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I quantum mechanics we can represent the state of a system $\vert\psi\rangle$ in some Hilbert space as a complete set of basis vectors $\vert n\rangle$;

\begin{equation} \vert\psi\rangle=\sum_n^Nc_n\vert n\rangle \end{equation}

having an inner product $c_n=\langle\psi\vert n\rangle$ from which we can determine the probability of the energy belonging to each basis vector as $P(E_n)=|c_n|^2$. Now, in quantum mechanics, we also have a state operator, $\rho$, that must fulfill three criteria that I will not list here. Having some operator $R$ representing some observable, we can determine the probability of each outcome of the operator $R$ in this state operator by

\begin{equation} \text{Prob}(R=r|\rho) = \langle r_n |\rho | r_n \rangle \end{equation}

with $r_n$ expressing the (non-degenerate) eigenstates belonging to specific eigenenergies of the operator $R$. My question is; is the method for determining the probability of each state using coefficients equivalent to using the state operator?

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  • $\begingroup$ Do you want to know whether knowing the state (operator/wave function) is equivalent to knowing "all probabilities"? If so, it is a duplicate: See here $\endgroup$ Oct 10, 2023 at 12:40

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The short answer is: Yes these two ways are equivalent. Let's look at the reason for why that is:

Let's assume we have some Observable $A$ with a discrete spectrum and a basis of eigenvectors $| a_n \rangle$ where the eigenvector $|a_n\rangle$ has the eigenvalue $a_n$. Let's also assume that our system is described by a pure state $|\psi\rangle$. Then we can write \begin{equation} |\psi\rangle = \sum_{n} c_n |a_n\rangle \end{equation} with some coefficients $c_n$. If we choose $A$ to be the Hamiltonian of the system, then $|a_n\rangle = |n\rangle$ are the energy eigenstates. In this case the equation is the same as your first equation. Now if we measure $A$, the probability for measuring the eigenvalue $a_m$, which corresponds to the eigenstate $|a_m\rangle$, is (in the nondegenerate case) \begin{equation} |\langle a_m|\psi \rangle|^2 = \left| \sum_n c_n \langle a_m | a_n \rangle \right|^2 = \left| \sum_n c_n \delta_{nm} \right|^2 = |c_m|^2. \end{equation} If we choose the Hamiltonian as our observable $A$, this is the probability of measuring the energy corresponding to $|a_m\rangle = |m\rangle$, which you also wrote in your post.

Now let's look at the formalism employing the density matrix. For our state $|\psi\rangle$ the density matrix is $\rho = |\psi\rangle\langle\psi |$. If we measure $A$, the probability for measuring the eigenvalue $a_m$ is \begin{equation} \langle a_m |\rho| a_m \rangle = \langle a_m |\psi\rangle \langle \psi| a_m \rangle = |\langle a_m |\psi \rangle |^2 = |c_m|^2. \end{equation} In the last step I used the fact that we already calculated $|\langle a_m |\psi \rangle |^2$ earlier in this post. So as you see, we get exactly the same result as before.

One last thing: Note that the density matrix formalism is more general in the sense that not every density matrix corresponds to a pure state. Therefore the second method can always be applied but the first method can only be applied if the state is pure. However, if the state is pure, the two methods are equivalent.

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  • $\begingroup$ Thank you for the very clear explanation. $\endgroup$ Oct 12, 2023 at 11:35
  • $\begingroup$ You're very welcome! Thank you for the feedback :) $\endgroup$
    – WillHallas
    Oct 12, 2023 at 12:26
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We can always write an operator $A$ in the form $$ A = \sum_{n,m} A_{n,m} | r_n \rangle\langle r_m | $$

Applying this to the density operator we have \begin{align} \mathrm{Prob}(R=r_i | \rho) &= \langle r_i | \rho | r_i \rangle\\ &= \sum_{n,m} \rho_{n,m} \langle r_i | r_n \rangle\langle r_m | r_i \rangle\\ &= \rho_{i,i} \end{align}

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