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I have been reading QFT for the Gifted Amateur, on page 110, it reads:

"For the case of the real scalar field...it is Hermitian. For our complex scalar field there is no reason why the field operator should be Hermitian."

Is this because, the Lagrangian Density for a real scalar field doesn't have any adjoints, and so when trying to use to calculate the Hamiltonian Operator (which is Hermitian), the scalar field would also need to be Hermitian? Whereas for a complex scalar field, it's Lagrangian Density has adjoints in it and so it is not required for the field to be Hermitian, as the no-Hermitian components can be 'cancelled out', and so the Hamiltonian operator would be Hermitian?

EDIT: What I mean is, the Lagrangian density for a complex scalar field would be something like: $$ L=(∂^{μ}ψ)^{†}(∂_{μ}ψ)-m^{2}ψ^{†}ψ $$ Since it is guaranteed that all the terms are Hermitian (since ψ† ψ is Hermitian), then there is no need for the field to be Hermitian. However, the Langrangian for a real scalar field, would be something like: $$L=\frac{1}{2}(∂_{μ}ϕ)^{2}-\frac{1}{2}(mϕ)^{2}$$ Since it isn't guranteed that the terms will be Hermitian due to the lack of terms adjoints, ϕ must be Hermitian.

My Question is: Is this reasoning correct?

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  • $\begingroup$ It's just because the complex scalar field involves creation and annihilation operators that are not adjoints of each other. Or are you asking why this is the case from "first principles"? $\endgroup$ Feb 26, 2021 at 3:43
  • $\begingroup$ Yes, without referring to creation and annihilation operators. $\endgroup$
    – Matrix001
    Feb 26, 2021 at 3:52
  • $\begingroup$ Possible duplicate: here. $\endgroup$ Feb 26, 2021 at 4:02
  • $\begingroup$ complex conjugation maps to the adjoint operator upon quantization. $\endgroup$
    – Prahar
    Feb 26, 2021 at 14:31

2 Answers 2

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I have been reading QFT for the Gifted Amateur, on page 110, it reads:

"For the case of the real scalar field...it is Hermitian. For our complex scalar field there is no reason why the field operator should be Hermitian."

Is this because...

You seem to have already accepted that a real scalar field (which is just a real number at every point in space) is mapped into a Hermitian operator at every point in space as we move from the classical to the quantum theory.

So, I will just show why the complex scalar field is not mapped to a Hermitian operator.

Let $\phi$ be a complex scalar field that gets mapped to some operator $\hat\phi$ as we go from classical to quantum field theory. We want to show that $\hat \phi$ is not Hermitian.

Since $\phi$ is complex, we can write it as $$ \phi = \phi_1 + i\phi_2\;, $$ where $\phi_1$ and $\phi_2$ are real.

So, going over to operator-valued fields, we have: $$ \hat\phi = \hat \phi_1 + i \hat \phi_2\;, $$ where $\hat\phi_1$ and $\hat \phi_2$ are Hermitian.

Taking the Hermitian conjugate of $\hat \phi$ we find: $$ \hat \phi^\dagger = \hat \phi_1^\dagger - i \hat\phi_2^\dagger $$

But, since $\hat \phi_1$ and $\hat \phi_2$ are Hermitian, we have $$ \hat \phi^\dagger = \hat \phi_1 - i \hat\phi_2\;, $$ which is clearly not equal to $\hat \phi$: $$ \hat \phi - \hat \phi^\dagger = 2i\hat\phi_2 $$ (unless $\hat\phi_2$ is zero everywhere, but in that case we are back at the case of a real scalar field)

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Quantization maps complex-valued functions over the phase space to operators acting on the Hilbert space.

Real functions over the phase space are a special case. They are mapped to Hermitian operators.

The value of the complex-valued field at a point is a complex-valued function over the phase space (the space of solutions to the eom). Hence it is mapped to a non-Hermitian operator.

(Caveat: to make this mathematically precise, the field value needs to be smeared with some test function rather than taken at a single point in spacetime, but this detail doesn’t matter for your question.)

The value of the real-valued field at a point is a real-valued function on the phase space. After quantization it becomes a Hermitian operator.

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