Hermitian adjoint of 4-gradient in Dirac equation

Question

I'm having issues deriving the Dirac adjoint equation, $$\overline{\psi}(i\gamma^{\mu}\partial_{\mu}+m)=0.\tag{1}$$ I started by taking the Hermitian adjoint of all components of the original Dirac equation, giving me $$\psi^{\dagger}(-i\gamma^{\mu\dagger}\partial_{\mu}^{\dagger}-m)=0.\tag{2}$$ The adjoint of the gamma matrices is defined to be $\gamma^{\mu\dagger}=\gamma^0\gamma^{\mu}\gamma^0$, so no issues there. Now intuitively, I would think that the adjoint of the 4-gradient would be $\partial_{\mu}^{\dagger}=-\partial_{\mu}$. In non-relativistic quantum mechanics, it can be shown that first derivative operators are anti-Hermitian, so for example, $\frac{d}{dx}^{\dagger}=-\frac{d}{dx}$. So I would think this would the same case for the 4-gradient, but apparently it isn't. Among the many derivations I've gone over, for example, on page 77 here, it claims that the 4-gradient is self-adjoint. Could someone please explain why my intuition is incorrect?

Answer 1

Expanding on my comment.

The basic idea is that what you mean by adjoint depends on the vector space being considered. For example, we might have $\mathbb{C}^n$ as our space, with the usual inner product; in that case, the adjoint of a vector or matrix is the transpose conjugate. Note that technically, taking adjoint of a vector doesn't return a vector, because row vectors and column vectors belong to different spaces.

We could also use $L^2(\mathbb{R}^n)$ as our vector space. Its elements are functions, and the inner product is defined by

$$(f,g) = \int d^n x\ f^* g$$

You can take adjoints here too, using the inner product defined above. It can be shown that the derivative operator, which is a linear transformation on $L^2$, is anti-Hermitian.

Now to the Dirac equation. The vector space (that is, spinor space) being considered here is $\mathbb{C}^4$, not $L^2$. That is, $\psi$ is a vector because it has four components, not because it's a function. The fact that its components are functions is irrelevant here. When we take adjoints, we transpose and conjugate vectors and matrices. The derivative is an operator if you think about what it does to functions, but it is not a $4\times4$ matrix; it does nothing to spinors. Therefore, the particular adjoint we're doing here doesn't affect it.

Answer 2

My answer is pretty much similar to what others have written here, but maybe in a slightly different fashion.

So let's first motivate the definitions of :-

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

Hermitian Conjugate (Adjoint) of "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.

Answer 3

Perhaps the following argument is more convincing:

1. The Dirac equation$^1$ $$ (i\gamma^{\mu}\stackrel{\rightarrow}{\partial}_{\mu}-m)\psi~=~0 \tag{A}$$ is by the fundamental lemma of variational calculus equivalent to $$ \forall \phi:\quad 0~=~\int d^4x~\bar{\phi}(i\gamma^{\mu}\stackrel{\rightarrow}{\partial}_{\mu}-m)\psi, \tag{B}$$ where $\phi$ is an arbitrary (off-shell) Dirac spinor.

2. Hermitian conjugation in Dirac spinor space leads to $$ \forall \phi:\quad0~=~\int d^4x~\bar{\psi}(-i\stackrel{\leftarrow}{\partial}_{\!\mu}~\gamma^{\mu}-m)\phi, \tag{C}$$ which is equivalent to $$ \bar{\psi}(i\stackrel{\leftarrow}{\partial}_{\!\mu}~\gamma^{\mu}+m)~=~0,\tag{D}$$ cf. the above comment by Javier.

3. On the other hand, if we also integrate (C) by parts, we get (after discarding boundary terms) $$ \forall \phi:\quad 0~=~\int d^4x~\bar{\psi}(i\gamma^{\mu}\stackrel{\rightarrow}{\partial}_{\mu}-m)\phi, \tag{E}$$ where the derivative now acts on $\phi$.

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$^1$ We use the following conventions: $$ \bar{\psi} ~=~ \psi^{\dagger}\gamma^0 , \qquad\gamma^{\mu\dagger}~=~\gamma^0\gamma^{\mu}\gamma^0, \qquad (\gamma^0)^2~=~{\bf 1}_{4\times 4}. \tag{F}$$

Answer 4

I totally agree with the answers above, but there is a problem when you try to prove that the Hamiltonian operator is Hermitian. In fact, in the hermitian adjoint of the Dirac hamiltonian, the alpha-matrices terms will have a minus sign because of the i factor, and the space derivative will stay unaffected. So at the end we will have to say that the alpha matrices should be anti-hermitian in order for the hamiltonian to be hermitian. Sorry for not writing the expressions. I'm new here and I don't know yet how to write equations.