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Given a qubit in state $\mid 0\rangle$, first find the probabilities of the outcomes from measuring it in the basis of $\mid a\rangle =\dfrac{1}{\sqrt{2}}(1,i)$ and $\mid b\rangle =\dfrac{1}{\sqrt{2}}(1,-i)$. Following the first measurement, the qubit is measured again in the basis of $\mid 0\rangle ,\mid 1\rangle$. What are the probabilities of the outcomes?

For the first half of the question, I applied Born's Law to find the two outcomes of $\mid a\rangle$ and $\mid b\rangle$, both with a probability of $50\%$. However, I am slightly confused by the second part. Am I supposed to apply Born's Law again? If so, how should I account for the previous measurement?

Random Person
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1 Answers1

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Assuming $|0\rangle=\begin{pmatrix}1\\0\end{pmatrix}$ and $|1\rangle=\begin{pmatrix}0\\1\end{pmatrix}$ the probabilities you are looking for are \begin{align}\tag{1} |\langle 0|a\rangle|^2,\quad|\langle 0|b\rangle|^2,\quad|\langle 1|a\rangle|^2,\quad|\langle 1|b\rangle|^2 \end{align} which all are equal to $\frac{1}{2}$. These are the probabilities of the outcomes of the first measurement.

After this measurement each state $|0\rangle$ and $|1\rangle$ has collapsed to either $|a\rangle$ or $|b\rangle$. If you do a second measurement in the basis $\{|0\rangle,|1\rangle\}$ the probabilities of the outcomes are again $\frac{1}{2}$ by symmetry of the expressions in (1).

Kurt G.
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