5
$\begingroup$

Is the definition of the adjoint spinor $\bar{\psi}=\psi^\dagger \gamma^0$ forcing a particular choice of representation of the Dirac matrices (or a subset of the possible choices)?

More precisely, I assume (possibly incorrectly) that the adjoint of $\psi$ can always be written as $\psi^\dagger$ times some linear combination of the gamma matrices $$\bar{\psi} = \psi^\dagger c_\mu \gamma^\mu.$$ Where the usual choice of gamma matrices leads to $c_\mu = (1,0,0,0)$.

So broken into smaller steps the confusion is:

  • Given a concrete choice of matrices to represent $\gamma^\mu$, how can we determine what is the adjoint spinor in terms of the spinor components?
  • Is it always linear, $\bar{\psi} = \psi^\dagger c_\mu \gamma^\mu$?
  • Is it always given by $c_\mu = (1,0,0,0)$ independent of the choice of gamma matrices?

This question is a follow up of comments on: spinor vs vector indices of Dirac gamma matrices

To keep this self-contained, here is a summary of thoughts for why $\bar{\psi}=\psi^\dagger \gamma^0$ may not be true for all choices of the gamma matrices.

The Dirac matrices (gamma matrices), are defined by the anticommutation relation $$\{ \gamma^\mu, \gamma^\nu \} = \gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 \eta^{\mu \nu} I_4 $$ where $\eta^{\mu \nu}$ is the Minkowski metric (using (+ - - -) convention) and $I_4$ is the 4x4 identity matrix.

Given one set of matrices $\gamma^\mu$ that satisfy this anticommutation relation, the matrices $\gamma'^\mu = \Lambda^\mu{}_\nu \gamma^\nu$ will also satisfy the anticommutation relation if $\Lambda^\mu{}_\nu$ is a Lorentz transformation :

$$\{ \gamma'^\mu, \gamma'^\nu \} = \{ \Lambda^\mu{}_\alpha \gamma^\alpha, \Lambda^\nu{}_\beta \gamma^\beta \} = \Lambda^\mu{}_\alpha \gamma^\alpha \Lambda^\nu{}_\beta \gamma^\beta + \Lambda^\nu{}_\beta \gamma^\beta \Lambda^\mu{}_\alpha \gamma^\alpha \\ = \Lambda^\mu{}_\alpha \Lambda^\nu{}_\beta (\gamma^\alpha \gamma^\beta+\gamma^\beta \gamma^\alpha) = \Lambda^\mu{}_\alpha \Lambda^\nu{}_\beta 2 \eta^{\alpha \beta} I_4 = 2 \eta^{\mu \nu} I_4 $$

If we assume that $\bar{\psi}\psi=\psi^\dagger \gamma^0\psi$ is a scalar for any choice of $\gamma^0$, the above appears to lead to the result that $\psi^\dagger \gamma^1\psi$ is also a scalar (or similarly with any gamma matrix). This is because we can use the above to make a new choice of gamma matrices with $\gamma'^0 = a \gamma^0 + b \gamma^1$ (where $a^2-b^2=1$). And thus would require $\psi^\dagger \gamma'^0\psi = a \psi^\dagger \gamma^0\psi + b \psi^\dagger \gamma^1\psi$ is also a scalar. Similarly then, $\psi^\dagger \gamma^\mu \psi$ would have to be a scalar for any choice of $\mu$. It seems more likely that the starting assumption, that $\bar{\psi}=\psi^\dagger \gamma^0$ independent of representation choice, is wrong.

$\endgroup$
  • $\begingroup$ Warning for answerers: OP thinks the Dirac adjoint and Hermitian adjoint are the same thing. $\endgroup$ – knzhou Mar 9 '18 at 16:54
  • $\begingroup$ @knzhou Yes I was struggling with terminology, but there is a non-degenerate bilinear form which takes in two spinors and yields a Lorentz scalar. I felt that this vaguely "inner product structure" like thing which gives us a scalar, could allow us to define the Dirac adjoint representation independent, and also gives us the sense in which the "Dirac adjoint" is an adjoint: it is the dual of the spinor in this bilinear form mapping spinor x spinor -> scalar. This is probably still incorrect terminology, but I'd appreciate if you post explanations instead of warnings. $\endgroup$ – Coconut Mar 10 '18 at 1:43
  • $\begingroup$ It was in that sense, which I felt by definition we should have $\text{DiracAdjoint}(\text{DiracAdjoint}(\psi)) = \psi$, and I distinguished this from $\psi^\dagger$ which is a "dual" in a different sense. So while I don't deny making mistakes (I need to learn hence asking questions), I don't feel your dismissive summary is appropriate or correct . $\endgroup$ – Coconut Mar 10 '18 at 1:50
4
$\begingroup$

