# Reading off probabilities for measurement outcome rather than using projection operator?

Let $$\alpha_{0}$$ = $$\alpha_{1}$$ = $$\frac{1}{\sqrt{2}}$$.

Suppose the state vector $$| \psi \rangle = \alpha_{0}| \psi_{0} \rangle + \alpha_{1} |\psi_{1}\rangle$$ describes a quantum mechanical system prior to observation.

The density operator $$\hat{\rho}$$ for the state $$| \psi \rangle$$: $$| \psi \rangle \langle \psi |$$ = $$|\alpha_{0}|^{2} | \psi_{0} \rangle \langle \psi_{0} | + |\alpha_{1}|^{2} || \psi_{1} \rangle \langle \psi_{1} |$$ = [1/2 0; 0 1/2] says that the probability for the state vector $$| \psi \rangle$$ collapsing to $$| \psi_{0} \rangle$$ or $$| \psi_{1} \rangle$$ upon measurement is probability $$\frac{1}{2}$$, $$\frac{1}{2}$$, respectively.

Taking $$trace(\hat{\rho}) = 1$$ says that the total probability associated with the outcome of measurement for state $$| \psi_{0} \rangle, | \psi_{1} \rangle$$ is $$1$$.

So let $$m = {m_{0}, m_{1}}$$ denote the measurement outcome of the state $$| \psi \rangle$$, then Born's rule says

$$prob(m_{i}|\psi\rangle) = trace(\hat{P_{m_{i}}} \hat{\rho}) = 1, \forall i \in {1, 2}$$.

Is it correct to understand $$\hat{P_{m_{0}}} \hat{\rho}, \hat{P_{m_{1}}} \hat{\rho}$$ as stating the probability of each outcome of measurement of $$\hat{\rho}$$? But if this is true, why not read the probabilities off rather than invoking the projection operator $$\hat{P_{m_{i}}}$$?

My hunch says that the projection operator is useful for computing the probabilities associated with state $$| \psi_{0} \rangle, | \psi_{1} \rangle$$ under a different set of basis vectors (using a different measurement basis to project the $$| \psi_{0} \rangle, | \psi_{1} \rangle$$ onto a subspace spanned by different set of basis vectors) but I am not sure how correct this is.

Any insights is appreciated.

Reading off the probabilities arises from the fact that you have a pure state that is expressed in the basis you measure in. So, in this case, $$P_i = |\psi_i\rangle\langle \psi_i|$$ and $$\rho = |\psi\rangle\langle \psi|$$. The probability of measuring $$i$$ is given by $$p(i)=\text{tr}(P_i\rho)=\text{tr}(|\psi_i\rangle\langle \psi_i||\psi\rangle\langle \psi|)=|\langle \psi_i|\psi\rangle|^2 = |a_i|^2$$ where we used the cyclic property of the trace. But when you're measuring in a different basis, or you're studying a mixed state, you cannot immediately get the probabilities without performing any calculation.