mu$ independent, unless $\mu$ enters into the $i\epsilon$ prescription). There are basically two ways to analyze this more carefully: 1) Start from the path integral representation of the partition function. This means that we do have to go to euclidean space and impose boundary conditions on $\psi$. We find that the path integral contains convergence factors that depend on $\mu$, which turn into $i\epsilon$ prescriptions. See, for example, the text book by Negele and Orland. This can also be seen by carefully discretizing the theory on the lattice. We find that $\mu$ enters as a gauge field, that means as a $\exp(\pm \mu a)$ factor for forward/backward going links with lattice spacing $a$ (see, for example, arXiv:hep-lat/0308016). 2) Use canonical quantization and start from the fact that the ground state of the non-interacting theory is a filled Fermi sphere. This is described in Fetter and Walecka or Abrikosov, Gorkov, Dzyaloshinskii. In the end we get the same $\mu$ dependent $i\epsilon$ prescription.

share|cite|improve this answer
This makes sense to me... I have to work it out. Now I see that the vacuum wave functional may be different and this may imply different $i\epsilon$ terms. One question: Why does one have to impose periodic boundary condition if the temperature is zero? – Diego Mazón Aug 21 '12 at 5:39
Your answer seems to me very interesting, could you please elaborate it a little bit more and be more explicit. Also, it would help me if you said which sections of the books you cited are the most relevant for this question. Thanks. – Diego Mazón Aug 21 '12 at 20:18
In order to derive a path integral representation you have to start from the finite temperature partition function (we need the trace of an exponential so that we can write it as an imaginary time evolution operator), and take the T->0 limit in the end. There are many details that are explained in standard text books, for example chapter 2,3 of Negele and Orland. – Thomas Aug 22 '12 at 1:05
Even if I'm not interested in statistical mechanism but in particle physics at finite density (scattering, etc)? I could not get Negele and Orland, but I have Walecka et al and Gorkov et. al. Thank you. – Diego Mazón Aug 22 '12 at 1:20
One more remark: The transformation you consider is a gauge transformation. $\mu$ enters as like the zero'th component of a gauge field, and this is why it appears that $\mu$ can be gauged away. In a theory with a U(1) gauge field this is indeed correct, there is no dependence on $\mu$. – Thomas Aug 22 '12 at 1:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.