3
$\begingroup$

In this post, I claimed in my answer that $\partial_{\mu}^{\dagger}=\partial_{\mu}$. The reason I claimed that is because $D^{\dagger}_{\mu}=\partial_{\mu}-iqA_{\mu}$ and I assumed that

$$(\partial_{\mu}+iqA_{\mu})^{\dagger}=\partial_{\mu}^{\dagger}+(iqA_{\mu})^{\dagger}=\partial_{\mu}-iqA_{\mu}.$$

(1) Is this assumption correct?

However, a commenter claimed that $\partial_{\mu}^{\dagger}=\partial_{\mu}$ is not true, since the Klein-Gordon field belongs to $L^2(\mathbb{R}^{1,3})$, and therefore

$$\langle f | \partial_\mu g \rangle=\langle f |\partial_\mu | g \rangle =-\langle \partial_\mu f | g \rangle=- \langle f |\partial_\mu^\dagger | g \rangle,$$

where in the second equality we did integration by parts.

(2) Is the commenter is right? According to this equation, does that mean that $\partial_{\mu}=-\partial_{\mu}^{\dagger}$? Unfortunately, I'm not familiar at all with the concept of $L^p$ spaces and therefore I can't judge this on my own.

Additionally, I found this post, in which the answer claims that it does not make sense to take the hermitian conjugate of $\partial_{\mu}$.

(3)If that's the case, how do you show that $D^{\dagger}_{\mu}=\partial_{\mu}-iqA_{\mu}$?

But then again, in this post, the first answer by Javier states that

The vector space (that is, spinor space) being considered here is $\mathbb{C}^4$

and therefore the hermitian adjoint does not affect $\partial_{\mu}$

(4) This seems to be in contradiction with the previous post. What am I missing? Does it make sense to take the hermitian adjoint of $\partial_{\mu}$ or not?

This is all too confusing!

(5) I would like to know which fields belong to which space (like $L^2$ or $\mathbb{C}^2$) and what the consequences of that are. Some that come to mind would be: a real scalar field, a complex scalar field, the Klein-Gordon field, the Dirac field, the vector field $A_{\mu}$ etc. I own a couple of QFT books and I've never seen this pure mathematical aspect.

$\endgroup$
5
  • 1
    $\begingroup$ I am far from an expert (since I just realised the problem from this question), but I find the answer by Souparna Nath to one of your linked questions go more in depth than the accepted answer. So, to answer the parts in turn (1) - the transformation of $\partial$ depends on what it is (an operator or a vector). (2) - In the case of writing a Lagrangian, which should be a scalar, where you have things like $\partial \phi$, these must be VECTORS, not operators, so $(\partial \phi)^\dagger = \partial \phi^\dagger$.... $\endgroup$
    – Vangi
    May 12 at 16:51
  • 1
    $\begingroup$ ... I think the commenter on your answer was trying to say that you can NOT always assume $\partial^ \dagger = \partial$, since, as in the answer by Souparna is shown, it is not valid if you're asking for the hermitian conjugate of the OPERATOR $\partial$. (3) - I think a way (maybe not correct!) of thinking about it is that in $\partial_\mu - iqA_\mu$, $iqA_\mu$ is just a vector, not an operator, so $\partial_\mu$ is not an operator either. (4) - I think this can be answered using the above (5) - I am afraid I can't help you with this, as I see $L^2$ for the first time... $\endgroup$
    – Vangi
    May 12 at 16:51
  • 1
    $\begingroup$ ... What I think the bottom line is that you need to be careful what context you are working in. In your answer, the assumption that you made was correct (although maybe people would rather see an expression for $(\partial \phi)^\dagger$ rather than only $(\partial)^\dagger$). The commenter was saying that you can't always assume what you wrote - ".., since $\partial^\dagger = \partial$", as this is dependent on what I wrote above. $\endgroup$
    – Vangi
    May 12 at 16:55
  • $\begingroup$ (2) is right. $\partial^\dagger= \overset{_\gets}\partial =-\partial$. (1) is flat wrong. Go to a reputable text. You may also investigate the hermiticity of the momentum operator. $\endgroup$ May 12 at 18:39
  • $\begingroup$ @Vangi Hi! Thank you very much, that helped a lot! Although I still have some questions. Let me see if I understood this correctly (after reading the answer by Souparna Nath). If we are asking what the hermitian conjugate of the operator $\partial_{\mu}$ is, the answer is $\partial_{\mu}^{\dagger}=- \partial_{\mu}$. However, when considering the hermitian conjugate of $\partial_{\mu} \phi$, which is a vector, the answer is $(\partial_{\mu} \phi)^{\dagger}=\partial_{\mu} \phi^{\dagger}$. My question then is, why is $\partial_{\mu} \phi$ a vector? $\endgroup$ May 13 at 11:52