Quick answer:

We can always define the Dirac adjoint of a spinor to be $\bar \psi := \psi^\dagger \gamma^0$. After that definition is set up, we have to additionally tell how it transforms under a Lorentz transformation and that depends on convention. For example, we could choose $\gamma'^\mu = \Lambda^\mu{}_\nu \gamma^\nu$ (note $\gamma'^0$ is no longer Hermitian) and define $\bar \psi := \psi^\dagger \gamma'^0 $, then we will see that $\bar\psi\psi$ no longer transforms as a scalar.

On the other hand, if we insist that $\bar\psi \psi$ should transform as a scalar, we should find a function of the gamma matrices $F(\gamma'^\mu)$ such that for any Lorentz transformation $\Lambda,\ \color{red}{S(\Lambda)^\dagger FS(\Lambda) \equiv F(\gamma'^\mu)}$. Then, defining $\bar \psi := \psi^\dagger F$ ensures that $\bar\psi\psi$ transforms as a scalar. The question is whether such a function always exists. With your example, you can convince yourself that the ansatz $F= c_\mu \gamma'^\mu$ will work if and only if $c_\mu \Lambda^\mu{}_j=0$ for all spatial indices $j$. This is a system of three equations $(j=1,2,3)$ with four unknowns $(c_\mu)$, and therefore possibly has many solutions. For example, if $\Lambda$ describes a boost in the $x$-direction with speed $\beta$, defining $\bar \psi := \psi^\dagger (\gamma'^0 + \beta \gamma'^1)$ makes $\psi^\dagger\psi$ a scalar. Note that $\gamma'^0 + \beta \gamma'^1 \propto \gamma^0$, so there is no contradiction.


Detailed answer:

A Clifford algebra over a $D$-dimensional spacetime equipped with the metric $g^{\mu\nu}$ is generated by $D$ hypercomplex numbers $\{\gamma^\mu\}, \mu\in \{0,\cdots,D-1\}$ defined by the following algebraic product:

$$ \{\gamma^\mu,\gamma^\nu\} = 2g^{\mu\nu} \mathbb I_n\,. \tag1 $$

The algebra is $2^D$-dimensional, meaning there is a list of $2^D$ linearly independent elements, closed under multiplication, formed by various products of the $D$ hypercomplex numbers. Moreover, there is always a representation of the algebra in terms of real $n \times n$ matrices where $n=2^{[D/2]}$, the operation $[\cdot]$ taking the integer part of the given number. If the dimension $D$ is even, then this representation is the one and only irreducible representation of the algebra (up to equivalences). Furthermore, if the representation is unitary, the $2^D$ basis elements can be chosen to be Hermitian.

Without loss of generality, take the metric to be of mostly negative signature, so that $(\gamma^0)^2 = +1$ and $(\gamma^i)^2=-1$. You already begin to notice why $\gamma^0$ is special when contrasted with the rest of $\gamma$-matrices. It is exactly the same way as time is special when contrasted with space, given that the metric has a Lorentzian signature. With this in mind, for D=4 a basis for $4\times4$ matrices is given by $\{\Gamma_j\}, j\in\{1,\cdots, 16\}$ where

\begin{align*} \Gamma_j \in \{&\mathbb I_4, \gamma^0,i\gamma^1,i\gamma^2,i\gamma^3,\\ &\gamma^0\gamma^1,\ \gamma^0\gamma^2,\ \gamma^0\gamma^3,\quad i\gamma^1\gamma^2,\ i\gamma^2\gamma^3,\ i\gamma^3\gamma^1,\ \\ &i\gamma^0\gamma^1\gamma^2,\quad i\gamma^0\gamma^1\gamma^3,\quad i\gamma^0\gamma^2\gamma^3,\quad i\gamma^1\gamma^2\gamma^3,\\ &i\gamma^0\gamma^1\gamma^2\gamma^3\}\,. \tag2 \\ \end{align*}

The imaginary number $i$ has been inserted in the above array so that for all $j, (\Gamma_j)^2=+1.$ Notice that this list is closed under multiplication. This allows you to write any $4\times4$ matrix $X$ as a sum over these matrices: $X = \sum_{i=1}^{16} x^i \Gamma_i$ where $x_i = \frac14 \text{Tr}(X\Gamma_i)$. Moreover, for all $j\ne1, \text{Tr}(\Gamma_j)= 0$. Using this information, you can prove the following lemma (Cf. Schwartz Ch. 4).

