# Understanding the probability of measurement w.r.t. density matrix

I am told that the probability of measuring $$\lambda$$ is $$p_\lambda = Tr(\hat{P}_\lambda\hat{\rho}) = Tr(\hat{P}_\lambda\hat{\rho}\hat{P}_\lambda)$$ where $$\hat{P}_\lambda = \sum_{n:\lambda_n = \lambda}|n\rangle\langle n|$$ is the projection operator for eigenstates $$n$$ with an eigenvalue $$\lambda$$.

I have no idea how this is derived or why $$Tr(\hat{P}_\lambda\hat{\rho}) = Tr(\hat{P}_\lambda\hat{\rho}\hat{P}_\lambda).$$

• It's simply definition. As such, there is no derivation. Whatever property is represented by the projector $\hat P$ then $\rm Tr (\rho\hat P)$ represents the probability of observing this property upon measurement. Commented May 1, 2023 at 1:40

If you are fine with the Born rule for pure states and wonder where this rule for mixed states comes from, write the mixed state as a mixture of pure states $$\rho = \sum_i p_i \lvert \psi_i\rangle\langle \psi_i\rvert$$ and consider that the probability to measure the eigenvalue $$\lambda$$ for some observable $$\Lambda$$ with eigenstates $$\lvert \lambda\rangle$$ in the state $$\lvert \psi_i\rangle$$ would be $$\lvert \langle \lambda\vert \psi_i\rangle \rvert^2$$ by the Born rule and $$\mathrm{tr}(\lvert \lambda\rangle \langle \lambda\vert \psi_i\rangle\langle \psi_i\rvert) = \lvert\langle \lambda\vert \psi_i \rangle\rvert^2,$$ so $$\mathrm{tr}(P_\lambda \rho) = \sum_i p_i \lvert\langle \lambda\vert \psi_i \rangle\rvert^2,$$ is just the probability to measure $$\lambda$$ in each of the $$\psi_i$$, weighted by their mixed probability $$p_i$$.
2. $$\mathrm{tr}(P\rho P) = \mathrm{tr}(P^2\rho) = \mathrm{tr}(P\rho)$$, where the first equality is due to the cyclicity of the trace ($$\mathrm{tr}(ABC) = \mathrm{tr}(CAB)$$ for any operators $$A,B,C$$) and the second because projections are idempotent ($$P^2 = P$$).
• So my understanding now is that the Born rule tells us that $p = \langle \psi | \lambda \rangle \langle \lambda | \psi \rangle$ But I am unsure on why this is equal to $tr( | \lambda \rangle \langle \lambda | \psi \rangle \langle \psi |)$. I see that $tr(| \phi \rangle \langle \psi |) = \langle \psi | \phi \rangle$, but I can't find a proof online. Commented May 1, 2023 at 10:04
• Also, one final addition. I understand now why it can be rewritten as $Tr(P_\lambda \rho P_\lambda)$ because it $P_\lambda$ is indempotent and the trace is cylic, thank-you. But why would we want to do this? Is it because $U \rho U^{\dagger}$ is the form of a unitary time evolution? Commented May 1, 2023 at 10:13
• @redpanda2236 Just write out the definition of the trace in an orthonormal basis $\lvert b_i\rangle$: $\mathrm{tr}(\lvert \lambda \rangle\langle \psi\rvert) = \sum_i \langle b_i\vert \lambda\rangle\langle \psi\vert b_i\rangle = \langle \psi\vert\left( \sum_i \vert b_i\rangle\langle b_i\vert\right)\vert \lambda\rangle$ and $\sum_i \vert b_i\rangle\langle b_i\vert = 1$ by definition of an orthonormal basis. As for why you'd want to add a $P_\lambda$, it might indeed be because $P_\lambda\rho P_\lambda$ is how the projector would act "normally" on $\rho$. Commented May 1, 2023 at 10:25