Making sure about the spreads in the Uncertainty Principle

[fog37]

Hello Forum,

Going through Griffith's book last night, I read a good paragraph on page 111 about the canonical uncertainty principle for position and momentum in the x direction:

$$\sigma_{x} \sigma_{p} \geq \frac{\hbar}{2}$$

If we could make 100,000 identical copies of a system in exactly the same state $|\Psi>$ and measured the position $x$ for each state, we would obtain a range of different position answers. The range of answers would be equal to $\sigma_{x}$, i.e. the uncertainty $\sigma_{x}$ represents the spread in these measurements.

After making these 100,000 position measurements, the 100,000 systems would not be the same state anymore. If we measured the momentum $p_x$ on each system we would surely obtain a range of momentum answers. But that range is not the $\sigma_{p_x}$ in the equation above, correct? I think the $\sigma_{p_x}$ in the uncertainty relation corresponds to the spread we would obtain if we had measured momentum instead of position at the very beginning when the 100,000 systems where in the same identical state. Is that correct?

So the two different spreads $\sigma_{x}$ and $\sigma_{p_x}$ surely relate to each other but we must consider the several identical systems in the same state. I don't think it matter if we first measure position or momentum as far as the two uncertainties go...

Thanks!

 

[PeroK]

I think the σpx\sigma_{p_x} in the uncertainty relation corresponds to the spread we would obtain if we had measured momentum instead of position at the very beginning when the 100,000 systems where in the same identical state. Is that correct?

Yes, that's correct. Or, out of your 100,000 systems you measure position in 50,000 of them and momentum in the other 50,000 to get $\sigma_x, \sigma_p$

 

[fog37]

Thank you!

 

[fog37]

I guess an example of many multiple systems prepared in the same state would be photons passed through a polarizing filter so they all have the same state of polarization. But how does the fact they are in the same specific state of polarization implies that they share the same exact state in its entirety?

 

[DrClaude]

I guess an example of many multiple systems prepared in the same state would be photons passed through a polarizing filter so they all have the same state of polarization. But how does the fact they are in the same specific state of polarization implies that they share the same exact state in its entirety?

It doesn’t. You need to take into account other things like frequency and direction of travel ($\mathbf{k}$). This can be achieved in the case of photons by using a laser as a source.

 

[fog37]

It doesn’t. You need to take into account other things like frequency and direction of travel ($\mathbf{k}$). This can be achieved in the case of photons by using a laser as a source.

Hello DrClaude, Thank you for your comment. It is important for me.

In general, we say that the wavefunction contain ALL the information about the quantum system. For example, in the case of a single electron whose state is a specific energy eigenstate, the energy eigenstate represents the wavefunction itself. The system has a well defined energy but the other observables and their values are not necessarily defined even if we perfectly know the state. The fact the electron is in that specific eigenstate does not make its wavefunction completely unique and specified, correct? Or does it? For instance, let's consider two different electrons in two different atoms both in the same energy eigenstate. That does automatically imply that he two electrons share the same identical wavefunction/state?

When we manage to create multiple systems in the same identical state (two photons in the same polarization, for example), the identical state they are in does not make the two systems have wavefunctions that are identical in their entirety. Am I correct?

 

[DrClaude]

The fact the electron is in that specific eigenstate does not make its wavefunction completely unique and specified, correct? Or does it?

If the electron is in an energy eigenstate, then the wave function of that electron is the corresponding eigenstate, up to a complex phase. The state of the electron is then completely known.

For instance, let's consider two different electrons in two different atoms both in the same energy eigenstate. That does automatically imply that he two electrons share the same identical wavefunction/state?

Assuming that you are talking about atomic orbitals, they have identical wave functions (again up to a complex phase) relative to their respective nuclei. The full wave function would take into account the position of the nuclei, so for two atoms in different locations, the wave function would be different.

When we manage to create multiple systems in the same identical state (two photons in the same polarization, for example), the identical state they are in does not make the two systems have wavefunctions that are identical in their entirety. Am I correct?

When you are talking about state preparation, that's mostly correct. However, there cases where the state is completely the same, as in Bose-Einstein condensation.