# Very basic question about quantum field operators

For a matrix $A$, the notation $A^\dagger$ implies the transpose of the complex conjugate of $A$ i.e., $A^\dagger=(A^*)^T$.

What does the symbol $\hat{\phi}^\dagger$ mean for a quantum operator corresponding to a classical field $\phi(x)$? Is it okay to think of $\hat{\phi}(x)$ as an infinite dimensional column vector and $\hat{\phi}^\dagger$ as a row vector with $\hat{\phi}^\dagger=(\hat{\phi}^*)^T$?

However, there are two problems that I can immediately see.

1. Operators in ordinary quantum mechanics are square matrices while (if my representation is valid) $\hat{\phi},\hat{\phi}^\dagger$ are column and row vectors.

2. For a complex scalar field $$[\hat{\phi}(t,\textbf{x}),\hat{\phi}^\dagger(t,\textbf{y})]=0\implies \hat{\phi}(t,\textbf{x})\hat{\phi}^\dagger(t,\textbf{y})=\hat{\phi}^\dagger(t,\textbf{y})\hat{\phi}(t,\textbf{x}).$$ If my representation is valid, this equation becomes meaningless because on one side we have number and on the other side we have a matrix.

What is is the correct way to visualize quantum field and interpret the commutation relation?

• You should give up on the matrix interpretation. It doesn't really hold in infinite dimensions. Quantum fields are operators, and $\dagger$ stands for the (Hermitian) adjoint of an operator, under the assumption (which for quantum fields is always valid) that the domain of the dagger-less operator is dense everywhere in the Fock space. – DanielC Oct 10 '17 at 10:04
• Why would you think of the field operator as a vector, when it's an operator? – knzhou Oct 10 '17 at 10:22
• In not so rigorous way, Quantum field operators can be thought of as matrices in Fock space. Then, Hermitian conjugation can easily be visualized. The source of confusion seems to be your interpretation of $x$ as a label for vector components, this quickly disappears if it is thought of as labeling for operators in some algebra. – Sunyam Oct 10 '17 at 14:28

Don't think about matrices. Let $\mathcal{H}$ be a Hilbert space, then one linear operator in $\mathcal{H}$ is a function $A : D(A)\subset \mathcal{H}\to \mathcal{H}$ which satisfies

$$A(\alpha v+\beta w)=\alpha A(v)+\beta A(w),\quad\forall\alpha,\beta\in\mathbb{C},v,w\in\mathcal{H}.$$

This isn't a matrix. The adjoint $A^\dagger$ is defined as the unique linear operator in $\mathcal{H}$ satisfying:

$$\langle Av,w\rangle=\langle v,A^\dagger w\rangle,\quad v,w\in \mathcal{H}.$$

It is also not a matrix. When you introduce a basis on $\mathcal{H}$ you can associate a matrix to $A$ and $A^\dagger$ inasmuch as every other operator, but matrices are matrices and operators are operators. It is them a result that the matrices of $A$, $A^\dagger$ are related by

$$[A^\dagger]=([A]^\ast)^T.$$

Now, in QFT field are operator valued distributions. For now think of them as actual functions on spacetime. Thus for any event $x\in M$, a quantum field $\phi$ associates an operator in some Hilbert space $\mathcal{H}$ denoted $\phi(x)$.

Since $\phi(x)$ is an operator in some Hilbert space it has one adjoint $\phi(x)^\dagger$ in the exact same manner as I said above.

By the way, when you use the Fock space representation (which is the standard one in QFT textbooks) you have that by definition

$$\phi(x)=\int \dfrac{d^3k}{(2\pi)^3}\dfrac{1}{\sqrt{2\omega_k}}(a_ke^{-ikx}+a_k^\dagger e^{ikx})$$

where $a_k$ and $a_k^\dagger$ are creation and annihilation operators in a certain Fock space (which is a space of states of a system of a variable number of particles).

Interestingly now $a_k$ has the meaning of removing a particle with momentum $k$ from a state, $a_k^\dagger$ has the meaning of creating a particle with momentum $k$ and this is due to the commutation relations they should satisfy in order for the Quantum Field satisfy the Canonical Commutation relations, namely

$$[a_k,a_q]=0,\quad [a_k^\dagger,a_q^\dagger]=0,\quad [a_k,a_q^\dagger]=(2\pi)^3 \delta(k-q).$$

But this is a particular case. Adjoints are adjoints, as defined above in the mathematical setting of Hilbert spaces.

The dagger $\dagger$ can have two meanings.

First, there is the linear algebra meaning. We use $\dagger$ to denote a Hermitian conjugate, that is transposition and complex conjugation.

Second, there is the quantum field operator meaning. Here, $\dagger$ is used to change a creation operator into an annihilation operator $-$ and vice versa.

To answer your question: if $\phi(x)$ is a scalar quantum field, then $\phi^\dagger(x)$ has the QFT-meaning.

Fun fact: the only ambiguity arises when dealing with spinors, $$\chi = \begin{pmatrix} \chi_1 \\ \chi_2\end{pmatrix},$$ because what does $\chi^\dagger$ mean? Is it the quantum field operator interpretation, $$\chi^{\dagger,LA} = \begin{pmatrix} \chi_1^\dagger \\ \chi_2^\dagger \end{pmatrix},$$ or is the Hermitian conjugate applied in both the operator and the linear algebra sense, as in $$\chi^{\dagger,QFT+LA} = \begin{pmatrix} \chi_1^\dagger , \chi_2^\dagger \end{pmatrix}?$$ This depends on the convention chosen by the author.

• This is wrong, there are no different meanings - the "quantum field operator meaning" is exactly the same as the "linear algebra meaning", both denote the Hermitian adjoint. – ACuriousMind Oct 10 '17 at 14:15
• Then how would you explain the ambiguity with spinors? – Stephan Oct 10 '17 at 21:47
• That's just a difference in convention as to whether when applying the adjoint to a vector of operators, you additionally transpose the vector or not. It has nothing to do with the adjoint itself. – ACuriousMind Oct 12 '17 at 11:31

That for a matrix the dagger denotes the transpose conjugate is really just a special (namely the finite-dimensional) case of the general definition of the Hermitian adjoint:

For any operator $A$ on a Hilbert space $H$, the adjoint $A^\dagger$ is the operator such that $$\langle v, Aw\rangle = \langle A^\dagger v,w\rangle$$ for all $v,w$ in the domain of definition of $A$.

Since for any quantum field $\phi$, $\phi(x)$ is an operator (neglecting the case where we treat the field as an operator-valued distribution rather than a function, replace $\phi(x)$ with $\phi(f)$ for some test function $f$ in that case), there is no problem in applying this definition to a quantum field.