Do we really mean Hermitian conjugate here?

1. Feb 23, 2016

ShayanJ

When people want to find a conserved current which is constructed from a Dirac spinor, they consider the Dirac equation and its "Hermitian conjugate". But the equations they consider are $(i\gamma^\mu \partial_\mu -m)\psi=0$ and $\bar{\psi}(i\gamma^\mu \overleftarrow{\partial_\mu}+m)=0$ where $\bar\psi=\psi^\dagger \gamma^0$.
But its obvious that they're assuming that $(i\partial_\mu)^\dagger=-i\partial_\mu$. But this is actually the naive way of finding the Hermitian conjugate of an operator because we know that the definition of the Hermitian conjugate of an operator is $\langle A x,y\rangle=\langle x,A^\dagger y \rangle$ and we also know that by this definition, the operator $i\partial_k$ is actually Hermitian. But I don't know about $i\partial_0$ because the inner product doesn't contain a time integral.
Can anybody clarify this?
Thanks

2. Feb 23, 2016

dextercioby

But nobody is questioning the hermiticity of a derivative operator, but the hermitean adjoint of a sum of 4x4 matrices $i \gamma^0 \partial_0 + ...$. To find this, you need to find the complex conjugate matrix and then transpose it.

3. Feb 23, 2016

ShayanJ

The point is, the operator $\partial_k$ by itself is not Hermitian. Its the operator $i\partial_k$ that is Hermitian.

4. Feb 24, 2016

Demystifier

To talk about hermiticity one first needs to specify the vector space on which hermiticity is defined. The confusion comes from the fact that here we have two vector spaces. One is the 4-dimensional spinor space, another is the infinite-dimensional Hilbert space of square-integrable functions. The derivative operator $\partial_k$ is anti-hermitian (so that $i\partial_k$ is hermitian) on the latter space, but not on the former one. The hermitian conjugate of the Dirac equation refers to the hermitian conjugate in the former space.

I hope this explanation will not create another confusion: Lorentz spinors are not Lorentz vectors, so how can spinors be objects in a vector space? If someone finds it confusing, here is the hint: 4-dimensional spacetime is one thing, n-dimensional space on which SO(1,3) group is represented is another. Spinor is not a vector in the former space, yet it is a vector in the latter space (with n=4 for the Dirac equation).

It seems that a mistake of confusing different vector spaces can happen even to authors of well-known QFT textbooks:

Last edited: Feb 24, 2016
5. Feb 24, 2016

stevendaryl

Staff Emeritus
The relevant $\dagger$ operator here is just $A^\dagger = (A^*)^T$. So it's not the same as the Hermitian conjugate used in Hilbert space.

In the derivation of the conjugate equation for a Dirac field, you don't need Hermitian conjugates, you just need $(A^*)^T$.

6. Feb 24, 2016

PeroK

To be even more careful mathematically, whether an operator is Hermitian also depends on the inner product you're using.

7. Feb 24, 2016

stevendaryl

Staff Emeritus
Right. If your Hilbert space is just 4-component complex column matrices, then $(A^*)^T$ is the Hermitian conjugate. But then you can't ask about whether $i \partial_\mu$ is Hermitian, because that is not an operator on that particular Hilbert space.

8. Feb 24, 2016

ShayanJ

That was really enlightening. Thanks man!

9. Feb 24, 2016

Demystifier

You are welcome.

If that was enlightening, perhaps the following will also be:
http://lanl.arxiv.org/abs/1309.7070 [Eur. J. Phys. 35, 035003 (2014)]