Associative product of two Anti-Linear(/Unitary) Operator

Question

An operator is said to be linear if it obeys the distributive law and commutes with the constant i.e.

$\hat{A}(a_1 |\psi_1\rangle + a_2|\psi_2\rangle)=a_1\hat{A}|\psi_1\rangle +a_2\hat{A}|\psi_2\rangle$ and same when we act in from bra. The product of two operator are not commutative (not all case) but their product are associative.

So in evaluating $\langle\phi|\hat{A}|\psi\rangle$, it does not matter if one first applies $\hat{A}$ to the ket and then takes the bra-ket or one first applies $\hat{A}$ to the bra and then takes the bra-ket i.e.

$(\langle\phi|\hat{A})|\psi\rangle=\langle\phi|(\hat{A}|\psi\rangle)$.

My question is does this associative property holds for linear operator too as in many text i have read they never talked about how anti-linear operator commutes as such and does it matter as stated above for linear operators that the order dosent matter as such.

Edit:- I know the property of anti-linear operator that it makes the constant complex when an operator applies on it. Would love answer's if the discussion is based on the notation of bra and ket as i am comfortable on this notation.

Answer 1

Recall that in the Dirac bra-ket notation $\langle \phi |=|\phi\rangle^{\dagger}$ and $\langle \phi |\hat{A}=(\hat{A}^{\dagger}|\phi\rangle)^{\dagger}$, so what OP calls associativity $(\langle \phi |\hat{A})|\psi\rangle=\langle \phi |(\hat{A}|\psi\rangle)$ is really the defining property of the adjoint operator $\hat{A}^{\dagger}$.

Similarly, the associativity $(\langle \phi |\hat{A})\hat{B}=\langle \phi |(\hat{A}\hat{B})$ is equivalent to the rule $(\hat{A}\hat{B})^{\dagger}=\hat{B}^{\dagger}\hat{A}^{\dagger}$.

See also this related Phys.SE post. For the existence of the adjoint operator for an antilinear operator, see this Phys.SE post.

Answer 2

Does $(\langle{\phi}\vert \hat{A})\vert \psi \rangle=\langle{\phi}\vert (\hat{A}\vert \psi \rangle)$ hold for linear operators? Yes it does. A simple way to convince yourself is to think in terms of vectors, forms and matrices and use a simple example like this:

$\begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 2 & 0 \end{pmatrix} \begin{pmatrix} 3 \\ -1 \end{pmatrix}= \begin{pmatrix} 1 & -1 \end{pmatrix} \begin{pmatrix} 3 \\ -1 \end{pmatrix}= 4$

Or

$\begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 2 & 0 \end{pmatrix} \begin{pmatrix} 3 \\ -1 \end{pmatrix}= \begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} 4 \\ 6 \end{pmatrix}= 4$

And of course $A=\begin{pmatrix} 1 & -1 \\ 2 & 0 \end{pmatrix}$ is a linear operator.

Conerning the sentence "how anti-linear operator commutes as such and does it matter as stated above for linear operators that the order dosent matter as such." Both linear and anti-linear operators do not commute in general. Finally let us see if $(\langle{\phi}\vert \hat{B}\vert) \psi \rangle=\langle{\phi}\vert (\hat{B}\vert \psi \rangle)$ holds for anti-linear operator $B$. Supose that $\psi$ and $\phi$ can be written as (finite) linear combination of basis vectors ${\vert i \rangle}$ of a $n$-dimensional Hilbert space with complex coeficients i.e $\vert\psi\rangle=\displaystyle\sum_{i=1}^n c_i \vert i\rangle$ and $\vert\phi\rangle=\displaystyle\sum_{j=1}^n d_j \vert j\rangle$. Then we have: $(\langle{\phi}\vert \hat{B})\vert \psi \rangle=\left(\displaystyle\sum_{j=1}^n \langle j\vert d_j^* B\right)\displaystyle\sum_{i=1}^n c_i \vert i\rangle= \displaystyle\sum_{i=1}^n\displaystyle\sum_{j=1}^n c_id_j \langle j\vert B \vert i\rangle$. On the other hand: $\langle{\phi}\vert (\hat{B}\vert \psi \rangle)=\displaystyle\sum_{j=1}^n \langle j\vert d_j^* \left(B\displaystyle\sum_{i=1}^n c_i \vert i\rangle\right)= \displaystyle\sum_{i=1}^n\displaystyle\sum_{j=1}^n c_i^*d_j^* \langle j\vert B \vert i \rangle$. So in general the answer is no.