Suppose I have a quantum state $\frac{1}{\sqrt{2}}|0\rangle+\frac{1}{\sqrt{2}}|1\rangle$. Also I have a mixture of two quantum states $S_{1} = |0\rangle$ and $S_{2} = |1\rangle$. In this mixture $50\%$ chance of getting $S_{1}$ and $50\%$ chance of getting $S_{2}$.

  1. If I give you these two situation one by one, is there anyway you can tell which is what?
  2. Also what is the difference between them?

1 Answer 1


The pure state $|\psi \rangle = \dfrac{|0\rangle + |1\rangle}{\sqrt{2}} $ is only in superposition in the computational basis, $\bigg\{ |0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, |1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \bigg\}$, but it is not in superposition with respect to another basis. In other words, it is a definite state. To see this, note that if we measured $|\psi \rangle$ in the $\{ |+\rangle, |-\rangle \} $ basis then the result is not probabilistic, where we defined $|+\rangle = \dfrac{|0 \rangle + |1\rangle}{\sqrt{2} }$ and $|-\rangle = \dfrac{|0 \rangle - |1\rangle}{\sqrt{2} }$, then we will see that it is in the state $|+\rangle$ 100% of the time.

And also note that if I applied a unitary transformation defined as $H = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix} $ to the state $|\psi \rangle$ then I will get back the $|0\rangle$ state. That is, $ H |\psi \rangle = |0\rangle $. Now upon measuring in the regular computational basis, $\{ |0\rangle, |1\rangle \}$, I will see that 100% I will see the state $|0\rangle$. All pure states are superpositions with respect to some basies and not with respect to others. Because the result of measuring a state in superposition is probabilistic (like the case $|\psi\rangle$ being measured in the computational basis), it is tempted to think the state $|\psi \rangle = a|0\rangle + b|1\rangle$ as a probabilistic mixture of $|0\rangle$ and $|1\rangle$ when it is NOT. The state, $|\psi\rangle$, is a definite state, and when we measured in certain bases, we will have deterministic results, while in other bases we will have random results.

Now, if I have a mixed state $\rho = \begin{pmatrix} 1/2 & 0\\ 0 & 1/2 \end{pmatrix}$ like what you described in your second case, then no matter what basis I measure it in, either the computational basis $\{ |0\rangle, |1\rangle \}$ or $\{ |+\rangle, |-\rangle \} $ the result is the same.

  • 2
    $\begingroup$ Thanks ...Now it's crystal clear to me. $\endgroup$
    – user27286
    Feb 8, 2021 at 8:01
  • $\begingroup$ Is it really fair to call the second example a mixed state? I would rather consider that an ensemble of states, where each of the states in the ensemble has a probability that is entirely independent from the state measurement. The 50% is his case are the probability of which of the states in the ensemble you are in. $\endgroup$
    – midor
    Feb 8, 2021 at 23:44

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