Adjoint of the derivative in Minkowsky space.

[carllacan]

I'm sorry if it is a silly question, but I've looked elsewhere and I haven't found the answer.

Please tell me if any of the following is true:

∂_μ^† = -∂_μ
∂_i^† = -∂_i
∂^μ† = -∂^μ
∂^i† = -∂ⁱ

I'm studying QFT and I don't know how to proceed when I have to take the adjoint of a derivative. Sometimes it looks like one of these has to be true, sometimes it doesn't, and I don't know what to think anymore.

Thanks for your time

 

[stevendaryl]

I'm sorry if it is a silly question, but I've looked elsewhere and I haven't found the answer.

Please tell me if any of the following is true:

∂_μ^† = -∂_μ
∂_i^† = -∂_i
∂^μ† = -∂^μ
∂^i† = -∂ⁱ

I'm studying QFT and I don't know how to proceed when I have to take the adjoint of a derivative. Sometimes it looks like one of these has to be true, sometimes it doesn't, and I don't know what to think anymore.

Thanks for your time

The adjoint is defined with respect to a notion of a scalar product. Confusingly, there is more than one scalar product involved in QFT. One is the scalar product of two states: Given $|\psi\rangle$ and $|\phi\rangle$, you can form the scalar $\langle \psi|\phi \rangle$. A second notion of scalar product is the scalar product of two column matrices, which comes up in theories of particles with spin: If $\psi = \left( \begin{array} \\ \psi_1 \\ \psi_2 \\ ... \\ \psi_n \end{array} \right)$ and $\phi = \left( \begin{array} \\ \phi_1 \\ \phi_2 \\ ... \\ \phi_n \end{array} \right)$, then there is a scalar product $\psi^\dagger \phi = \psi_1^* \phi_1 + \psi_2^* \phi_2 + ... + \psi_n^* \phi_n$

So before you start talking about what the adjoint is, you need to specify what scalar product you're talking about. Sorry for not directly answering, but it's important to be clear about this. If you have a notion of scalar product, $\langle \psi | \phi \rangle$, and $A$ is an operator, then $A^\dagger$ is defined via:

$\langle A(\psi) | \phi \rangle = \langle \psi | A^\dagger (\phi) \rangle$

In non-relativistic, spinless QM, states are square-integrable functions on space. The scalar product of two states $\psi$ and $\phi$ is given by:

$\langle \psi| \phi \rangle = \int dV \psi^* \phi$ (where $dV$ means integrating over all spatial coordinates). For this notion of scalar product, $\frac{\partial}{\partial x^j}$ is an operator whose adjoint is $- \frac{\partial}{\partial x^j}$:

$\langle \frac{\partial \psi}{\partial x^j}| \phi \rangle = \int dV (\frac{\partial \psi}{\partial x^j})^* \phi = \int dV \psi^* (- \frac{\partial \phi}{\partial x^j})$

 

[carllacan]

Ok, so what about QFT?

Would the answer depend on which fields ∂_μ is acting on? Now that I think about it I'm not sure how the scalar product is defined in QFT. Is it the same as in non-relativistic QM?

 

[haushofer]

You should give an example. In QFT, usually the fields are regarded as operators and the coordinates as just labels. So taking the conjugate of these kinds of expressions (products of fields and derivatives of fields) does not affect the derivative. The conjugate in that case only acts on the operators, which are the fields.

E.g., are you referring to the action leading to the Dirac eqn, or something similar?

 

[carllacan]

You should give an example. In QFT, usually the fields are regarded as operators and the coordinates as just labels. So taking the conjugate of these kinds of expressions (products of fields and derivatives of fields) does not affect the derivative. The conjugate in that case only acts on the operators, which are the fields.

E.g., are you referring to the action leading to the Dirac eqn, or something similar?

Yes, I was working with the Dirac equation.

 

[stevendaryl]

Yes, I was working with the Dirac equation.

As I said in my first post, there are two different notions of "adjoint" involved in the Dirac equation, because there are two different notions of a scalar product. Given two Dirac spinors $\psi$ and $\phi$, you can form $\psi^\dagger \phi$. With this notion of scalar product, the operators are 4x4 matrices. But there is also the integrated scalar product:

$\langle \psi | \phi \rangle = \int dx dy dz \bar{\psi} \phi$ (where $\bar{\psi} \equiv \psi^\dagger \gamma^0$).

With this notion of scalar product, $\partial_x, \partial_y$ and $\partial_z$ are Hilbert-space operators with adjoints, but $\partial_t$ is not.

 

[stevendaryl]

With this notion of scalar product, $\partial_x, \partial_y$ and $\partial_z$ are Hilbert-space operators with adjoints, but $\partial_t$ is not.

This is a little bit of a confusing point, and maybe somebody can explain it better than I can. The claim is that $\partial_x$ is a Hilbert-space operator with an adjoint, but $\partial_t$ is not. This seems wrong, because Schrödinger's (or Dirac's) equation has the form: $i \partial_t \psi = H \psi$. So it seems as if $\partial_t = -i H$. So $\partial_t$ should have an adjoint, since $H$ does. But that reasoning is not quite correct:

$\partial_t \psi = -i H \psi$

This is not an operator identity for all elements of the Hilbert space. It's just a constraint on $\psi$ The Hilbert space basis elements are time-independent, so they don't satisfy this constraint.