Question

heisenberg uncertanity principle

Answer 1

what about it?

Answer 2

Wondering about the same thing, but I'm just a slave so here we go:
\[\huge \Delta \Omega _{1} \Delta \Omega _{2} \ge \frac{ 1 }{ 2 } \left| \langle \left[ \hat{\Omega}_{1}, \hat{\Omega}_{2} \right] \rangle \right|\]

Answer 3

To someone who isn't familiar with quantum mechanics, let alone chemistry in High school, will not understand that. I dont even understand that.

Answer 4

it says Both position and momentum of electron cannot be determined simultaneously :)
if u find one accurately then other will have uncertain value (NOT real ) :)

Answer 5

I explain it even more then.
The general uncertain principle is what I wrote. Heisenberg just past the special case with momentum and position. So the uncertainty for momentum and position most be:
\[\large \Delta p ~ \Delta q \ge \frac{ 1 }{ 2 } \left| \langle \left[ \hat{x}, \hat{p} \right] \rangle \right|\]
Delta p is the uncertainty of linear momentum and Delta q is the uncertainty of position.
We now evaluate commutator of the position and linear momentum operators:
\[\large \hat{x} ~\hat{p _{x}} \Psi = x \times \frac{ \hbar }{ i } \frac{ d \Psi }{ dx }\]
\[\large \hat{p _{x}} ~ \hat{x} ~ \Psi = \frac{ \hbar }{ i }\frac{ d }{ dx }x \Psi = \frac{ \hbar }{ i }\left( \Psi + x\frac{ d \Psi }{ dx } \right)\]
The commutator become:
\[\large \left[ \hat{x} ,\hat{p _{x}} \right] =\hat{x} ~\hat{p _{x}} - \hat{p _{x}} \hat{x}\]
\[\large \left( \hat{x} ~\hat{p _{x}} - \hat{p _{x}} \hat{x} \right) \Psi=\frac{ \hbar }{ i }\left( x \frac{ d \Psi }{ dx } \right)- \frac{ \hbar }{ i }\left( \Psi+x \frac{ d \Psi }{ dx } \right)= i \hbar \Psi\]
So we now got:
\[\large \Delta p ~ \Delta q \ge \frac{ 1 }{ 2 } \left| i \hbar \right|\]
We take the modulus of the complex number and get:
\[\large \Delta p ~ \Delta q \ge \frac{ 1 }{ 2 }\hbar\]
The expression above is known as the Heisenberg uncertainty principle.

Answer 6

So what does this mean?
We simply use the equation to tell us:
Lets say we know a particles position with complete certainty (Delta p = 0) the only way the Heisenberg uncertainty principle can be satisfied then is to set Delta q = infinity. (meaning complete uncertainty)

Answer 7

what does the accent circonflexe (for lack of mathematical notation knowledge) mean?

Answer 8

what the funk does any of this mean?!

Answer 9

The accent circonflexe means "operator" of the observable (postural III)

Answer 10

hahah

what does the operator do in this case?
what's postural III? lol

Answer 11

hahaha simmer down, kids, simmer down.

Answer 12

<^>\(^2\)

Answer 13

I only vaguely remember this from quantum, but I dont remember the formula being that confusing.

Answer 14

Right well now when I got the latex right I can begin writing:

Postulate III states that for each observable property \(\Omega \) there is a corresponding operator \(\hat{\Omega} \) built from the position and linear momentum operator, which I have summarized below:
\[\large \hat{x}=x \times\]
\[\large \hat{p _{x}}=\frac{ \hbar }{ i }\frac{ d }{ dx }\]
To take an example would be to take the well known hamiltonian operator \(\hat{H}\) which has the observable property energy (kinetic energy + potential energy) \(E\). Therefor the Schrödinger equation:
\[\hat{H} \Psi=E \Psi\]
In general terms, the eigenvalue equation:
\[\large \sf \left[ operator ~ for ~ the ~ observable ~ \Omega \right] \Psi=\left[ value ~of ~the ~observable ~ \Omega \right] \times \Psi\]
The expected value of an operator is given by the equation:
\[\large \langle \Omega \rangle = \int\limits_{}^{} \Psi ^{*} ~ \hat{\Omega} \Psi ~ d \tau\]
\(\Psi^*\) is the complex conjugated wavefunction, as the expected value is for a normalized wavefunction.

The uncertainty principle build among others on complementary observables
\[\large \hat{\Omega}_{1}\hat{\Omega}_{2} \Psi \neq \hat{\Omega}_{2}\hat{\Omega}_{1} \Psi\]
When the expression above is true we say two operators do not commute, while two operators that commute follow:
\[\large \left[ \hat{\Omega}_{1},\hat{\Omega}_{2} \right]=\hat{\Omega}_{1}\hat{\Omega}_{2} -\hat{\Omega}_{2}\hat{\Omega}_{1} =0\]
\(\left[ \hat{\Omega}_{1},\hat{\Omega}_{2} \right]\) I called the commutator.

So you could in fact rewrite postulate III as:
For every observable property \(\Omega\) there is a corresponding operator \(\hat{\Omega}\) built from the position and linear momentum operators that ensure the commutator for position and linear momentum is true and given by:
\[\large \left[ \hat{x},\hat{p _{x}} \right]=i \hbar\]

Thanks Abb0t for the latex advise! You are the man.

Answer 15

The uncertainties are defined as the root mean square form their mean values:
\[\large \Delta p=\left( \langle p ^{2} \rangle - \langle p \rangle ^{2} \right)^{1/2}\]
\[\large \Delta q=\left( \langle q ^{2} \rangle - \langle q \rangle ^{2} \right)^{1/2}\]