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Old Jan 27th 2016, 04:45 PM   #11
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Originally Posted by MBW View Post
My previous post was leading up toward asking if UP always involves a time verses space combination,
or to put it another way, is UP fundamentally an issue related to space-time combinations?
Heisenberg's UP is a direct consequence of the (pure) mathematics of Schrodinger's equation or the paired operators.

For two observables A and B which have representative operators a and b in the governing equation (a and b are what PMB means by the momentum wave function) there are two possibilities.

a and b commute (may be taken in any order) in the equation

Then the two observables may be measured simultaneously.

If they do not commute the equation does not possess a complete set of eigenfunctions in common to both observables.
Then measurement of one observable will affect the value of the other.

This is all very general and leads to the Schwarz inequality in pure maths (there are other equations with this property even in classical mechanics).

Heisenberg's UP can be derived from the Schwarz inequality, if you want to look this up.

Remember that Schrodinger's equation is an equation of motion not directly momentum, position, time or energy.
Look at the dimensions of the wavefunction psi.
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Old Jan 27th 2016, 08:24 PM   #12
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Originally Posted by studio
For two observables A and B which have representative operators a and b in the governing equation (a and b are what PMB means by the momentum wave function) there are two possibilities.
That doesn't make sense to me. You defined a and be to be operators and as such they're not wave functions. I will say this: when choosing a representation to work with one must also use the corresponding representation of the operator.

Originally Posted by studio
a and b commute (may be taken in any order) in the equation
Why do you say this? Only certain pairs of operators commute. Not all operators. E.g. The operator x does not commute with the operator p_x whereas the operators x and y do commute.

Originally Posted by studio
If they do not commute the equation does not possess a complete set of eigenfunctions in common to both observables.
What equation are you referring to?

Originally Posted by studio
Then measurement of one observable will affect the value of the other.
l very general and leads to the Schwarz inequality in pure maths (there are other equations with this property even in classical mechanics).

Heisenberg's UP can be derived from the Schwarz inequality, if you want to look this up.

Originally Posted by studio
Remember that Schrodinger's equation is an equation of motion not directly momentum, position, time or energy.
Look at the dimensions of the wavefunction psi.
The Schrodinger's equation (SE) is only an equation of motion in the position representation. The SE describes how Psi evolves in time and in doing so describes how the probability of finding an observable changes with time. The dimensions of Psi depends on the representation.
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Old Jan 27th 2016, 08:28 PM   #13
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Originally Posted by MBW View Post
My previous post was leading up toward asking if UP always involves a time verses space combination,
or to put it another way, is UP fundamentally an issue related to space-time combinations?
No. Not at all. There is a general uncertainty inequality for any two incompatible observables. First go to:

https://en.wikipedia.org/wiki/Uncertainty_principle

Then scroll down to read Robertson–Schrödinger uncertainty relations
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Old Feb 9th 2016, 06:51 AM   #14
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There is a general uncertainty inequality for any two incompatible observables.
I think this ambit is too wide.

For instance two (what does incompatible mean? ) observables might be volume and distance from the centre of the Moon.
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Old Feb 9th 2016, 07:36 AM   #15
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Originally Posted by studiot View Post
I think this ambit is too wide.

For instance two (what does incompatible mean? ) observables might be volume and distance from the centre of the Moon.
What does "ambit" mean and why is it too large?

Two observables A and B are incompatible is [A, B] is not zero.
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Old Feb 9th 2016, 09:32 AM   #16
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Originally Posted by Pmb View Post
What does "ambit" mean and why is it too large?

Two observables A and B are incompatible is [A, B] is not zero.
https://www.google.co.uk/?gws_rd=ssl#q=ambit

Because it literally allows me to select any two observables at opposite ends of the universe and apply your statement.

Your statement [A, B] is a standard inner product notation.
If it is meant to be an inner product how is

the inner product [Force, Distance] incompatible?

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Old Feb 9th 2016, 02:52 PM   #17
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I've been looking at some of the links included in this thread,
Most of the descriptions seem either rather vague and woolly or involve mathematical descriptions that I am not able to (completely) follow.

However would I be correct that the indeterminacy issue is an inescapable consequence of using a wave-form description?
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Old Feb 9th 2016, 03:49 PM   #18
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However would I be correct that the indeterminacy issue is an inescapable consequence of using a wave-form description
Basically, yes.


Schrodinger's equation is a not quite wave equation but is a similar equation also connecting three variables. (in one dimension).


The dependent variable is usually denoted phi for the true (classical) wave equation and psi for the Schrodinger one.

The two independent variables in both equations are position and time.

Since these are independent variables, they must be mathematically (not temporally) simultaneous in position and time.

Mathematically simultaneous means that the equation leads to a pair of simultaneous equations between the dependent variable and each independent variable separately.

This leads to the Schwartz inequality (Heisenberg as a special case for Schrodinger) for these two variables.

This inequality basically happens because we obtain a different result if we solve the equation between phi or psi and time first and substitute the result into the second between phi (or psi) and position from if we do it the other way round. The magnitude of this difference is given by the inequality.

This is what is meant by non commutativity of operators.

Last edited by studiot; Feb 9th 2016 at 03:52 PM.
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Old Feb 13th 2016, 02:15 AM   #19
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Does the non-commutativity have an analogue in macroscopic waves (sound waves, water waves etc.)?
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Old Feb 14th 2016, 04:38 AM   #20
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Originally Posted by MBW View Post
Does the non-commutativity have an analogue in macroscopic waves (sound waves, water waves etc.)?
Yes indeed there are examples.

Here is an obvious one without the maths.

There is a minimum bandwidth dw that an amplifier must possess in order to successfully amplify input pulses of duration dt.
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