Heisenberg uncertainty principle

In summary: It seems like it varies from book to book but in general shouldn't this be the same? Say someone posed a question like this online, how would I know what to use?When you have:\Delta x \Delta p = \hbarThe \Delta's do not mean the standard deviation, but instead the deviation which covers 50% of the possible values. When we have:\Delta x \Delta p = \frac{\hbar}{2}It requires that \Delta be the standard deviation.AFAIK, both of these are only actually correct when x and p are normally distributed.Hope this clears that up for you.
  • #1
ovoleg
94
0
Hey guys I have a few questions...

My book defines the Heisenberg uncertainty principle by

∆x∆Px >= aitch-bar

All other resources I have have it stated as
∆x∆Px >= aitch-bar/2. They mention that ∆Px and ∆x represent the rms values of independent measurements.

My book represents ∆x and ∆Px as the standard-deviation uncertanties right..

So say for instance you get a general question like x-cordinate of a proton is measured with uncertainty of 1.3mm. What is the xcomponent of velocity to the minimum percentage of uncertainty of 33%.

would you take ∆x as the standard deviation uncertanties or rms values ?

It seems like it varies from book to book but in general shouldn't this be the same? Say someone posed a question like this online, how would I know what to use?
 
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  • #2
When you have:

[tex]\Delta x \Delta p = \hbar[/tex]

The [itex]\Delta[/itex]'s do not mean the standard deviation, but instead the deviation which covers 50% of the possible values. When we have:

[tex]\Delta x \Delta p = \frac{\hbar}{2}[/tex]

It requires that [itex]\Delta[/itex] be the standard deviation.

AFAIK, both of these are only actually correct when x and p are normally distributed.

Hope this clears that up for you.
 
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  • #3
Jheriko said:
When you have:

[tex]\Delta x \Delta p = \hbar[/tex]

The [itex]\Delta[/itex]'s do not mean the standard deviation, but instead the deviation which covers 50% of the possible values. When we have:

[tex]\Delta x \Delta p = \frac{\hbar}{2}[/tex]

It requires that [itex]\Delta[/itex] be the standard deviation.

AFAIK, both of these are only actually correct when x and p are normally distributed.

Hope this clears that up for you.

I'm still a bit confused as to how both of them could be correct?

Makes sense that if delta(x) is one standard deviation then in the first one it covers half the values, but then how are these two equivalent?

Also, rms does not equal a standard deviation as far as I know...

I'm confused :/
 
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  • #4
ovoleg said:
Makes sense that if delta(x) is one standard deviation then in the first one it covers half the values, but then how are these two equivalent?

Also, rms does not equal a standard deviation as far as I know...

I'm confused :/

First off I made some nasty typos in my tex and the equalses should be greater than or equal tos.

[tex]\Delta x \Delta p & \ge \hbar[/tex]


[tex]\Delta x \Delta p & \ge \frac{\hbar}{2}[/tex]

Secondly, the important point is that in the first equation the delta does NOT represent a standard deviation. A standard deviation covers some percentage of the values that I can never remember... ~68% apparently. The required deviation for the first form uses the size of the range which covers
50% of the values. Its just confusing that the same symbol gets used, it might be better to have

[tex]\Delta_{50\%} x \Delta_{50\%} p & \ge \hbar[/tex]


[tex]\Delta x \Delta p & \ge \frac{\hbar}{2}[/tex]

This is not standard though, and just something I made up on the spot.

I don't know where the root mean square is coming from...
 
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  • #5
Jheriko said:
First off I made some nasty typos in my tex and the equalses should be greater than or equal tos.

[tex]\Delta x \Delta p & \ge \hbar[/tex]


[tex]\Delta x \Delta p & \ge \frac{\hbar}{2}[/tex]

Secondly, the important point is that in the first equation the delta does NOT represent a standard deviation. A standard deviation covers some percentage of the values that I can never remember... ~68% apparently. The required deviation for the first form uses the size of the range which covers
50% of the values. Its just confusing that the same symbol gets used, it might be better to have

[tex]\Delta_{50\%} x \Delta_{50\%} p & \ge \hbar[/tex]


[tex]\Delta x \Delta p & \ge \frac{\hbar}{2}[/tex]

This is not standard though, and just something I made up on the spot.

I don't know where the root mean square is coming from...
if it was 50% of each then it would be aitch-bar/4?

