# Keeping the spin constant during an experiment?

In summary, quantum computing relies on keeping the electron in a fixed state long enough to achieve a desired result. However, this is more demanding than simply creating a lot of particles and assuming that statistically a certain percentage will be in the desired state.

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In texts about quantum computing, one prepares electrons or other particles in definite spins (up or down), then pass them through various gates, etc. But since the probability that an electron changes spin state within a very short time, then how can this be realistically done? Is there a way to keep the electron in a fixed state long enough? Or does one simply create a lot of particles and assume that statistically a certain percentage will be in the desired state? Am I just stating the problem called decoherence?

But since the probability that an electron changes spin state within a very short time
Why should it?
Also, what is a very short time for you?

This has some relation to decoherence, but "keeps the spin orientation" is more demanding than "does not lead to decoherence"

If the initial spin configuration of your system is an eigenstate of the Hamiltonian then it should remain the same.

Thank you, mfb and Einj, for your replies.
First, to mfb
mfb said:
Why should it?
As I understand it, the electron has a definite spin value only upon being measured. So, after measurement, it would seem that it would have a high probability of being in some other state than upon being measured. Apart from that, the system evolves with time, doesn't it?
mfb said:
Also, what is a very short time for you?
From one point of view, right after being measured; it would of course also depend on its environment: for example, however, in terms of flipping from an indefinite state to a definite one, in the environment of a human brain, Max Tegmark comes up with a value of about 10-13 seconds.
mfb said:
This has some relation to decoherence, but "keeps the spin orientation" is more demanding than "does not lead to decoherence"
Yes, I see that now. Thank you for pointing that out.

To Einj
Einj said:
If the initial spin configuration of your system is an eigenstate of the Hamiltonian then it should remain the same.
Alas, I do not completely understand. If the eigenstate of the Hamiltonian has two eigenvalues, then supposedly each one has a non-zero expectation value, so why should it continue to be one rather than the other? I have no doubt that your answer is correct, but I am missing the connection between the premise and the conclusion. I would be grateful if you could spell it out for me.

As I understand it, the electron has a definite spin value only upon being measured.
No. It has to have one when it is measured, but it does not lose it afterwards (or at least not necessarily). This definite spin refers to a single axis only, and that is a key point if you want to make other measurements with it afterwards.
Apart from that, the system evolves with time, doesn't it?
So does the solar system system and we are still revolving around sun in the same direction we always did, because all influences on it are tiny.

Alas, I do not completely understand. If the eigenstate of the Hamiltonian has two eigenvalues, then supposedly each one has a non-zero expectation value, so why should it continue to be one rather than the other? I have no doubt that your answer is correct, but I am missing the connection between the premise and the conclusion. I would be grateful if you could spell it out for me.

You are not considering how a state evolves with time in Quantum Mechanics. Suppose that your system is described by an Hamiltonian H and that at the initial time t=0 it is in a state ##|\psi(t=0)\rangle=|\psi_0\rangle##. Then your state for any time is given by:

$$|\psi(t)\rangle=e^{iHt/\hbar}|\psi_0\rangle.$$

Now, suppose that at t=0 your system is in a definite eigenstate of the Hamiltonian, say for example spin-up such that ##H|up\rangle=E_{up}|up\rangle##, where ##E_{up}## is the energy of the spin-up state. Then at any given time you'll have:
$$|\psi(t)\rangle=e^{iHt/\hbar}|up\rangle=e^{iE_{up}t/\hbar}|up\rangle,$$
so, you state only picks up a phase, i.e. it will remain in the spin-up configuration forever.