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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 $$Tr(\hat{P}_\lambda\hat{\rho}) = Tr(\hat{P}_\lambda\hat{\rho}\hat{P}_\lambda).$$

Qmechanic
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redpanda2236
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  • 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. – joseph h May 01 '23 at 01:40

1 Answers1

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  1. This is not derived, this is the Born rule or the Lüders-von Neumann measurement rule. Either way, it is a fundamental axiom of quantum mechanics that measurements work this way. See e.g. this, this or this question for more discussion of possible motivations.

    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 ($P^2 = P$).

ACuriousMind
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  • 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. – redpanda2236 May 01 '23 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? – redpanda2236 May 01 '23 at 10:13
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    @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 $\rho$. – ACuriousMind May 01 '23 at 10:25