:( I'm lost
 
  • #6
Jheriko said:
I don't know where the root mean square is coming from...

We often call it that because we define the uncertainty in x as

[tex]\Delta x \equiv \sqrt{<(x - <x>)^2>}[/tex]

i.e. as the square root of the mean of the square of the difference between x and its mean. In QM of course we speak of "expectation value" rather than "mean" in this context.

I think we define [itex]\Delta x[/itex] this way because it has nice mathematical properties as compared to other ways that we could define it.
 
  • #7
jtbell said:
We often call it that because we define the uncertainty in x as

[tex]\Delta x \equiv \sqrt{<(x - <x>)^2>}[/tex]

i.e. as the square root of the mean of the square of the difference between x and its mean. In QM of course we speak of "expectation value" rather than "mean" in this context.

I think we define [itex]\Delta x[/itex] this way because it has nice mathematical properties as compared to other ways that we could define it.

Well as the example that I posed initially, how would you determine what the delta values are? Are they standard deviations or are they rms ?

and it doesn't seem to be as nice of a relationship when you compare rms and a standard deviation(1/2)
 
  • #8
First, what are your definitions of "standard deviation" and "rms"? To me, they mean basically the same thing. But the important thing is the mathematical definition, not the words.

For the definitions of [itex]\Delta x[/itex] and [itex]\Delta p[/itex] that your book is using, you had best look for them in the book itself. Whatever they're using, it isn't the standard definition. Mathematically, the standard definition (in terms of expectation values) is the one that I gave, and it leads to [itex]\Delta x \Delta p \ge \hbar/2[/itex].

Either your book's [itex]\Delta x \Delta p \ge \hbar[/itex] is a typographical error, or the author is being a bit sloppy by neglecting numerical constants, or he's using a non-standard definition of [itex]\Delta x[/itex] and [itex]\Delta p[/itex] that is different from the standard one by a factor of [itex]\sqrt{2}[/itex].
 
  • #9
jtbell said:
First, what are your definitions of "standard deviation" and "rms"? To me, they mean basically the same thing. But the important thing is the mathematical definition, not the words.

For the definitions of [itex]\Delta x[/itex] and [itex]\Delta p[/itex] that your book is using, you had best look for them in the book itself. Whatever they're using, it isn't the standard definition. Mathematically, the standard definition (in terms of expectation values) is the one that I gave, and it leads to [itex]\Delta x \Delta p \ge \hbar/2[/itex].

Either your book's [itex]\Delta x \Delta p \ge \hbar[/itex] is a typographical error, or the author is being a bit sloppy by neglecting numerical constants, or he's using a non-standard definition of [itex]\Delta x[/itex] and [itex]\Delta p[/itex] that is different from the standard one by a factor of [itex]\sqrt{2}[/itex].

Thanks, actually I had to look up rms and "standard deviation" and they do differ...but not nicely for me to rearrange the equation to remove the constant.

It does seem like he has a different defition for Delta(x) and Delta(p) but the problems are worded in a way that can be interpreted with either inequality which is misleading because I am not sure which one to use...

I was just curious on who else would see this type of definition.

Thanks again!
 

1. What is the Heisenberg uncertainty principle?

The Heisenberg uncertainty principle is a fundamental principle in quantum mechanics that states that it is impossible to simultaneously know the exact position and momentum of a subatomic particle. This means that the more precisely we know the position of a particle, the less we know about its momentum, and vice versa.

2. Who discovered the Heisenberg uncertainty principle?

The Heisenberg uncertainty principle was first proposed by German physicist Werner Heisenberg in 1927.

3. Why is the Heisenberg uncertainty principle important?

The Heisenberg uncertainty principle is important because it fundamentally changes our understanding of the physical world at the subatomic level. It also has practical applications in quantum technology and has implications for our understanding of the universe.

4. How does the Heisenberg uncertainty principle relate to measurement?

The Heisenberg uncertainty principle states that the act of measuring one property of a particle will inevitably affect the measurement of another property. This is because the very act of measurement disturbs the particle, making it impossible to know both properties with absolute certainty.

5. Is the Heisenberg uncertainty principle a proven law?

Yes, the Heisenberg uncertainty principle has been extensively tested and has been shown to hold true in numerous experiments. It is considered a fundamental principle of quantum mechanics and has a significant impact on our understanding of the physical world.

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