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I want to find the hermitian conjugate of 4-derivative $\partial_\mu$ for the real scalar Lagrangian defined as

$$\mathcal{L} = \frac{1}{2} (\partial_\mu \phi)^\dagger(\partial^\mu\phi) - \frac{1}{2}m^2(\phi^\dagger\phi)-V(\phi^\dagger\phi),$$ where I use the Minkowski sign convention $(+,-,-,-)$.

I am confused that when I find the hermitian conjugate of $\partial_\mu$ by the usual procedure of calculating expectation value, I get $\partial_\mu^\dagger = -\partial_\mu$, but in Dirac equation we used $\partial_\mu^\dagger = +\partial_\mu$. So I want to know that when will be there plus sign with hermitian conjugate and when will be there minus sign and what what will be the hermitian conjugate for the above given Lagrangian.

I found a similar question Hermitian adjoint of 4-gradient in Dirac equation , but there is no satisfactory answer to the more general question I asked.

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Short answer : In QFT, $\partial_\mu$ is not an operator in the quantum mechanics sense. It does not really make sense to take its hermitian adjoint.

Long answer : In non-relativistic mechanics, the Hilbert space is the space of square integrable wave-function on euclidean space (or on more general configuration spaces). The gradient $\nabla$ is an operator on this space of wave function. Since the Hermitian product is : $$(\psi,\phi) = \int \psi^*\phi \text dx$$

The usual integration by part shows that the hermitian adjoint of $\nabla$ is $-\nabla$.

Now, in QFT, the field are not wave-functions. Rather $\phi(x)$ is an operator-valued (more precisely a distribution). It's gradient is defined just like for any function : $$h^\mu \partial_\mu \phi(x) = \lim_{\epsilon \to 0} \epsilon^{-1}(\phi(x+\epsilon h) - \phi(x))$$

For a real scalar field, $\phi(x)$ is a hermitian operator for every $x$. Therefore, the formula above gives : $$(\partial_\mu \phi(x)) ^\dagger = \partial_\mu \phi(x)$$

Note that, had we chosen $\phi(x)$ to be anti-hermitian (as in a purely imaginary field), we would have $\phi(x)^\dagger = -\phi(x)$ and $(\partial_\mu \phi(x))^\dagger = -\partial_\mu\phi(x)$.

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  • $\begingroup$ Can the derivative of operator valued distributions actually be defined rigorously? $\endgroup$
    – Charlie
    May 24, 2021 at 12:52
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    $\begingroup$ Sure, $\int \nabla \phi f = - \int \phi \nabla f$ for any test function $f$. The result is the same $\endgroup$ May 24, 2021 at 13:01
  • $\begingroup$ Ah sure, makes sense ty. $\endgroup$
    – Charlie
    May 24, 2021 at 13:06

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