1 Answer 1

3
$\begingroup$

First, let's consider the derivative $\partial$ as an operator on the Hilbert space $\mathcal{H} = L^2(\mathbb{R}^n)$ of square-integrable functions on $\mathbb{R}^n$ (the space typically encountered in quantum mechanics). Because not all $L^2$-functions are differentiable, we restrict $\partial$ to the (dense) subspace $C_c^\infty(\mathbb{R}^n) \subset L^2(\mathbb{R}^n)$ of smooth compactly supported functions. A simple integration by parts argument leads to the following identity for all $f,g \in C_c^\infty(\mathbb{R}^n)$: $$\langle f, \partial g \rangle = \int_{\mathbb{R}^n} \overline{f(x)}\partial g(x) \mathrm{d} x = -\int_{\mathbb{R}^n} \overline{\partial f(x)} g(x) \mathrm{d} x = - \langle \partial f, g \rangle. $$ The boundary terms appearing after integrating by parts vanish because $f,g$ are compactly supported. In this sense, $\partial^\dagger f = -\partial f$ for all $f \in C_c^\infty(\mathbb{R}^n)$.

Remark: The momentum operator $P_0=-\mathrm{i} \partial$ is symmetric on $C_c^\infty(\mathbb{R}^n)$, and has a self-adjoint extension $P$ (i.e. $P^\dagger = P$) that is defined on the Sobolev space $H^1(\mathbb{R}^n) \subset L^2(\mathbb{R}^n)$ (basically the space of all $L^2$-functions for which the first derivative exists in a weak sense).

A quantum field $\phi(x)$ is, according to the Wightman axioms, an operator-valued distribution. This means that for every test-function $f$ (usually a Schwartz space function), $$\phi(f) := \int_{\mathbb{R}^n} f(x) \phi(x) \ \mathrm{d} x$$ defines an operator on $\mathcal{H}$. The operator-valued distribution $\partial \phi$ is defined as the map that sends a test function $f$ to $\partial\phi(f) = -\phi(\partial f)$. Thus, $\partial \phi$ is not the composition of two operators (the derivative and the quantum field), but another operator-valued distribution.

Lecture notes that discuss some aspects of the mathematical framework of QFT: https://wdybalski.faculty.wmi.amu.edu.pl/Notes-QFT41.pdf

$\endgroup$
6
  • 1
    $\begingroup$ Could you explicitly type what $(\partial \phi)^\dagger$ can be rewritten as? $\endgroup$
    – Vangi
    May 13 at 1:29
  • 1
    $\begingroup$ If the field is hermitean, then, for every test function $f$, we have $\phi(f)^\dagger = \phi(\overline{f})$. Thus, $\partial \phi(f)^\dagger = - \phi(\overline{\partial f})$. $\endgroup$
    – Janik
    May 13 at 8:51
  • 1
    $\begingroup$ @Vangi Just follow the rules: $(\partial \phi)^\dagger= \phi^\dagger \partial^\dagger =\phi^\dagger \overset{_\gets}\partial = \partial \phi^\dagger$. $\endgroup$ May 13 at 13:30
  • $\begingroup$ @CosmasZachos So, you do agree with the answer which OP gave in his first link. The context he was talking in was $(\partial \phi)$, but gave (wrong) justification by writing down the transformation of $\partial$, which you have corrected in your comment on the question. $\endgroup$
    – Vangi
    May 13 at 14:42
  • $\begingroup$ @Vangi Sorry, I don't wish to get involved in cross-link "you know what I mean" conversations. I stand by all my formal statements. You obviously appreciate how misframed and misbegotten the entire conversation is. Any decent QM text clears up the hermiticity and application rules for the momentum operator, which only differs from the gradient by a -i factor... $\endgroup$ May 13 at 14:54

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.