# Lagrange Multiplier as chemical potential in Lagrangian Density

Within Matsubara formalism, we often add one chemical potential term to our Lagrangian density: $$\mu\phi^\dagger\phi$$, claiming that the chemical potential $$\mu$$ is one Lagrange multiplier. But at this stage, it doesn’t work that the equation of motion $$\delta\mathcal{L}/\delta\mu=0$$ of the chemical potential imposes the constraint, rather we want to impose: $$\int\phi^\dagger\phi=N.$$ For example we may consider Bogoljubov’s model of dilute Bosonic gas, equipped with $$U(1)$$ global internal symmetry, as discussed in the book by e.g. Schakel (section 2.2):

$$$$\mathcal{L} = \imath\hbar\phi^\dagger\partial_t\phi - \frac{\hbar^2}{2m}\nabla\phi^\dagger\nabla\phi+\mu\phi^\dagger\phi,$$$$

where we impose $$\int\phi^\dagger\phi=N$$, with $$N$$ being the total particle number. At this stage I would rather write the following Lagrangian: $$$$\mathcal{L} = \imath\hbar\phi^\dagger\partial_t\phi - \frac{\hbar^2}{2m}\nabla\phi^\dagger\nabla\phi+\mu(\phi^\dagger\phi-\rho),$$$$ with $$\int\rho = N$$, but certainly no one does that. I understand that in statistical mechanics, $$\mu$$ is indeed the Lagrange multiplier when we try to extremize the entropy. But I haven’t heard of functional representation of entropy, my question would be in what sense does the chemical potential act as a Lagrange multiplier in Matsubara formalism?

It seems that, the term to which the so called ‘Lagrange multiplier’ couples is constant of motion, but if we want to be extremely honest and say we want to extremize the action with respect to the ‘Lagrange multipliers’, then the constant of motion is set to zero. In some other part of theoretical physics, one would like to add non-dynamical auxiliary field to ensure the symmetry of a Lagrangian (e.g. BRST formalism), I also try to view them as extra constraints. This also partially constitutes my intuition behind the question.

• A Lagrangian multiplier is basically a barrier method trying to impose a constraint by increasing the cost when the constraint is violated (a penalty function). Not sure if that is what you mean as well?
– Emil
Commented Jun 29 at 7:59
• @Emil I totally agree, and I think that‘s the intuition behind such procedures. But if I want to argue that our Lagrangian completely specifies our system, the action is minimized when the total particle number is set to zero. I’ve adjusted my question further:).
– JinH
Commented Jun 29 at 8:12

So I think that this is just a misunderstanding of Lagrange multipliers. But maybe I am mistaken. I find looking back at the strict mathematical definition of the Lagrange multiplier for real valued functions is helpful. Of course this definition must be extended to functional spaces using the language of variations but I think the basic idea should be the same.

The definition is:

Let $$U \subset \mathbb{R}^{m}$$ be an open set, let $$f: U \rightarrow \mathbb{R}$$ and $$g = (g_1,\ldots,g_l): U \rightarrow \mathbb{R}^{l}$$, with $$l < m$$ be differentiable real valued functions, with g furthermore being of full rank, in other words the rank of $$g^{\prime}$$ is equal to $$l$$ for all $$x \in U$$. Let $$$$M = \{x \in U: g_j(x) = 0, j = 1, \ldots, l\}.$$$$ if $$f|_M$$ has a local extremum in $$x^{*} \in M$$, then there exists a $$\lambda^{*} = (\lambda_1^{*}, \ldots, \lambda_l^{*}) \in \mathbb{R}^{l}$$ with the following properties:

• The gradient of f is a linear combination of the gradient of the components of g: $$$$\nabla f(x^{*}) + \sum_{j = 1}^{l} \lambda_j^* \nabla g_j(x^*) = 0$$$$

• The function $$F: U \times \mathbb{R}^{l} \rightarrow \mathbb{R}$$, $$$$F(x, \lambda) := f(x) + \sum_{j = 1}^{l}\lambda_j^*g_j(x)$$$$

has a stationary point in $$(x^*, \lambda^*)$$, i.e. $$\nabla F(x^*, \lambda^*) = 0$$

In other words what this tells us is that well formulated functional constraints can be enforced by minimizing a modified function over the set of inputs and over the set of possible lagrange multipliers. If one were to naively extend this to a functional spaces, it seems like the expression of the constraint in the initial lagrangian is correct. I see that in your question you only consider $$\frac{\delta \mathcal{L}}{\delta \mu} = 0$$, but I would think a functional extension of this theorem would require $$(\frac{\delta \mathcal{L}}{\delta \mu}, \frac{\delta \mathcal{L}}{\delta \phi}) = (0,0)$$.

To directly answer your question I would say that the chemical potential is a lagrange multiplier by definition in our mathematical abstraction of the physical world. We don't have to associate the lagrange multiplier that we multiply with the constraint $$\int \phi^\dagger \phi = N$$ with the chemical potential at all, it's functional relationship and effect on our physical formalism would be the same. That is, $$\mu$$ could be called $$\lambda$$, but regardless it would still be a unique value who's value is minimized at all stationary points in the set of points where the constraint is fulfilled, which is why WE IDENTIFY IT WITH the chemical potential.

Taking this one step further towards the example in your question and minimizing a lagrangian modified similarly by some constraints: $$$$\delta \int_C \left[L(\vec{q}, \dot{\vec{q}}, t) - \sum_j \lambda_j g_j(\vec{q})\right] = 0 \Leftrightarrow \frac{\partial L}{\partial q_k} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_k} - \sum_j \lambda_j \frac{\partial g_j}{\partial q_k} = 0$$$$

So there does seam to be an extension for functional spaces, where the gradient is now replaced with Euler-Lagrange equations. I believe it can be shown that $$\phi$$ and $$\phi^{\dagger}$$ are the generalized coordinates for your system, so the above maybe applies directly, but honestly a bit outside my scope of knowledge at this point. Hope this helps.

• Thanks for the answer, so the chemical potential is chosen as such that they minimize the action, but the term which couples to it in the initial Lagrangian does not impose any constraint? (I feel the function $g$ in your answer would correspond to $\phi^\dagger\phi - \rho$
– JinH
Commented Jun 29 at 11:17
• $(\mu, \phi, \phi^{\dagger})$ are all chosen to minimize the action AND to respect a constraint (Here N particles in the system). The theorem I wrote shows that modifying the action in the way that we do with the lagrange multipliers means that stationary points of the modified function are those where both the action is stationary and the constraint is respected (opposed to just minimizing the action). The $g$ would correspond to $\phi^{\dagger}\phi$ or $\phi^{\dagger}\phi - N$ if you want, that won't change the stationary points Commented Jun 29 at 11:28
• I had some notation mistakes that I just fixed Commented Jun 29 at 11:35
• although I still think the stationary point is obtained by minimizing the action, I agree with you that adding $N$ or not does not influence the stationary points. But that push me further in the direction of thinking my Lagrangian as written in the second form. In a sense, adding this extra term only modifies my partition function by a constant, which does not make a difference. I appreciate the discussion:)
– JinH
Commented Jun 29 at 11:42
• I think practically speaking you are right. I think we generally just tend to minimize the action, and don't have do anything in particular with the chemical potential, because we assume that there is some unique value for the chemical potential since we are describing a physical system. From a purely mathematical perspective, things are more complicated and more considerations have to be made. I am just thinking off the top of my head though. Commented Jun 29 at 12:38