How is a Qubit in two states under Superposition?

Question

I have read a little about Quantum computing.

From what I understand, Quantum Superposition is when a qubit is in a state $\alpha|0\rangle$ + $\beta|1\rangle$, where $\alpha$ and $\beta$ are probability amplitudes. Now when we try to measure or see the qubit it collapses to a 0 or 1.

Why don't we consider that the electron is in either the 0 or 1 state in superposition but we don't know which state because we haven't measured it yet.

I'll give you an example. I ask a friend to go inside a closed room and flip a coin. Now I know that it's either heads or tails with equal probability. So, can I say that the coin is in the superposition state $\sqrt0.5|heads\rangle$ + $\sqrt0.5|tails\rangle$ ?

I then ask my friend to dope the coin so it has a higher chance (0.75) of falling heads up. So, can I say that the coin is in a superposition $\sqrt0.75|heads\rangle$ + $\sqrt0.25|tails\rangle$ ?

Then I go inside the room and suddenly I see the coin. So can I say it's state collapsed to a heads or tails because I measured it?

Answer 1

Suppose I have an electron. There are various states it could be in. The states where we have a definite spin component are

$|z_{UP}\rangle$ z component of its spin is pointing upwards

$|z_{DOWN}\rangle$ z component of its spin is pointing downwards

$|x_{RIGHT}\rangle$ x component of its spin is pointing to the right

$|x_{LEFT}\rangle$ x component of its spin is pointing to the left

(we'll ignore the y direction). Now a Stern Gerlach apparatus gives us a means for measuring these spin components - if we orient the apparatus in the z plane its powerful magnets separate incoming electrons so that the ones it's measured as z-spin-up go upwards and the ones as z-spin-down go downwards.

Now suppose I have an incoming beam of electrons, where I've prepared them so that 50% of them are z-spin-up and 50% are z-spin-down. (I could do this by using a previous Stern Gerlach apparatus, measuring the z spins then recombining the beam). This is a mixture, not a superposition. If I now do my z measurement, I find 50% in z-spin-up and 50% in z-spin-down as I expect.

If, instead of the mixture, my electrons are in a true superposition state, say $$ \frac{1}{\sqrt{2}}(|z_{UP}\rangle + |z_{DOWN}\rangle)$$ Then when I do my measurement, I again measure 50% as z-spin-up and 50% as z-spin-down. These, however are not the same thing:

Suppose I reorient my Stern Gerlach apparatus to this time measure the x spin. To see what will happen you need the relations $$ |z_{UP}\rangle = \frac{1}{\sqrt{2}}(|x_{RIGHT}\rangle + |x_{LEFT}\rangle)$$ $$ |z_{DOWN}\rangle = \frac{1}{\sqrt{2}}(|x_{RIGHT}\rangle - |x_{LEFT}\rangle)$$ For the case where the incoming particles are in the z mixture, I'll get 50% x-spin-right and 50% x-spin-left.

For the superposition case though $$ \frac{1}{\sqrt{2}}(|z_{UP}\rangle + |z_{DOWN}\rangle)$$ $$=\frac{1}{\sqrt{2}} \{ \frac{1}{\sqrt{2}}(|x_{RIGHT}\rangle + |x_{LEFT}\rangle) + \frac{1}{\sqrt{2}}(|x_{RIGHT}\rangle - |x_{LEFT}\rangle)\}$$ $$=|x_{RIGHT}\rangle $$

So I'll get 100% x-spin-right.

In the coin example, we have a mixture, not a superposition. The reason macroscopic objects behave as mixtures rather than superpositions is generally explained by the phenomenon of decoherence.