Error : what is n in quantum mechanics

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Suppose I have an operator A. Its average is <A> and the standard deviation $$\sigma=\sqrt {<A^2>-<A>^2} $$.
I now want the standard error which is $$\sigma/\sqrt {n} $$.

I wondered what n is in quantum mechanics ? The wsvefunction is supposed to describe a single particle so it should be 1 ? Or shall we take an ensemble interpretation ?
 

Demystifier

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n is the same as in standard statistics: the number of observations.
 
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So it is not possible to compute the error from the axioms, it is an experimental data ?
 

stevendaryl

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So it is not possible to compute the error from the axioms, it is an experimental data ?
Yes, the standard error is about the error you get from using only finitely many examples to compute your statistics. For example, suppose we are flipping coins, and we give "Heads" the value +1 and "Tails" the value -1. You'd expect the average of many flips to be 0, because you'd get the same number of heads as tails. But you won't get precisely zero, typically. You'll get something like [itex]0 \pm \sigma/\sqrt{n}[/itex] where [itex]\sigma[/itex] is the standard deviation (for this experiment, I think the standard deviation is 1). So the bigger [itex]n[/itex] is, the closer the average will be to the theoretical average of 0.
 
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How do we derive this square root ?

In quantum mechanics if we repeat a measurement we then get always the same outcome because the state is an eigenstate. But practically we cannot repeat a measurement because the particle like a photon is absorbed ?
 

Nugatory

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How do we derive this square root ?
In quantum mechanics if we repeat a measurement we then get always the same outcome because the state is an eigenstate. But practically we cannot repeat a measurement because the particle like a photon is absorbed ?
You start with an ensemble of identically prepared systems and perform a single measurement on each member of the ensemble.

(As an aside, repeated measurements produce the same result only if the observable in question commutes with the Hamiltonian. Immediately after the measurement the system will indeed be in an eigenstate of that observable, but it will only stay there if the observable commutes with the Hamiltonian).
 

stevendaryl

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How do we derive this square root ?

In quantum mechanics if we repeat a measurement we then get always the same outcome because the state is an eigenstate. But practically we cannot repeat a measurement because the particle like a photon is absorbed ?
The [itex]\frac{\sigma}{\sqrt{n}}[/itex] doesn't have anything to do with quantum mechanics; it's just statistics.

Some facts about variances, which I'll give you without proof:
  1. If [itex]X[/itex] is a real-valued random variable, and [itex]c[/itex] is a real-valued constant, then [itex]var(c X) = c^2 var(X)[/itex]
  2. Let [itex]X_1, ..., X_n[/itex] be [itex]n[/itex] independent measurements of the same random variable, [itex]X[/itex]. Let [itex]T = X_1 + ... + X_n[/itex] be the sum of all the results. Then [itex]var(T) = n\ var(X)[/itex].
  3. The standard-deviation is just the square-root of the variance.
So putting these facts together: Let [itex]A = T/n[/itex] be the average of [itex]n[/itex] independent measurements of the same random variable [itex]X[/itex]. Then

[itex]var(A) = var(T/n) = var(T)/n^2 = (n\ var(X))/n^2 = var(X)/n[/itex]

Take the square root to get the standard deviation:
[itex]\sigma(A) = \sqrt{var(A)} = \sqrt{var(X)}/\sqrt{n} = \sigma(X)/\sqrt(n)[/itex]
 
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Maybe the question goes for simple probabilities too : the probabilities obtained by quantum mechanics are thus not in the form n (a)/N ? It is not a frequentist approach ?
 
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Maybe the question goes for simple probabilities too : the probabilities obtained by quantum mechanics are thus not in the form n (a)/N ? It is not a frequentist approach ?
Frequentest, Bayesian - it makes no difference - the strong law of large numbers connects the two. Personally I take the axiomatic view but this is not the place to go into it.

Its an ensemble of identically prepared systems and the usual laws of probability are applied. The classic standard text is by Feller:
https://www.amazon.com/dp/0471257117/?tag=pfamazon01-20

Also has some very good comments about mathematics and its relation to applications at the start - worth seeking out for that alone.

Thanks
Bill
 
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560
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Frequentest, Bayesian - it makes no difference - the strong law of large numbers connects the two. Personally I take the axiomatic view but this is not the place to go into it.

Its an ensemble of identically prepared systems and the usual laws of probability are applied. The classic standard text is by Feller:
https://www.amazon.com/dp/0471257117/?tag=pfamazon01-20

Also has some very good comments about mathematics and its relation to applications at the start - worth seeking out for that alone.

Thanks
Bill
So the number of identical systems tends to infinity and the standard error is always zero ?
 
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vanhees71

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So the number of identical systems tends to infinity and the standard error is always zero ?
This is misleading. The point is that the preparation of any true state of a particle implies that the standard deviation of position is ##\Delta x=\sqrt{\langle x^2 \rangle -\langle x \rangle ^2}>0##.

