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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.

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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.

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  • $\begingroup$ I read this in a text that "This rule is easiest to remember in words: we say that in the case of an antilinear operator, it does matter whether the operator acts to the right or to the left in a matrix element, and if we change the direction in which the operator acts, we must complex conjugate the matrix element. In the case of antilinear operators, parentheses are necessary to indicate which direction the operator acts." Can you help me on it $\endgroup$ – Anshul Sharma May 21 at 14:28
  • $\begingroup$ @Anshul Sharma Well look at what I did above, $c_i$ and $d_j$ can be taken as the entries of matrix, and I got different matrix elements when $B$ acted on the right ket $\vert \psi\rangle$ or on the left bra $\langle \phi \vert\ $. $\endgroup$ – vin92 May 21 at 18:28
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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.

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  • $\begingroup$ I saw this expression to write antilinear operator as they dont behave like linear operator that $(\bra{\phi}B)\ket{\psi}=[\bra{\phi}(B\ket{\psi})]*$.Is it correct and what does it signify for an antilinear operator $\endgroup$ – Anshul Sharma May 21 at 14:31

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