I apologise in advance if I say something very stupid, I feel a bit out of my depth here. So according to wikipedia if I have a system prepared in a state given by a density matrix $$\rho=\sum_{j} p_{j}\left|\psi_{j}\right\rangle\left\langle\psi_{j}\right|$$ and I want to measure some observable $A$ then I can measure the probability of getting a measurement $m$ for the observable $A$ using the following:

$$p(m)=\sum_{j} p_{j}\left\langle\psi_{j}\left|\Pi_{m}\right| \psi_{j}\right\rangle=\operatorname{tr}\left[\Pi_{m}\left(\sum_{j} p_{j}\left|\psi_{j}\right\rangle\left\langle\psi_{j}\right|\right)\right]$$

where $\Pi_{m}$ is the projector. Now I know that when $\Pi_{m}$ acts on a basis state $\mid \psi_j \rangle$ it gives $\langle\phi_m \mid \psi_j\rangle$ where $\mid \phi_m \rangle$ is the eigenstate of $A$ with eigenvalue $m$. So if I wanted to find the matrix which gives this projector $\Pi_{m}$ then I just construct a matrix whose rows are the conjugate transpose of $\mid \phi_m \rangle$... Is this correct?


1 Answer 1


Not quite: the projector is given by the outer product $|\phi_m\rangle\langle \phi_m|$. In vector notation, if we know that $$|\phi_m\rangle=\sum_{j=1}^N c_j |\psi_j\rangle$$ then we can write the former in the $|\psi_j\rangle$ basis as $$|\phi_m\rangle=\begin{pmatrix}c_1\\c_2\\\vdots\\c_{N}\end{pmatrix}$$ and similarly $$\langle\phi_m|=\begin{pmatrix}c_1^*&c_2^*&\cdots&c_{N}^*\end{pmatrix}.$$ The outer product in that basis looks like $$|\phi_m\rangle\langle \phi_m|=\begin{pmatrix}c_1\\c_2\\\vdots\\c_{N}\end{pmatrix}\begin{pmatrix}c_1^*&c_2^*&\cdots&c_{N}^*\end{pmatrix}=\begin{pmatrix} c_1c_1^*&c_1c_2^*&\cdots&c_1c_{N}^*\\ c_2c_1^*&c_2c_2^*&\cdots&c_2c_{N}^*\\ \vdots&\vdots&\ddots&\vdots\\ c_Nc_1^*&c_Nc_2^*&\cdots&c_Nc_{N}^* \end{pmatrix}.$$ So yes, the rows deal with the conjugate transpose of $|\phi_m\rangle$, but you must multiply each by the correct element from the column vector version of $|\phi_m\rangle$.


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