Let $e$ be a finitely matrix representable operator.

In physics, specially in quantum mechanics (QM), it is customary to define the conjugate operator $e^{\dagger}$, as the adjoint or the Hermitian conjugate which in this case is just \begin{equation} e^{\dagger}=\left(\bar{e}\right)^{T} \tag{1} \end{equation} where $^{\bar{}}$ represents complex conjugation and $^T$ matrix transposition.

However, in some abstract algebras, including Clifford algebras the "conjugate" of an element of the algebra is defined in such a way that they obey the relation \begin{equation} \frac{e^*}{\|e\|^2}=e^{-1} \tag{2} \end{equation} where $e^*$ is the conjugate operator of $e$ in such algebras.

But, if $e$ is matrix representable with elements in $\Bbb{R}$ or $\Bbb{C}$, then $e$ must also obey the following matrix identity \begin{equation} \frac{\text{adj}(e)}{\text{det}(e)}=e^{-1} \tag{3} \end{equation} where $\text{adj}(e)$ is the adjugate of $e$ and $\text{det}(e)$ is the determinant of $e$.

By direct comparison of $(2)$ and $(3)$, it follows \begin{equation} e^*=\text{adj}(e) \tag{4} \end{equation} and \begin{equation} \|e\|^2=\text{det}(e)=e^*e \tag{5} \end{equation} If one takes the inspiration for the definition of "conjugate" from complex conjugation, using the representation \begin{equation} i = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \end{equation} Both gives a valid equivalent result $i^{\dagger}=i^*=-i$. However, the adjugate and the adjoint don't usually agree with each other. For example, for every Pauli matrix, we have $\sigma_i=\sigma_i^{\dagger}=-\sigma_i^*$.

So, my question is: For matrix representable operators like $\sigma$-matrices, is there a physical reason to prefer the adjoint over the adjugate?

This question may be seeing as trivial, however, in classical QM there was no ambiguity of what it meant by "conjugating" an operator. Yet, in more recent years, treating QM under Clifford formulation is becoming more and more common, and it may be confusing when to use one or the other.

PS: "Adjugate" is equivalent not only to complex conjugation but also split-complex conjugation, which match spacetime distance.


2 Answers 2


You may be reading too much into it: There are two completely different terms involved, growing out of historical accidents and usage in two completely different communities.

  • The physics adjoint (1) is what you started with, and heavily used in QM, an easy mnemonic, even though, of course, you may use it in classical mechanics.

  • The obsolete math term "classical adjoint" is just the adjugate, (3), and never the twain shall meet!

Just accept that two different communities use adjoint for different things! Fortunately, the second definition is rushing to complete obsolescence. Pauli matrices are hermitian, so exponentiating them with an i prefactor yields unitary matrices.

This time, physicists serve the role of Mathematicians in J W v Goethe's (1826) quote:

“Mathematicians are like Frenchmen: whatever you say to them they translate into their own language and forthwith it is something entirely different.”

  • $\begingroup$ Plutôt vrai; mais ici ce sont les physiciens qui jouent ce rôle, pas les mathématiciens... $\endgroup$ Feb 7 at 21:45

In physics we use whatever maths gives insight into the physical world. Hermitian operators have real eigenvalues and orthogonal eigenvectors. Both properties make them well-suited to play a role in basic physics.

This does not at all rule out employing other forms of conjugate wherever they are useful.

(As it happens I was not taught the term 'conjugate' on its own; when I learned quantum physics the terminology was always either 'Hermitian conjugate' or 'complex conjugate' as the case may be. We also learned 'adjoint' as a synonym for 'Hermitian conjugate'.)


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