This seems to be an absurd question, but bear with me.

In quantum field theory, the Dirac fermion mass Lagrangian term reads $$ m\bar\psi \psi = m(\bar\psi_L \psi_R + \bar\psi_R \psi_L) = m(\psi_L^\dagger \gamma_0\psi_R + \psi_R^\dagger \gamma_0\psi_L) $$ Assuming this fermion mass Lagrangian term is used as the integrand (in the exponential) in functional integral formalism of Dirac-fermion-related QFT, in the following we will regard $\psi$ components as anti-commuting Grassmann variables given the anti-symmetric nature of the fermions, rather than classical commuting complex variables.

Note that $m$ is supposed to be real throughout this post. The question here is about whether the mass Lagrangian term $m\bar\psi \psi$ is real or imaginary, not about $m$ parameter itself. (Note that there could be legitimate pseudoscalar mass term as in $m\bar{\psi} i\gamma_5\psi$. But we will not discuss about pseudoscalar mass in the current post. The interested readers can see here for more details.)

However, if you look under the hood of the familiar mass term $m\bar\psi \psi$, there are some surprises lurking around. Let's look at a simple example in the Weyl basis $$ \psi = (\xi, 0, \chi, 0)^T $$ where $\xi$ and $\chi$ are real Grassmann numbers ($\xi^*= \xi$, $\chi^*= \chi$, they are not 2 component columns, we will get to the general case of complex Grassmann numbers towards the end of the post). In the Weyl basis, $(\xi, 0, 0, 0)^T$ and $(0, 0, \chi, 0)^T$ represent left ($\psi_L$) and right ($\psi_R$) handed part of the Dirac spinor, respectively.

Let's calculate the mass term: $$ m\bar\psi \psi = m(\psi_L^\dagger \gamma_0\psi_R + \psi_R^\dagger \gamma_0\psi_L) = m(\xi^*\chi + \chi^*\xi) = m(\xi\chi + \chi\xi) = 0 $$ Oops, it's identical to zero since $\xi$ and $\chi$ are anti-commuting Grassmann numbers! Note that the $\gamma_0$ matrix just flip the left-handed components to the right-handed components and vice versa in the Weyl basis.

Now, let's tweak the the trial spinor to make one of its component imaginary Grassmann (multiplying $\chi$ by an $i$) $$ \psi = (\xi, 0, i\chi, 0)^T $$ Low and behold (I will omit the $m$ parameter hereafter): $$ \bar\psi \psi = \psi_L^\dagger \gamma_0\psi_R + \psi_R^\dagger \gamma_0\psi_L = \xi^*(i\chi) + (i\chi)^*\xi = i\xi\chi - i\chi\xi = 2i\xi\chi \neq 0 $$ The mass term is non-zero.

The remarkable and weird feature is that the mass term is imaginary with an $i$!

The interested reader can try all sorts of $\psi$ configurations in any representation (Weyl basis or not), and you will end up with the same result of imaginary mass. Any efforts of building a real mass term will be a wild goose chase, since the $\psi_L^\dagger \gamma_0\psi_R$ portion will always cancel out the $\psi_R^\dagger \gamma_0\psi_L$ portion.

Let's double check whether the imaginary mass term is Hermitian: $$ (2i\xi\chi)^\dagger = -2 i\chi^*\xi^* = -2i\chi\xi = 2i\xi\chi $$ So the Hermitian property is secured.

On the other hand, a real mass term (if it existed) $$ (2\xi\chi)^\dagger = 2 \chi^*\xi^* = 2\chi\xi = -2\xi\chi $$ would be non-Hermitian.

The key point here is that the Hermitian operator is by definition $$ (AB)^\dagger = B^\dagger A^\dagger $$ Note that there is NO minus sign, even if both $A$ and $B$ are Grassmann odd. (As a side note, transpose is defined as: $(AB)^T = -B^T A^T$, if $A$ and $B$ are Grassmann valued. Note that there is a minus sign! See here.)

