# representing the actual state of the particle by adding up functions?

## Recommended Posts

What does it mean when it is saying representing the actual state of the particle by adding up a bunch of functions? This is in reference to eigenstates.

That particle has allowed states called eigenstates. Well it is not that particle is in any one of those states, but you can represent its actual state by combining all of them together. That the part of eigenstates I do not understand.

Where a general state of a particle is a mixture of these eigenstates. This is the part I do not understand.

##### Share on other sites
6 minutes ago, nec209 said:

Where a general state of a particle is a mixture of these eigenstates. This is the part I do not understand.

Eigenstates are like overtones or harmonics on a musical instrument.
They all add up to the create the individuality of the sound of a given note played on that particular instrument.

##### Share on other sites
Posted (edited)

Also, eigenstates are always associated to observables = properties that you can measure. When the particle has a definite value for that observable (dispersionless, it's called) then it is in one of those eigenstates, which is German for "proper state." But in a general situation, observables have dispersion (statistically speaking; they are "blurry" in that observable) so what quantum mechanics assigns to those instances is a linear superposition of all those eigenstates with a statistical "probability amplitude," which is the corresponding coefficient.  By squaring the absolute value of all the amplitudes in a range, you get the probability that the observable falls within that certain range.

I hope between Studiot's beautiful analogy and my attempt at telling what QM is in a nutshell, we've shed some light.

But QM is very abstract, that's for sure.

Edited by joigus

##### Share on other sites
13 hours ago, joigus said:

By squaring the absolute value of all the amplitudes in a range,

##### Share on other sites

"Dispersionless',  is that what it's called ?

I always thought it was the difference between 'independent' and 'dependent' probabilities.
Independent get multiplied, while dependent probabilities get added and the probability that multiple events happen is subtracted from that. And since in this case there is only one particle, there are no multiple ' event' probabilities to subtract.

##### Share on other sites
Posted (edited)
3 hours ago, MigL said:

"Dispersionless',  is that what it's called ?

I always thought it was the difference between 'independent' and 'dependent' probabilities.
Independent get multiplied, while dependent probabilities get added and the probability that multiple events happen is subtracted from that. And since in this case there is only one particle, there are no multiple ' event' probabilities to subtract.

Maybe the terminology is more popular among physicists who study the fundamentals of QM. I mean statistical dispersion is zero for that observable in than state. Dispersion, or mean square deviation, understood as,

$\sigma^{2}\left(x\right)=\left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}$

Well, also, independent events factor in their probabilities. There are crossed aspects here dependence/independence plus the AND and OR logical operators let's call'em.

So prob(E_1 OR E_2) = prob(E_1) + prob(E_2) if events E_1, E_2 are independent

and,

prob(E_1 AND E_2) = prob(E_1)xprob(E_2) if events E_1, E_2 are independent

Let's say E_1, E_2 are energies.

For 1 particle, it doesn't really make sense to say prob(particle has energy E_1 AND E_2) (at the same time)

But it makes perfect sense to say prob(particle has energy E_1 OR E_2). In that case, the amplitudes sum:

Amp(E_1 OR E_2) = Amp(E_1) + Amp(E_2)

and the probabilities in quantum mechanics have an interference term:

prob = Amp^2

so that,

prob(E_1 OR E_2) = |Amp(E_1)|^2 + |Amp(E_2)|^2 + 2Amp*(E_1)Amp*(E_2)

2Amp*(E_1)Amp*(E_2) is called the 'interference term.'

Edited by joigus

##### Share on other sites
Posted (edited)
1 hour ago, joigus said:

prob(E_1 OR E_2) = |Amp(E_1)|^2 + |Amp(E_2)|^2 + 2Amp*(E_1)Amp*(E_2)

2Amp*(E_1)Amp*(E_2) is called the 'interference term.'

I meant,

prob(E_1 OR E_2) = |Amp(E_1)|^2 + |Amp(E_2)|^2 + 2Re(Amp*(E_1)Amp(E_2))

2Re(Amp*(E_1)Amp(E_2)) is called the 'interference term.'

without the 'star' for complex conjugate in the second term and taking the real part.

Edited by joigus

##### Share on other sites
20 hours ago, joigus said:

By squaring the absolute value of all the amplitudes in a range, you get the probability that the observable falls within that certain range.

And this is why sycamore is incoherent so often. Can't even work without a giant fridge and even then it's not spot on, whether it's calculating an equation in 3 minutes that would take a normal computer 10,000 years, there is still no way to tell if the solution was accurate or not.

##### Share on other sites
2 hours ago, NineTwentyEight said:

And this is why sycamore is incoherent so often. Can't even work without a giant fridge and even then it's not spot on, whether it's calculating an equation in 3 minutes that would take a normal computer 10,000 years, there is still no way to tell if the solution was accurate or not.

Well...

1) I don't know who sycamore is

2) I haven't the faintest idea what super-computing power has to do with the OP topic or the follow-ups

3) You must be careful with what I said --in case it's anything at all to do with what you're saying--, because probabilities equal the sum of the probabilities is only true for different (orthogonal) eigenstates. You sound a bit hasty and shouldn't run away with any idea, especially from one thread to another.

##### Share on other sites
1 minute ago, joigus said:

Well...

1) I don't know who sycamore is

2) I haven't the faintest idea what super-computing power has to do with the OP topic or the follow-ups

3) You must be careful with what I said --in case it's anything at all to do with what you're saying--, because probabilities equal the sum of the probabilities is only true for different (orthogonal) eigenstates. You sound a bit hasty and shouldn't run away with any idea, especially from one thread to another.

Uncertainty always leads to decoherence in qubits. That's all I was trying to say. The quantum is a code which I do believe can be cracked, otherwise we HAVE to use probability mechanics until then. Sycamore is google's latest quantum computer breakthrough:

##### Share on other sites
3 minutes ago, NineTwentyEight said:

Uncertainty always leads to decoherence in qubits. That's all I was trying to say. The quantum is a code which I do believe can be cracked, otherwise we HAVE to use probability mechanics until then. Sycamore is google's latest quantum computer breakthrough:

Uncertainty is not the same thing as dispersion and in any case I think this discussion is wandering off topic.

## Create an account

Register a new account