I have a question about the self-adjointness of the gradient in spinor space.

In the derivation of the Dirac adjoint equation, as in Hermitian adjoint of 4-gradient in Dirac equation , it has been pointed out that the adjoint does not influence the gradient $\partial_\mu$ since the vector space we are considering is $\mathbb{C}^4$, and not ${L^2[\mathbb{C}]}$, so $\partial_\mu$ is hermitian.

Nevertheless, the condition for the Dirac Lagrangian \begin{equation} \mathcal{L}_D = \bar{\Psi}(i \gamma^\mu{\partial_\mu} - m)\Psi \end{equation} to be hermitian seems to imply that $\partial_\mu$ is anti-hermitian.

I am a little confused about this: are we applying two different adjoint transformations, in two different spaces? If so, what is the criterion to choose which one to apply?


2 Answers 2


So let's first motivate the definitions of :-

Hermitian Conjugate of an Operator - VS - Hermitian Conjugate of a Vector

Hermitian Conjugate (Adjoint) of an "Operator" :-

A Hermitian Conjugate (technically known as Adjoint ) of an operator $\hat{\mathrm A}$ is defined via the rule

$\langle\phi|\hat{\mathrm A}|\psi\rangle\ =\langle\hat{\mathrm A}^\dagger\phi|\psi\rangle$, where $\ \hat{\mathrm A}^\dagger$ denotes the Adjoint of the operator $\mathrm{\hat{A}}$.

CONCLUSION 1 :- In order to find the Adjoint of an "Operator", one needs to consider the expectation value of the operator, that is, to evaluate an Integral over all Space-Time Coordinates.

$\langle\phi|\hat{\mathrm A}|\psi\rangle\ =\displaystyle \int_{Entire\\ Domain} \phi^*(t,\mathbf{r})\ \hat{\mathrm A}\ \psi(t,\mathbf{r})\ dt\ d^3\mathbf{r}\ $

SIDE NOTE:- For Operators that can be written as Finite Dimensional Matrices such as \begin{bmatrix} f(x) & g(x) \\ h(x) & k(x) \end{bmatrix} where $x \in (-\infty,\infty)$, we can find the Adjoint as $$\langle\phi|\hat{\mathrm A}|\psi\rangle = \displaystyle \int_{-\infty}^\infty \begin{bmatrix} a^*(x) & b^*(x) \end{bmatrix} \begin{bmatrix} f(x) & g(x) \\ h(x) & k(x) \end{bmatrix} \begin{bmatrix} c(x) \\ d(x)\end{bmatrix}\ dx$$ $$ = \displaystyle \int_{-\infty}^\infty \left(\begin{bmatrix} f^*(x) & h^*(x) \\ g^*(x) & k^*(x) \end{bmatrix} \begin{bmatrix} a(x) \\ b(x) \end{bmatrix}\right)^\dagger \begin{bmatrix} c(x) \\ d(x)\end{bmatrix}\ dx = \langle{\hat{\mathrm A}^\dagger}\phi|\psi\rangle $$ Implying $\hat{\mathrm A}^\dagger = \begin{bmatrix} f^*(x) & h^*(x) \\ g^*(x) & k^*(x) \end{bmatrix}$.

Therefore, one can easily find the Adjoint simply by taking the Complex Conjugate Tranpose of the Operator represented as a matrix. This technique however does NOT apply to Operators that CAN NOT be represented as Finite Dimensional Matrices. Keep that in mind for it is important.

Hermitian Conjugate of a "Vector" :-

If $\ |\psi\rangle$ is a state vector, then the Hermitian Conjugate is defined as $\ |\psi\rangle^\dagger\ =\langle\psi|$.

A finite dimensional vector helps us to visualise this better.

Take for example, $|\psi\rangle = \begin{pmatrix} a\\ b\\ c \end{pmatrix}$

Then, $\langle\psi| = \begin{pmatrix} a^* & b^* & c^* \end{pmatrix}$

CONCLUSION 2 :- In order to find the Hermitian Conjugate of a "Vector", one needs to consider only the complex conjugate transpose of the vector, and NOT any Integral over Space-Time Coordinates.

