# Superposition State in Coin Toss

I was reading the following lines from Quantum Computation and Quantum Information by Nielsen and Chuang on page 278 of chapter 7.

A coin has two states and makes a good bit but a poor qubit because it cannot remain in a superposition state(of 'heads' and 'tails') for very long

My question is how do I define a general superposition state in case of a coin toss?

• $|\text{coin} \rangle = \frac{1}{\sqrt{2}} \left(|\text{heads} \rangle + |\text{tails} \rangle\right)$ – innisfree Dec 22 '15 at 5:58
• but will the probability amplitudes be same all the time? – Subhadip Roy Dec 22 '15 at 6:02
• Obviously no. As you evolve you act it upon the states by $e^{iHt}$ and since it is a classical object you need to take $\hbar \rightarrow 0$ limit. – Jaswin Dec 22 '15 at 6:04
• @Jaswin: $\hbar$ is a constant. Nature doesn't care about us changing it. Such a change in theory doesn't lead to a correct physical description of reality, even though older books are (unfortunately) trying to suggest that it's a useful manipulation. What makes the behavior of the coin classical is that it entangles with its environment, which leads to constant weak measurements of its position/attitude, whether we are actively observing it, or not. We simply can't keep it from suffering decoherence. – CuriousOne Dec 22 '15 at 6:08
• @Jaswin Actually I meant to say whether the initial state will be an equal superposition of heads and tails to begin with. – Subhadip Roy Dec 22 '15 at 6:09

You can write the superposition state as $$\psi = a\uparrow + b\downarrow$$where $a^2 + b^2 = 1$. But on evolving with a large Hamiltonian, going from one state to another becomes negligible.
• Sorry for being lazy, I meant $|H| >> \hbar$, the magnitude of a classical hamiltonian times t is way bigger than $\hbar$. – Jaswin Dec 22 '15 at 6:39
• But $H$ and $\hbar$ don't even have the same unit – innisfree Dec 22 '15 at 11:17