# How one arrives at the Lüder's Rule?

When working with density matrices the collapse axiom of quantum mechanics says that a measurement of $S$, with result $s_i$, transforms the original density matrix $\sigma_{0}$ into the conditional density matrix $\sigma_{1}$ according to the Lüder's rule:

$$\sigma_{0}\rightarrow\frac{P_{s_i} \sigma_{0} P_{s_i}}{Tr(P_{s_i}\sigma_{0})}=\sigma_{1}(s_i)$$

where $P_{s_i}$ is the projector onto the state $|s_i\rangle$, $P_{s_i}=|s_i\rangle\langle s_i|$.

Now, I've been trying to understand the derivation of this rule but could not find it. Can you explain to me how it's obtained or at least refer me to some online literature on it?

Thank you very much.

How one arrives at the Lüder's rule?

By looking closely at the pure state case and generalizing.

Say system S in pure state $|\Psi\rangle$ is subject to a projective measurement onto the states $|s_i\rangle$. Assume set $\Big\{ |s_i\rangle \Big\}_i$ is complete, such that $\sum_i{|s_i\rangle \langle s_i |} = \sum_i{P_i} = I$ for $P_i = |s_i\rangle \langle s_i |$. The probability of a result $|s_i\rangle$ is, as usual, $p_i = | \langle s_i| \Psi\rangle|^2$. That is, a measurement applied to a large ensemble of N identical copies of S, all prepared initially in state $|\Psi\rangle$, produces on average $p_i N$ copies in state $|s_i\rangle$, and leaves the ensemble in a mixed state described by the density matrix $$\rho_\text{out} = \sum_i{p_i |s_i\rangle \langle s_i | }$$ At the ensemble level, the measurement takes as input the pure state $\rho_\text{in} = | \Psi\rangle \langle \Psi |$ and produces the mixed state $\rho_\text{out}$.

But let us look closer at each term in the output state. We can always write $$p_i = \langle s_i | \Psi\rangle \langle \Psi |s_i\rangle = Tr\left( P_i | \Psi\rangle \langle \Psi | P_i \right) = Tr\left( P_i \rho_\text{in} P_i \right)$$ and $$p_i |s_i\rangle \langle s_i | = |s_i\rangle p_i \langle s_i | = |s_i\rangle \langle s_i | \Psi\rangle \langle \Psi |s_i\rangle \langle s_i | = P_i | \Psi\rangle \langle \Psi | P_i = P_i \rho_\text{in} P_i = p_i \frac{P_i \rho_\text{in} P_i}{Tr\left( P_i \rho_\text{in} P_i \right)}$$ Then substituting back into the output state gives us Lüder's rule: $$\rho_\text{out} = \sum_i{P_i \rho_\text{in} P_i } \equiv \sum_i{p_i \frac{P_i \rho_\text{in} P_i}{Tr\left( P_i \rho_\text{in} P_i \right)} }$$ By analogy, the measurement may be said to produce state $P_i \rho_\text{in} P_i \;/\; Tr\left( P_i \rho_\text{in} P_i \right)$, $Tr\left[\; P_i \rho_\text{in} P_i \;/\; Tr\left( P_i \rho_\text{in} P_i \right) \;\right] =1$, with probability $p_i$. But does it really hold in the same way when the measurement is applied to an initial state that is itself mixed?

Yes, because as far as the ensemble is concerned, the measurement operation is linear on $\rho_\text{in} = | \Psi\rangle \langle \Psi |$, and any mixed state is ultimately a linear, albeit convex, superposition of pure states. So we can generalize straightforwardly for $$\rho_\text{in} = \sum_k{\bar{p}_k | \Psi_k\rangle \langle \Psi_k |}$$ Exercise: Check this last statement explicitly, on the ensemble output.