Now, lets examine a difference between the two Hermitian Conjugates :-

The statements written in Conclusion 1 and 2 are very important. Why?

Consider the vectors

$|\chi_1\rangle = \begin{pmatrix} e^{-\frac{x^2}{2}} \\ \frac{1}{x^2+1} \end{pmatrix}$; $\quad$ $|\chi_2\rangle = \begin{pmatrix} x^2e^{-\frac{x^2}{2}} \\ \frac{1}{x^2+4} \end{pmatrix}$

Task 1:- To find the "Hermitian Conjugate Vector" of $\frac{\mathrm d}{\mathrm {dx}}|\chi_1\rangle$, where $x \in (-\infty,\infty)$

$\frac{\mathrm d}{\mathrm {dx}}|\chi_1\rangle = \begin{pmatrix} \frac{\mathrm d}{\mathrm {dx}}\left(e^{-\frac{x^2}{2}}\right) \\ \frac{\mathrm d}{\mathrm {dx}}\left(\frac{1}{x^2+1}\right) \end{pmatrix} = \begin{pmatrix} xe^{-\frac{x^2}{2}} \\ -\frac{2x}{(x^2+1)^2}\end{pmatrix}$

$\left(\frac{\mathrm d}{\mathrm {dx}}|\chi_1\rangle\right)^\dagger = \begin{pmatrix} xe^{-\frac{x^2}{2}} & -\frac{2x}{(x^2+1)^2}\end{pmatrix} = \begin{pmatrix} \frac{\mathrm d}{\mathrm {dx}}\left(e^{-\frac{x^2}{2}}\right) & \frac{\mathrm d}{\mathrm {dx}}\left(\frac{1}{x^2+1}\right) \end{pmatrix} = \frac{\mathrm d}{\mathrm{dx}} \begin{pmatrix} e^{-\frac{x^2}{2}} \\ \frac{1}{x^2+1} \end{pmatrix}^\dagger = \frac{\mathrm d}{\mathrm {dx}} \left(|\chi_1\rangle^\dagger\right)$

$\boxed{\left(\frac{\mathrm d}{\mathrm {dx}}|\chi_1\rangle\right)^\dagger = \frac{\mathrm d}{\mathrm {dx}} \left(|\chi_1\rangle^\dagger\right)}$

CONCLUSION 3 :- The Hermitian Conjugate ($\dagger$) did nothing to $\frac{\mathrm d}{\mathrm {dx}}$ simply because we are NOT computing any Integral over the spatial(and/or temporal) coordinates at all. $\frac{\mathrm d}{\mathrm {dx}}|\chi_1\rangle$ simply yields another "Vector" whose complex conjugate transpose is being calculated here. That's all. Nothing more.

Task 2:- To find the "Hermitian Conjugate Operator" of $\frac{\mathrm d}{\mathrm {dx}}$, where $x \in (-\infty,\infty)$

$\langle\chi_2|\frac{\mathrm d}{\mathrm {dx}}|\chi_1\rangle\ =\displaystyle \int_{-\infty}^\infty \begin{pmatrix} x^2e^{-\frac{x^2}{2}} & \frac{1}{x^2+4} \end{pmatrix} \begin{pmatrix} \frac{\mathrm d}{\mathrm {dx}}\left(e^{-\frac{x^2}{2}}\right) \\ \frac{\mathrm d}{\mathrm {dx}}\left(\frac{1}{x^2+1}\right) \end{pmatrix}\ dx$

$ = \displaystyle \int_{-\infty}^\infty x^2e^{-\frac{x^2}{2}}\frac{\mathrm d}{\mathrm {dx}}\left(e^{-\frac{x^2}{2}}\right)\ dx + \int_{-\infty}^\infty \frac{1}{x^2+4}\frac{\mathrm d}{\mathrm {dx}}\left(\frac{1}{x^2+1}\right)\ dx$