At the end of the day, physicists seem not troubled by the imaginary nature of the mass term as long as it's Hermitian. I have to underscore (in response to @octonion's comments) that being Hermitian and being real are two disparate notions.

You might wonder why the imaginary mass is not mentioned in the usual text books. It's because when we deal fermions, the common practice is to use complex Grassmann numbers $$ \xi = \xi_1 + i\xi_2 \\ \chi = \chi_1 + i\chi_2 $$ where $\xi_1$, $\xi_2$, $\chi_1$, and $\chi_2$ are real Grassmann numbers

And thus the mass term of $\psi = (\xi, 0, \chi, 0)^T$ is $$ \bar\psi \psi = \psi_L^\dagger \gamma_0\psi_R + \psi_R^\dagger \gamma_0\psi_L = \xi^*\chi + \chi^*\xi $$ the imaginary nature is hidden in plain sight. Only when we write out the explicit terms $$ \xi^*\chi + \chi^*\xi = (\xi_1 + i\xi_2)^*(\chi_1 + i\chi_2) + (\chi_1 + i\chi_2)^*(\xi_1 + i\xi_2) = 2i(\xi_1\chi_2 + \chi_1\xi_2) $$ the imaginary mass is manifest.

Added note in response to the wrong answer below (by @alexarvanitakis) saying that "The signs and/or presence of factors of i in fermion lagrangians is somewhat superfluous and convention dependent".

Of course, one can have a Dirac Lagrangian with no $i$. For example, one can simply changing the metric from (+, -, -, -) to (-, +, +, +), see more details here. However, using what ever convention you choose, you still end up with a imaginary mass term! This is because that each column element of the Dirac wave function is valued in complex Grassmann space. Whereas a real Grassmann valued Dirac wave function will imply zero mass.

The wrong answer further says that "For example you can work with a convention where complex conjugation does not reverse the order of a product of fermions which changes radically the appearances of factors of i".

Please note that in deriving the imaginary mass above, there is no reverse invoked for any product of fermions. So the mass term is still imaginary. The only place the complex conjugation convention in reverse of a product of fermions is relevant is the proof that the mass term is Hermitian, albeit imaginary. If one takes the wrong answer's convention, the mass term will be both non-Hermitian and imaginary!

Additionally, the wrong answer says that "You instead want to look at the Klein-Gordon equation satisfied by the fermion field...You need to correlate the sign of $m^2$ with the convention for the gamma matrices so that the above operator does not admit tachyonic-type solutions."

I am talking about $m\bar\psi \psi$ being imaginary, not $m$ being imaginary. The wrong answer's proof of $m$ being real (or $m^2$ being positive) is totally irrelevant to the question here!

More added note:

I am not talking about whether the expectation value $\langle m\bar\psi \psi\rangle$ being real. That is not the problem. What I am talking about is the Lagrangian/action in integrand before path integral, while the expectation value is after path integral. Actually, even-though the Lagrangian mass term is imaginary, the path integral expectation value $\langle m\bar\psi \psi\rangle$ IS real. I am not contesting about the expectation value.

