# Help with understanding the uncertainty principle

• sodper
In summary, the equation for the mean energy of an electron in a stationary state is: KE "= (delta p)^2/2m.
sodper
[SOLVED] Help with understanding the uncertainty principle

I may have posted this in the wrong forum. If so I am sure a moderator could move it to the correct one.

Problem statement:
I have an assignment in which I need to calculate the lowest possible mean of an electron's KE , <KE>, based on the uncertainty of the electron's moment.

Given data:
The radius in which the electron is allowed to move, d=2.81*10^-10

Attempt at solution:
The uncertainty of the electrons position corresponds to d, deltax=d

Using the uncertainty principle I got, deltap = hbar/(2*deltax), where deltap is the lowest possible deviation.

I got deltaKE from, deltaKE = deltap^2/(2m), (m is the electron mass)

A fellow student explained to me that because my problem involves a stationary system (an electron bound to an atom), <KE> should be zero, which gives me:

(deltaKE)^2 = <KE^2> - <KE>^2 => (deltaKE)^2 = <KE^2>

But how do I proceed from here?

Last edited:
sodper said:
I may have posted this in the wrong forum. If so I am sure a moderator could move it to the correct one.

Problem statement:
I have an assignment in which I need to calculate the lowest possible mean of an electron's KE , <KE>, based on the uncertainty of the electron's moment.

Given data:
The radius in which the electron is allowed to move, d=2.81*10^-10

Attempt at solution:
The uncertainty of the electrons position corresponds to d, deltax=d

Using the uncertainty principle I got, deltap = hbar/(2*deltax), where deltap is the lowest possible deviation.

I got deltaKE from, deltaKE = deltap^2/(2m), (m is the electron mass)

A fellow student explained to me that because my problem involves a stationary system (an electron bound to an atom), <KE> should be zero,
Ah... the good ol' "fellow student". Unfortunately, the fellow student is often not correct. For a stationary state $\delta <H>$ is zero, not <KE> (where H=KE + PE). The way you were proceeding previously by considering delta p as related to the given delta x looks fine to me.

for example, if the electron is confined to an atom of size 'a' then the KE is roughly
$$\frac{\hbar^2}{m a^2}$$
(c.f. particle in a box energy levels (\frac{\hbar^2 \pi^2}{2 m a^2}), etc).

I don't think I quite understand. Did you, as I did, derive the expression for the KE from KE = p^2/(2m) ?
I simply substituted p with deltap to get the min deviation in KE.

I'm not sure how I am supposed use the expression of H.

sodper said:
I don't think I quite understand. Did you, as I did, derive the expression for the KE from KE = p^2/(2m) ?
I simply substituted p with deltap to get the min deviation in KE.
yeah. that's right.

I'm not sure how I am supposed use the expression of H.
dont use it for anything; I was just pointing out that in a stationary state it is the *total* energy which has no variance, not the *kinetic* energy.

Ok, but is the expression (deltaKE)^2 = <KE^2> valid?

If so, can I just take the square root to get the mean KE?

sodper said:
Ok, but is the expression (deltaKE)^2 = <KE^2> valid?

no... but we are just getting rough estimates here. That's all that is required. That's why we are using the "equation"

KE "=" (delta p)^2/2m

where (delta p) is given by \hbar divided by (delta x).

I see. Thanks! You've made it a lot clearer.

I'll post again if I need more help.

## 1. What is the uncertainty principle?

The uncertainty principle, also known as Heisenberg's uncertainty principle, is a fundamental concept in quantum mechanics that states that it is impossible to simultaneously know the precise position and momentum of a particle.

## 2. Why is the uncertainty principle important?

The uncertainty principle is important because it places a fundamental limit on our ability to measure and understand the behavior of particles at the atomic level. It also plays a crucial role in many quantum phenomena and has implications for our understanding of the nature of reality.

## 3. How does the uncertainty principle affect our daily lives?

Although the uncertainty principle may seem like a concept that only applies to the microscopic world, it actually has implications for our daily lives. For example, it helps explain why we cannot precisely predict the outcome of certain events and why we must always account for a margin of error in our measurements and observations.

## 4. Can the uncertainty principle be proven?

The uncertainty principle is a fundamental principle in quantum mechanics and has been supported by numerous experiments and observations. While it cannot be proven in the traditional sense, its predictions have been consistently confirmed by scientific evidence.

## 5. Are there any exceptions to the uncertainty principle?

There are certain cases where the uncertainty principle does not apply, such as in macroscopic objects where the effects of quantum mechanics are negligible. However, in the realm of quantum mechanics, the uncertainty principle is a fundamental principle that applies to all particles and interactions.

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