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.


1 Answer 1


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.


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