Lemma:

Given two sets of gamma matrices $\{\gamma^\mu\}$ and $\{\gamma'^\mu\}$, correspondingly basis vectors $\{\Gamma_j\}$ and $\{\Gamma'_j\}$, there exists a non-singular matrix $S$ such that

$$ \boxed{\gamma'^\mu = S\gamma^\mu S^{-1}}\,, \qquad \text{ where } S=\sum_{i=1}^{16} \Gamma'_i F \Gamma_i\,, $$

and $F$ is so chosen that $S$ is non-singular.

Furthermore, $S$ is uniquely fixed (up to a numerical factor).

You have correctly identified that $\gamma'^\mu := \Lambda^\mu{}_\nu \gamma^\nu$ obeys the same commutation relation as $\gamma^\mu$, and it is precisely because of this that the above lemma tells us of the existence of a non-singular $S$ so that

$$ S^{-1} \gamma^\mu S = \Lambda^\mu{}_\nu \gamma^\nu\,, \tag3$$

which is required for the Lorentz invariance of the Dirac equation.

At this point, let us make a conventional choice (which @akhmeteli has been pointing out all along). Let us choose $\gamma^0$ such that it is Hermitian and the $\gamma^i$s such that they are anti-Hermitian. In other words, this means that $\gamma^\mu = \gamma^0 (\gamma^\mu)^\dagger \gamma^0$. Notice that we did not need any such assumption in the previous discussion. But this convention will simplify greatly what is about to follow.

Observe that because $\Lambda^\mu{}_\nu$ is real, and because we chose this convention,

\begin{align*} (S^\dagger\gamma^0) \gamma^\mu (S^\dagger \gamma^0)^{-1} = (S^{-1} \gamma^\mu S)^\dagger &\stackrel{(3)}{=} \Lambda^\mu{}_\nu (\gamma^\nu)^\dagger\\ \text{(Multiplying by $\gamma^0$ from both sides)} \Rightarrow (\gamma^0 S^\dagger\gamma^0) \gamma^\mu (\gamma^0 S^\dagger \gamma^0)^{-1} &= \Lambda^\mu{}_\nu \gamma^\nu \stackrel{(3)}{=}S^{-1} \gamma^\mu S\,. \\ \end{align*}

After rearranging we find that,

\begin{align*} &\Rightarrow (S\gamma^0 S^\dagger\gamma^0) \gamma^\mu (S\gamma^0 S^\dagger \gamma^0)^{-1} = \gamma^\mu \\ &\Rightarrow S\gamma^0 S^\dagger\gamma^0 = c\mathbb I_4 \\ &\Rightarrow S^\dagger\gamma^0 = c \gamma^0 S^{-1}\,, \tag4 \\ \end{align*}

where $c$ is some constant which you can convince yourself is real.

Now, if we normalize $S$ such that $\det(S)=1$, then $c^4 = 1$ or $c= \pm 1$. Let us see which situations correspond to $c=+1$ and $c=-1$. Observe the following identity.

\begin{align*} S^\dagger S = (S^\dagger\gamma^0)\gamma^0S &\stackrel{(4)}= c\gamma^0(S^{-1} \gamma^0 S)\\ &\stackrel{(3)}= c\gamma^0 \Lambda^0{}_\nu \gamma^\nu \\ &= c(\Lambda^0{}_0 \mathbb I_4 - \sum_{k=1}^3 \Lambda^0{}_k\gamma^0\gamma^k)\\ \Rightarrow \text{Tr}(S^\dagger S) &= 4c\Lambda^0{}_0 \,. \tag5 \end{align*}

Since $S^\dagger S$ has real eigenvalues (it is Hermitian) and positive-definite (since S is non-singular), its trace must be positive. This means that when $ \Lambda^0{}_0 \le -1, c=-1$ and when $ \Lambda^0{}_0 \ge +1, c=+1$.

Conclusion:

Defining $\bar \psi := \psi^\dagger \gamma^0$, we see that under a Lorentz transformation (taking $\psi \to S\psi$) that does not involve time reversal ($\Lambda^0{}_0 \ge +1$), $\bar \psi \to +\bar\psi S^{-1}$, whereas for Lorentz transformations that do reverse time ($\Lambda^0{}_0 \le -1$), $\bar \psi \to -\bar\psi S^{-1}$.

If we had not normalized $S$, we would write $\bar\psi \to c\bar\psi S^{-1}$. Then this extra factor of $c$ will need taking care of in the Lagrangian through perhaps a field redefinition.

