Understand Uncertainty Principle in a Week

[pmb_phy]

Gentlemen;
Listening to you talk is better than any course I've had at University,(peer to my best course in Quantum ). Thanks

You're welcome. We're glad to be of help.

Pete

 

[kahoomann]

I learned about this in class and some parts have been bugging me for a while. I am to understand that because were dealing with such small and minute measurements, that if one were to attempt to measure values that the experiment would mess up the values read. That it is "physically" impossible to measure both values at once.

This is a common misconception of uncertainty principle that if one were to attempt to measure values that the experiment would mess up the values read.
It happens to classical systems too, Not just quantum systems. For example, if you measure the temperature of water, its temperature would change due to the measurement process.
This measurement has nothing to do with uncertainty principle.

 

[pmb_phy]

This is a common misconception of uncertainty principle that if one were to attempt to measure values that the experiment would mess up the values read.
It happens to classical systems too, Not just quantum systems. For example, if you measure the temperature of water, its temperature would change due to the measurement process.
This measurement has nothing to do with uncertainty principle.

It appears that another common misconception is the definition of the uncertainty of an observable. By definition the uncertainty in the physical observable A is given by

\Delta A =\sqrt{\langle(A-\langle A\rangle)^2\rangle}

Therefore the value of the uncertainty of a physical observable is therefore completely determined by the state that the system is in. In the case of the physical observable x we have

\Delta X =\sqrt{\langle(X-\langle X\rangle)^2\rangle}

where X is the operator corresponding to the position eigenvalue x. Notice that this has absolutely nothing to do with the error \delta x in the measured position of a particle. There is no reason why \delta x can't be zero. Currently there is no known lower bound for \delta x. Note that \delta x and \Delta x represent very different quantities and that they are unrelated to each other in all but the most cursory way (i.e. both address position, otherwise they are not related).

Pete

 

[pmb_phy]

It appears that another common misconception is the definition of the uncertainty of an observable. By definition the uncertainty in the physical observable A is given by

\Delta A =\sqrt{\langle(A-\langle A\rangle)^2\rangle}

Therefore the value of the uncertainty of a physical observable is therefore completely determined by the state that the system is in. In the case of the physical observable x we have

\Delta X =\sqrt{\langle(X-\langle X\rangle)^2\rangle}

where X is the operator corresponding to the position eigenvalue x. Notice that this has absolutely nothing to do with the error \delta x in the measured position of a particle. There is no reason why \delta x can't be zero. Currently there is no known lower bound for \delta x. Note that \delta x and \Delta x represent very different quantities and that they are unrelated to each other in all but the most cursory way (i.e. both address position, otherwise they are not related).

Pete

I forgot to mention that uncertainty also goes by other names, namely root-mean-squared deviation and standard deviation. It should be noted that uncertainty only has a statistical meaning and has no relevance to sinlge observations other than the fact that one uses single data to compile a set of data which is then used to calculate the uncertainty.

Pete

 

[pmb_phy]

It appears that another common misconception is the definition of the uncertainty of an observable. By definition the uncertainty in the physical observable A is given by

\Delta A =\sqrt{\langle(A-\langle A\rangle)^2\rangle}

Therefore the value of the uncertainty of a physical observable is therefore completely determined by the state that the system is in. In the case of the physical observable x we have

\Delta X =\sqrt{\langle(X-\langle X\rangle)^2\rangle}

where X is the operator corresponding to the position eigenvalue x. Notice that this has absolutely nothing to do with the error \delta x in the measured position of a particle. There is no reason why \delta x can't be zero. Currently there is no known lower bound for \delta x. Note that \delta x and \Delta x represent very different quantities and that they are unrelated to each other in all but the most cursory way (i.e. both address position, otherwise they are not related).

Pete

I must be getting old because it took me until now to realize that there is a very simple example which will illustrate the point I've been trying to make.

Consider an electron which goes through a spin analyzer and which is initially in the state (let a = 1/sqrt(2))

|\Psi> = a|+> + a|->

Where |+> is an eigenket corresponding to the operator S_z, i.e. it represents an electron which is the "up" spin state. Similary the |-> eigenket is also an eigenket corresponding to the operator S_z, but which represents an electron in the "down" spin state. The eigenvalue corresponding to |+> is hbar/2 and that corresponding to |-> is -hbar/2. IT is to be noted that a measurement of S_z can only yield two possible values, i.e. hbar/2 and -hbar/2, each of which is measured exactly (no error in measured value of single electron). The uncertainty in S_z is found to be hbar/sqrt(2). This is a clearcut example of where there is no error in the measured physical observable but for which there is a finite, non-zero, value of the uncertainty in S_z.

Pete

 

[pmb_phy]

It appears that another common misconception is the definition of the uncertainty of an observable. By definition the uncertainty in the physical observable A is given by

\Delta A =\sqrt{\langle(A-\langle A\rangle)^2\rangle}

In case someone wishes to know where this comes from I scanned and uploaded the definition from Quantum Mechanics, Cohen-Tannoudji et al. The two pages are at

http://www.geocities.com/physics_world/uncertainty_01.jpg
http://www.geocities.com/physics_world/uncertainty_02.jpg

This same definition appears in ever single text that I've ever read in which \Delta X is explicitly defined. It is evident from those pages that the Heisenberg Uncertainty Principle is really a misnomer since a principle is something which, by definition, cannot be derived from more elementary postulates. However the uncertainty relation is a derivable relationship.

Pete