To measure it you have to prepare very many particles independently from each other in this state and measure the position much more precisely (i.e., with a much higher position resolution) than given by the standard deviation due to the state. Then making the ensemble size very large your measured standard deviation, will tend to the quantum mechanical ##\Delta x##.
 
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So we don't have the quantum mechanical $$\Delta x $$ divided by $$\sqrt {n} $$ as the number of identical system increases ? The error is simply the quantum mechanical delta x ?

It seems to me it is like if we had two ways of calculating the error either delta x by quantum mechanics or take an average over a sample of size n and calculate the standard deviation delta x divided by square root of n ?
 

stevendaryl

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So we don't have the quantum mechanical $$\Delta x $$ divided by $$\sqrt {n} $$ as the number of identical system increases ? The error is simply the quantum mechanical delta x ?

It seems to me it is like if we had two ways of calculating the error either delta x by quantum mechanics or take an average over a sample of size n and calculate the standard deviation delta x divided by square root of n ?
I think you're misunderstanding the relationship between the two numbers. Nothing you're talking about is specific to quantum mechanics. Let's look at something much simpler, coin flips. If we assign "heads" the value +1 and "tails" the value -1, then for a fair coin, we have:

[itex]\langle V \rangle = 0[/itex] (on the average, you'll get as many heads as tails, so the average value is zero)
[itex]\langle V^2 \rangle = 1[/itex] (the square of the value is always 1)
[itex]\sigma(V) = \sqrt{\langle V^2 \rangle - \langle V \rangle^2} = 1[/itex] The standard deviation is 1.

So we have a theoretical random variable with average 0 and standard deviation 1.

Now, if we flip lots of coins, and compute the values for each, then we can come up with an experimental estimate of the average: Add up the values, and divide by the number of coin flips.

This experimental average will not be precisely zero. It will deviate from zero by a certain amount. If the coin is fair, then the deviation will be

[itex]\Delta(V)_{exp} = \sigma(V)/\sqrt{n} = 1/\sqrt{n}[/itex]

where [itex]n[/itex] is the number of coin flips.

[itex]\Delta(V)_{exp}[/itex] and [itex]\sigma(V)[/itex] are related, but they aren't the same. If you wanted to figure out, empirically, what [itex]\sigma(V)[/itex] is, you could use the estimate:

[itex]\sigma(V) = \sqrt{n} \Delta(V)_{exp}[/itex]
 
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So in quantum mechanics we can compute $$\sigma (V) $$ but not $$\Delta (V) $$ ?
 

stevendaryl

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So in quantum mechanics we can compute $$\sigma (V) $$ but not $$\Delta (V) $$ ?
In what I wrote, [itex]\Delta(V)[/itex] is an experimental result. You have to actually do an experiment to get it.
 
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In your calculation you get Delta as a function of n without doing an experiment.
 

stevendaryl

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In your calculation you get Delta as a function of n without doing an experiment.
I'm sorry if I was confusing. Let me go through it once more.

We have a theory (such as quantum mechanics) that allows us to compute the mean and standard-deviation of some quantity, [itex]V[/itex]:
[itex]\langle V\rangle_{theory}[/itex] is the mean and [itex]\sigma(V)[/itex] is the standard deviation.

Now, we perform an experiment to test that theory.
  • We measure [itex]V[/itex] [itex]n[/itex] times and get measurement results [itex]v_1, v_2, ..., v_n[/itex].
  • We compute an experimental mean: [itex]\langle V \rangle_{exp} = \frac{1}{n} (v_1 + v_2 + ... + v_n)[/itex]
  • If the theory is correct, the experimental mean should be close to the theoretical mean. Specifically, we should expect that [itex]\langle V \rangle_{exp} = \langle V \rangle_{theory} \pm \Delta(V)[/itex] where [itex]\Delta(V) = \frac{\sigma(V)}{\sqrt{n}}[/itex]
  • If the experimental mean is not in that range, then the theory is not looking too good. We can try increasing [itex]n[/itex] and seeing if it gets better, but if it consistently fails the above test, then the theory is likely wrong.
 
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So here in a quantum experiment we mix theoretical sigma with an experimental n ?
 

stevendaryl

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So here in a quantum experiment we mix theoretical sigma with an experimental n ?
As I said, it doesn't have anything specifically to do with quantum mechanics. If you have a theory that allows you to compute a theoretical mean and standard deviation, then you test that theory by computing an experimental mean, and comparing it with the theoretical mean.
 

jtbell

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We have a theory (such as quantum mechanics) that allows us to compute the mean and standard-deviation of some quantity, VV:
⟨V⟩theory\langle V\rangle_{theory} is the mean and σ(V)\sigma(V) is the standard deviation.
To be more explicit, one could call these the "expected mean" and "expected standard deviation" (which are calculated from theory), as opposed to the "experimental mean" and "experimental standard deviation" (which are calculated from experimental data). Note that <V> is usually called the "expectation value of V" in QM.
 

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