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    $\begingroup$ Hi @MadMax: This question is about classical (super-commuting) variables. The QM & QFT tags do not seem to apply. $\endgroup$
    – Qmechanic
    Commented Feb 6, 2020 at 19:33
  • 3
    $\begingroup$ Hi, we've noticed that you have made a large number of minor edits to this post. Please be mindful that every edit bumps the post in the "active" tab of the site and try to make your edits substantial. If you foresee improving this post repeatedly, maybe collect several edits and make them in one go instead of submitting them individually. $\endgroup$
    – Chris
    Commented Jan 26, 2021 at 6:06
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    $\begingroup$ @MadMax I'm trying to understand the question... You're asking whether the term $m\bar\psi\psi$ in $L$ is real or imaginary, in the context of the integrand of a path integral, right? You know that the components of $\psi$ and $\bar\psi$ are not numbers (real or complex), and products of those components are not numbers either. They're elements of an abstract vector space with a special kind of product. Do you have a basis-independent definition of complex conjugation in mind? Or are you asking what other people really mean when they say the Lagrangian must be "real"? $\endgroup$ Commented Jan 28, 2021 at 23:40
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    $\begingroup$ @MadMax Yes, that's what I meant. I should have said it like this: the components of the matrices $\psi$ and $\bar\psi$ are elements of a Grassmann algebra, not numbers. (They can be multiplied by complex numbers, of course.) $\endgroup$ Commented Jan 29, 2021 at 0:02
  • 1
    $\begingroup$ @MadMax "Grassmann number" is just the name for an element (or specifically a generator) of a Grassmann algebra. I searched briefly for a good concise online overview and didn't find one, but here's the rough idea: you have a list of things $\psi_k$ called "generators," some of which may be denoted $\bar\psi_k$ instead, but they're all linearly independent. They can be added, mutliplied by complex numbers, and multiplied by each other, and the generators anticommute with each other. [part 1 of 2...] $\endgroup$ Commented Jan 29, 2021 at 0:56

3 Answers 3


The signs and/or presence of factors of $i$ in fermion lagrangians is somewhat superfluous and convention dependent. (For example you can work with a convention where complex conjugation does not reverse the order of a product of fermions which changes radically the appearances of factors of $i$. Most of the time this is not a convention you want to use though.)

You instead want to look at the Klein-Gordon equation satisfied by the fermion field. Assume your fermion EOM is $$ (\gamma^\mu{\partial_\mu}+m)\psi=0 $$ Then it follows that $$ (\gamma^\mu{\partial_\mu}+m)^2\psi=0\implies \left(\frac{1}{2}\{\gamma^\mu,\gamma^\nu\}\partial_\mu\partial_\nu +m^2\right) \psi=0 $$ so the equation of motion for the fermion implies also a Klein-Gordon type equation for each of the components of $\psi$. You need to correlate the sign of $m^2$ with the convention for the gamma matrices so that the above operator does not admit tachyonic-type solutions.

  • $\begingroup$ This answer is totally wrong! See added notes in my question. $\endgroup$
    – MadMax
    Commented Jan 26, 2021 at 4:30
  • 1
    $\begingroup$ No, my answer looks fine. Naive reality of the kinetic term in the lagrangian doesn't matter; it only matters whether solutions propagate within the lightcone and whether the operator is formally self-adjoint and has a nice spectrum of (real) eigenvalues etc. $\endgroup$
    – user21299
    Commented Jan 26, 2021 at 22:43
  • $\begingroup$ $m$ is real for sure. But that is NOT what I am talking about. I am talking about $m\bar\psi \psi$. $\endgroup$
    – MadMax
    Commented Jan 26, 2021 at 22:46

I am going to consider the problem in $d=0+1$ dimensions to simplify the notation. For higher $d$ the discussion is identical except that you also include momentum modes, which are not relevant to the question.

Take a free Dirac fermion. The most general lagrangian is $$ L=i\psi^\dagger\dot\psi+m\psi^\dagger\psi $$ This lagrangian is hermitian iff $m\in\mathbb R$.

The associated hamiltonian, obtained by the usual (constrained) Legendre transform, is $$ H=m\psi^\dagger\psi $$ This is an hermitian operator, naturally.

The quantum theory is obtained by canonical quantization, giving the (again, constrained) Dirac-Poisson brackets $$ \{\psi,\psi\}=\{\psi^\dagger,\psi^\dagger\}=0,\qquad\{\psi,\psi^\dagger\}=1 $$

The Hilbert space is two-dimensional: $$ |0\rangle\quad\text{and}\quad\psi^\dagger|0\rangle $$ where the vacuum $|0\rangle$ is defined as $$ \psi|0\rangle=0 $$ It is trivial to compute the action of the Hamiltonian on these states, with the result $$ m\psi^\dagger\psi=\begin{pmatrix}0&0\\0&m\end{pmatrix} $$ (with respect to the basis $|0\rangle$, $\psi^\dagger|0\rangle$).

