# A Basic Quantum Question

## Recommended Posts

I'm having trouble understanding a basic concept of quantum theory, in particular the uncertainty principle: why does when one tries to figure out the mass of a particle is one not then able to find the velocity or likewise?

##### Share on other sites

I think that you got confused here. There are several uncertainty relationships, the most famous being momentum (which is mass times velocity) and position. One explanation for that is that due to the wave nature of matter, and the relationship between wavelength and momentum (wavelength = Plank's constant / momentum), and that to measure something small you need a wavelength that is proportionately small. But to get a small wavelength, you need a lot of momentum, and if you have a lot of momentum, you bump what you are measuring. The more accurately you want to determine the position, the more momentum you need and the bigger the bump.

##### Share on other sites

I think that you got confused here. There are several uncertainty relationships, the most famous being momentum (which is mass times velocity) and position. One explanation for that is that due to the wave nature of matter, and the relationship between wavelength and momentum (wavelength = Plank's constant / momentum), and that to measure something small you need a wavelength that is proportionately small. But to get a small wavelength, you need a lot of momentum, and if you have a lot of momentum, you bump what you are measuring. The more accurately you want to determine the position, the more momentum you need and the bigger the bump.

That's not entirely accurate. It's not just the measuring equipment that causes the uncertainty, it's quantum mechanics itself.

I could try explaining, but all you'd get is a bunch of nonsense with the word "wavefunction" scattered through it. When I'm more awake I may try checking one of my books to see if it has a sensical explanation.

##### Share on other sites

I think that you got confused here. There are several uncertainty relationships, the most famous being momentum (which is mass times velocity) and position. One explanation for that is that due to the wave nature of matter, and the relationship between wavelength and momentum (wavelength = Plank's constant / momentum), and that to measure something small you need a wavelength that is proportionately small. But to get a small wavelength, you need a lot of momentum, and if you have a lot of momentum, you bump what you are measuring. The more accurately you want to determine the position, the more momentum you need and the bigger the bump.

What do you mean by "bump what you are measuring"

##### Share on other sites

What do you mean by "bump what you are measuring"

In order to make a measurement of a particle, you have to interact with it. Usually this means scattering a photon, which will necessarily impart some momentum to it, i.e. bump it.

However, as Cap'n Refsmmat has pointed out, the HUP is not the same as the observer effect — the uncertainty is inherent in the system.

You can prepare an ensemble of atoms in the same quantum state, and measure the position of half of them, and momentum of the other half. That way the measurement of one attribute does not change the other. You will still have an uncertainty in each, and the product will be larger than $\hbar/2$

Edited by swansont
##### Share on other sites

Thanks swansont: that's what I needed to know.

##### Share on other sites

However, as Cap'n Refsmmat has pointed out, the HUP is not the same as the observer effect — the uncertainty is inherent in the system.

What do you mean here when you say that the uncertainty is inherent in the system? The value calculated for the uncertainty is uniquely determined by the quantum state that the system is in. E.g. $\Psi$ determines $\Delta X$. Is that what you meant?

##### Share on other sites

What do you mean here when you say that the uncertainty is inherent in the system? The value calculated for the uncertainty is uniquely determined by the quantum state that the system is in. E.g. $\Psi$ determines $\Delta X$. Is that what you meant?

I meant that it is not a limitation of measurement.

##### Share on other sites

• 4 weeks later...

uhm may i ask what your credentials are mr skeptic all due respect?

##### Share on other sites

uhm may i ask what your credentials are mr skeptic all due respect?

My most important credential is that I am skeptical of everything, including myself. And that I love science. And my PhD degrees in Quantum Physics, General Relativity, and Hawking Aerodynamics.

##### Share on other sites

My most important credential is that I am skeptical of everything, including myself. And that I love science. And my PhD degrees in Quantum Physics, General Relativity, and Hawking Aerodynamics.

I'm skeptical of that.

## Create an account

Register a new account