More generally, if we did not assume the hermiticity properties of the gamma matrices, things would be much more complicated, but a definition is a definition. After defining $\bar \psi := \psi^\dagger \gamma^0$, our task would be to find its transformation rules and then implement it accordingly in the Lagrangian. Alternatively, if you wish to preserve the transformation rules, you will have to change the definition accordingly.

$\endgroup$
  • $\begingroup$ Please reread what I wrote in the question. You seem to have misunderstood me. I can answer the questions you ask in "Quick answer" section, but after more thinking, I feel I have a simpler argument that using $\psi^\dagger \gamma^0$ as the adjoint of $\psi$ is not representation independent. The bar notion doesn't work well with multi-symbols, so I'll use $\text{Ad}(\psi)$ to mean the adjoint of $\psi$. ... $\endgroup$ – Coconut Mar 9 '18 at 4:17
  • $\begingroup$ By definition of what it means to be an adjoint, $\text{Ad}(\text{Ad}(\psi)) = \psi$. If $\text{Ad}(\psi)=\psi^\dagger \gamma^0$ was independent of the representation, then $\psi = \text{Ad}(\text{Ad}(\psi)) = \text{Ad}(\psi^\dagger \gamma^0) = (\gamma^0)^\dagger \gamma^0 \psi$. Which requires $(\gamma^0)^\dagger = \gamma^0$, but we already know that isn't representation independent. $\endgroup$ – Coconut Mar 9 '18 at 4:18
  • $\begingroup$ @Coconut You are confusing names. When we say Dirac adjoint, we do not mean the usual Hermitian adjoint of a matrix (that's why psi-bar is a better name than Dirac adjoint). The adjoint of a spinor is simply $\psi^\dagger$. If you have answered yourself by a confused misunderstanding of words, I cannot help you. $\endgroup$ – Nanashi No Gombe Mar 9 '18 at 6:23
  • $\begingroup$ If we represent a spinor $\psi$ as a complex valued column vector, then $\psi^\dagger$ is the conjugate transpose and a row vector. The adjoint instead is referring to the Hilbert space of these objects. Here, the conjugate transpose of a vector is not the same as the adjoint. However, due to linearity of the inner product, if we define $\text{Ad}(\psi) = \psi^\dagger \gamma^0$ then for any matrix $M$ we have $\text{Ad}(M\psi) = (M\psi)^\dagger \gamma^0 = \psi^\dagger M^\dagger \gamma^0$ $\endgroup$ – Coconut Mar 9 '18 at 6:34
  • $\begingroup$ @Coconut Could you please support your claim with a reference? Also, it'd be helpful if you read the Wikipedia article once. $\endgroup$ – Nanashi No Gombe Mar 9 '18 at 6:38
0
$\begingroup$

As I tried to explain in the comments to your original question, $\bar{\psi}=\psi^\dagger\gamma^0$ is not correct for all choices of the representation of the $\gamma$-matrices, but it is correct for "conventional" choices of $\gamma$-matrices, where $\gamma^0$ is hermitian and all other $\gamma$-matrices are anti-hermitian. If $\gamma$-matrices satisfy this condition, your transformed $\gamma$-matrices $\gamma'^\mu$ do not necessarily satisfy it, because Lorentz transforms are not necessarily unitary. Therefore, the expression for the Dirac-adjoint spinor is not necessarily the same.

$\endgroup$
  • $\begingroup$ For any choice, is there only one linear combination of $\gamma^\mu$ that is Hermitian? I'm not sure how to prove that, but if that is the case, does that then fix how $\bar{\psi}$ is related to $\psi$ and $\gamma^\mu$ given the representation choice? $\endgroup$ – Coconut Mar 5 '18 at 5:01
  • $\begingroup$ @Coconut: No. For example, for a "conventional" choice, you can replace $\gamma^0$ by $-\gamma^0$. $\endgroup$ – akhmeteli Mar 5 '18 at 5:12
  • $\begingroup$ You say $\bar{\psi} = \psi^\dagger \gamma^0$ is not correct for all choices, and despite all discussion, and even making a new question, it is still unclear to me what IS the correct way to write it for all choices. You keep giving partial answers. It would be one thing if you also gave a suggestion of how to complete the answer, but you leave me to guess, and with each guess instead of pointing me in the right direction, you just point out how I am wrong. Instead of continually making me guess how to fish + telling me my guesses for how to fish are wrong -- answer questions about how to fish. $\endgroup$ – Coconut Mar 5 '18 at 6:15
  • $\begingroup$ Given a concrete choice of matrices to represent $\gamma^\mu$, how can we determine what is the adjoint spinor in terms of the spinor components? $\endgroup$ – Coconut Mar 5 '18 at 6:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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