# What is the difference between $|0\rangle+|1\rangle$ and a balanced mixture of $|0\rangle$ and $|1\rangle$? [duplicate]

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?

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.