Integrating By Parts would give:-

$\left[\left(x^2e^{-\frac{x^2}{2}}\right)\left(e^{-\frac{x^2}{2}}\right)\right]_{-\infty}^\infty - \displaystyle \int_{-\infty}^\infty \frac{\mathrm d}{\mathrm {dx}}\left(x^2e^{-\frac{x^2}{2}}\right)e^{-\frac{x^2}{2}}\ dx + \left[\left(\frac{1}{x^2+4}\right)\left(\frac{1}{x^2+1}\right)\right]_{-\infty}^{\infty} - \int_{-\infty}^\infty \frac{\mathrm d}{\mathrm {dx}}\left(\frac{1}{x^2+4}\right)\frac{1}{x^2+1}\ dx$

Surely, the boundary terms are 0. This yields

$\displaystyle \int_{-\infty}^\infty \begin{pmatrix} -\frac{\mathrm d}{\mathrm {dx}}\left(x^2e^{-\frac{x^2}{2}}\right) & -\frac{\mathrm d}{\mathrm {dx}}\left(\frac{1}{x^2+4}\right) \end{pmatrix} \begin{pmatrix} e^{-\frac{x^2}{2}} \\ \frac{1}{x^2+1} \end{pmatrix}\ dx\ = \int_{-\infty}^\infty \left[-\frac{\mathrm d}{\mathrm {dx}}\begin{pmatrix} x^2e^{-\frac{x^2}{2}} & \frac{1}{x^2+4} \end{pmatrix}\right] \begin{pmatrix} e^{-\frac{x^2}{2}} \\ \frac{1}{x^2+1} \end{pmatrix}\ dx$

Thus, $\langle\chi_2|\frac{\mathrm d}{\mathrm {dx}}|\chi_1\rangle\ = \langle -\frac{\mathrm d}{\mathrm {dx}}\chi_2|\chi_1\rangle$ , yielding

$\boxed{\left(\frac{\mathrm d}{\mathrm {dx}}\right)^\dagger = -\frac{\mathrm d}{\mathrm {dx}}}$

CONCLUSION 4 :- The Hermitian Conjugate ($\dagger$) does act on $\frac{\mathrm d}{\mathrm {dx}}$ this time, simply because we are ARE computing any Integral over the spatial(and/or temporal) coordinates.

FINAL CONCLUSION :- It was just the confusion of notation that led to the trouble. While finding the Hermitian Conjugate of an Operator, it has a completely different evaluation procedure than finding the Hermitian Conjugate of a Vector. Unfortunately, the same notation ($\dagger$) is used for both, which is where potentially lies the confusion.

P.S. You may wonder how does one know which definition one should use when? Well that is simple. In the context of Dirac equation

  • If you are supposed to find the Adjoint of Dirac Operator $(i\hbar\gamma^{\mu}\partial_{\mu} - mc)^\dagger$, you will simply have to use the definition applicable to operators, that is the Expectation Value of this Operator.

  • If you are supposed to find a Dirac-like Equation that would be satisfied by Hermitian Conjugate Vector $|\psi\rangle^\dagger$, then this is even simpler, for you do not need to find any Expectation Value. Just the Complex Conjugate Transpose of the entire equation. So, ALL Space-Time dependent Operators like $\frac{\mathrm d}{\mathrm {dx}}$ will remain unaffected.


Ultimately, it is Hermitian conjugation in both senses: Both wrt. Dirac indices and as differential operators. In particular we still have to integrate by parts. Also note that the complex conjugate of a product two supernumbers $z,w$ is by definition the opposite order: $(z w)^{\ast}=w^{\ast} z^{\ast}$.

  • $\begingroup$ If I understood that correctly, this operation does the conjugation only wrt Dirac indices: $$ \mathcal{L}_D^\dagger = \Psi^\dagger(-i {\gamma^\dagger}^\mu \overleftarrow{\partial_\mu} - m)\gamma^0 \Psi = \dots = \bar\Psi (-i\gamma^\mu \overleftarrow{\partial_\mu} - m) \Psi $$ and then, after an integration by part, we find the original $\mathcal{L}_D$ with the right (no pun intended) verse of the derivative? And the reason why we don't integrate by part in the adjoint Dirac equation is that we want the derivative to act on the adjoint spinor, which is on the left? $\endgroup$
    – Carlo
    May 1, 2020 at 19:45
  • $\begingroup$ It seems you understand correctly. $\endgroup$
    – Qmechanic
    May 1, 2020 at 20:23

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