We learn several lessons:

  • In this basis, the mass term is real and hermitian.

  • In other bases, the mass term is still hermitian but need not remain real. In an arbitrary basis $m\psi^\dagger\psi=U\begin{pmatrix}0&0\\0&m\end{pmatrix}U^\dagger$ for some unitary $U$.

  • There is no basis-independent notion of being real. Only hermiticity is basis independent. And the mass term is definitely hermitian, almost by definition (thou shall always take a hermitian lagrangian).

In higher $d$ the result is the same, except that there are some extra gamma matrices, and the Hilbert space is obtained by acting with the momentum modes $\psi_{\vec k}$ on the zero-modes. This does not change the fact that the mass term is hermitian, and that it is meaningless to ask whether it is real.

That being said, one can also ask about the system as a purely classical theory, i.e., $\psi$ is an $a$-number, not an operator. In this case the Lagrangian is $$ L=i\psi^*\dot\psi+m\psi^*\psi $$ and the mass term is still real: $$ (\psi^*\psi)^*=\psi^*\psi $$ because $(ab)^*=b^*a^*$ for $a$-numbers.

Equivalently, if you expand $\psi$ into its Majorana components, $$ \psi=\chi_1+i\chi_2 $$ the mass term becomes $$ (\chi_1-i\chi_2)(\chi_1+i\chi_2)=\chi_1^2+\chi_2^2+i(\chi_1\chi_2-\chi_2\chi_1) $$

Classically, $\chi_i^2=0$ and therefore, indeed, $$ m\psi^\dagger\psi=2im\chi_1\chi_2 $$ which is again real: $$ (\chi_1\chi_2)^*=\chi_2\chi_1=-\chi_1\chi_2 $$

So, in short: the mass term $i\chi_1\chi_2$ is real, because $i$ and $\chi_1\chi_2$ are separately purely imaginary.

In any case, ascribing reality properties to $a$-numbers is very much convention-dependent (they are not observable). And if you insist, you may run into some quasi-paradoxical results. Consider for example the Poisson-bracket of two real $a$-numbers, $$ \{\chi_i,\chi_j\}=i\delta_{ij} $$ which is purely imaginary even though $\chi_i,\chi_j$ are both individually real.

This all stays true in higher dimensions, but keep in mind that the reality properties of spinors depends sensitively on $d\mod8$. So for example in $d=4$ the reality condition $\psi=\psi^*$ is only valid as written in the Majorana basis, where $\gamma^0$ is purely imaginary.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Chris
    Commented Jan 30, 2021 at 21:40

Given a Grassmann algebra $\Lambda_{\infty}$, its generators are anti-commuting elements $\zeta^{i}$ such that $$\zeta^{i}\zeta^{j}+\zeta^{j}\zeta^{i}=0.$$

A supernumber $z\in\Lambda_{\infty}$ should take the form $$z=z_{B}+z_{S},$$

where $z_{B}\in\mathbb{C}$ is the body, and $$z_{S}=\sum_{k=1}^{\infty}\frac{1}{k!}C_{i_{1}\dots i_{k}}\zeta^{i_{1}}\cdots\zeta^{i_{k}},\quad C_{i_{1}\dots i_{k}}\in\mathbb{C}$$

is its soul.

The involution (complex conjugation) in $\Lambda_{\infty}$ is defined as

  1. $(\zeta^{i})^{\ast}=\zeta^{i}$.
  2. $(\alpha z)^{\ast}=\bar{\alpha}z^{\ast}$, where $\alpha\in\mathbb{C}$.
  3. $(z+w)^{\ast}=z^{\ast}+w^{\ast}$.
  4. $(zw)^{\ast}=w^{\ast}z^{\ast}$.
  5. $(z^{\ast})^{\ast}=z$.

So under involution (aka complex conjugation), you always flip the order of $z$ and $w$. But they are purely classical numbers, not Hermitian quantum operators.

You call it "Hermitian conjugation", and that's fine. But the rest of the world call it complex conjugation (or involution). For example, in this wikipedia page:

enter image description here

For a Dirac spinor, the Lagrangian $$\mathcal{L}=\bar{\Psi}(i\partial\!\!\!/-m)\Psi$$

is not real (Grassmann-number-valued). To find the real "symmetrized" Lagrangian, consider $$\mathcal{L}^{\ast}=[\bar{\Psi}(i\gamma^{\mu}\overset{\rightarrow}{\partial}_{\mu}-m)\Psi]^{\dagger}=\Psi^{\dagger}(-i\gamma^{\mu\dagger}\overset{\leftarrow}{\partial}_{\mu}-m)\gamma^{0\dagger}\Psi=\Psi^{\dagger}(-i\gamma^{0}\gamma^{\mu}\gamma^{0}\overset{\leftarrow}{\partial}_{\mu}-m)\gamma^{0}\Psi=\bar{\Psi}(-i\gamma^{\mu}\overset{\leftarrow}{\partial}_{\mu}-m)\Psi.$$

Then, define a new Lagrangian $$\mathcal{L}_{\mathrm{S}}=\frac{1}{2}(\mathcal{L}+\mathcal{L}^{\ast})=\frac{1}{2}\bar{\Psi}(i\overset{\leftrightarrow}{\partial}\!\!\!\!\!/-2m)\Psi.$$

It follows that the new Lagrangian $\mathcal{L}_{\mathrm{S}}$ is real (Grassmann-valued), and so the functional integrand $e^{iS}$ is well-defined.

The mass term is NOT imaginary. Take your equation as an example, the term $2i\xi\chi$, whatever weird representation you choose, is real, because $$(2i\xi\chi)^{\ast}=-2i\chi^{\ast}\xi^{\ast}=2i\xi\chi,$$

when $\chi$ and $\xi$ are real.

For complex Grassmann numbers $\chi$ and $\xi$, you have $$(\xi^{\ast}\chi + \chi^{\ast}\xi)^{\ast}=\chi^{\ast}\xi+\xi^{\ast}\chi.$$

There's nothing tachyon or imaginary here. You thought it was imaginary because you don't flip the order in complex conjugation.

Reference: Ideas and Methods of Supersymmetry and Supergravity: Or a Walk Through Superspace, section 1.9.1

  • $\begingroup$ Could you please read through the whole post? This is just a specific example for warming up excise. The most general case can be found at the end of the post, where $\xi = \xi_1 + i\xi_2$, and $\chi = \chi_1 + i\chi_2$ $\endgroup$
    – MadMax
    Commented Jun 30, 2022 at 14:48
  • $\begingroup$ @MadMax Let's focus on your arming up exercise at the moment. Why should $\xi^{\ast}=\xi$? This is not true. $\endgroup$
    – Valac
    Commented Jun 30, 2022 at 14:52
  • $\begingroup$ It is very basic, let's say you have a complex number $z= x+iy$, it can be anywhere on the complex plane. One specific example is that it's on the real axis, which means $y=0$ and $z = x$. The same goes for complex Grassmann number. $\endgroup$
    – MadMax
    Commented Jun 30, 2022 at 14:56
  • $\begingroup$ @MadMax Since you have no tags for your equations, do you mind explaining the equation after "Low and behold (I will omit the m parameter hereafter):"? $\endgroup$
    – Valac
    Commented Jun 30, 2022 at 15:05
  • $\begingroup$ Please point out which part of the equation you need explanation: $\bar\psi \psi = \psi_L^\dagger \gamma_0\psi_R + \psi_R^\dagger \gamma_0\psi_L = \xi^*(i\chi) + (i\chi)^*\xi = i\xi\chi - i\chi\xi = 2i\xi\chi \neq 0$. $\endgroup$
    – MadMax
    Commented Jun 30, 2022 at 15:08

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