Position of an electron with exactly 0 momentum

[LostConjugate]

If we know the exact momentum of the electron then the position of the electron could be found to be anywhere, all probabilities equal.

So if the exact momentum of the electron was 0, does that mean the electron can be found anywhere?

It does not seem logical that an object with no momentum would ever be found somewhere else.

 

[Feldoh]

What do you mean "somewhere else"? You measured the momentum of an electron you don't know where it is. It can move somewhere else if there was no somewhere.

 

[Drakkith]

If you measure that electron with 0 momentum, you no longer know where it is after you measure it because of interactions with your measurement device I believe.

 

[unchained1978]

From what I understand, the uncertainty principle states that product of the uncertainties of position and momentum always exceed a certain value. To measure the momentum to be exactly zero isn't a practical phenomenon, with your measuring devices there will always be an associated error, and from what I understand this idea gives rise to an uncertainty principle. I'm not quite satisfied with my answer and I doubt you will be, but perhaps it will offer some insight into the problem.

 

[zaybu]

The uncertainty principle states:

ΔxΔp ≥ ℏ

If you would know precisely the momentum of a particle (Δp =0), then the uncertainty in its position (Δx) would be infinite, meaning the particle can be equally anywhere in space.

 

[dextercioby]

We can't know exactly the position of a particle, because, at least according to the statistical interpretation of QM, an infinite number of infinitely-precise (in the sense of accuracy of the measurement apparatus) measurements of position will never yield the same numerical value. The statistical spread (mean square deviation from the average value) cannot be 0, as per the equation

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

 

[alxm]

The expectation value of the momentum of an electron in an atom or molecule is exactly zero though.
(Anything else would imply a net momentum in some direction, meaning it'd leave the atom!)

 

[vanhees71]

The very simple proof that there's no state of sharp momentum is very simple. Take the position representation. Then the eigen-value equation for the momentum operator reads

$$-\mathrm{i} \vec{\nabla} u_{\vec{p}}(\vec{x})=\vec{p} u_{\vec{p}}(\vec{x})$$,

which can be easily integrated by separation of variables to give the plane-wave solution

$$u_{\vec{p}}=N_{\vec{p}} \exp(\mathrm{i} \vec{p} \cdot \vec{x})$$,

with $$N_{\vec{p}}=\text{const}$$. This is not a square-integrable function, and thus there's no eigenvalue and no eigenvector of $$\hat{\vec{p}}$$.

What we found there belongs to the space of distributions of a dense subspace of Hilbert space where the momentum operator is defined. As a distribution it's normalized to a $$\delta$$-distribution,

$$\langle \vec{p}_1 |\vec{p}_1 \rangle=N_{\vec{p}_1}^* N_{\vec{p}_2} \int \dd^3 \vec{x} \exp[\mathrm{i} \vec{x} \cdot (\vec{p}_2-\vec{p}_1)]=N_{\vec{p}_1}^* N_{\vec{p}_2} (2 \pi)^3 \delta^{(3)}(\vec{p}_1-\vec{p}_2).$$

Thus, choosing the arbitrary phase such as to make the normalization factor real and positive, we have the normalized generalized momentum-eigenfunction

$$u_{\vec{p}}(\vec{x})=\left (\frac{1}{2 \pi} \right)^{3/2} \exp(\mathrm{i} \vec{p} \cdot \vec{x})$$.

The spectrum of $$\hat{\vec{p}}$$ is whole $$\mathbb{R}^3$$.

Since there are thus no true eigenvectors but only generalized ones and the spectrum is everywhere continuous, a particle cannot have a sharp momentum. However, you can make the width of the momentum distribution arbitrarily small, and this is to the cost of the position accuracy of the particle. That's the meaning of Heisenberg's uncertainty relation, which correctly reads (in my units, where $$\hbar=1$$)

$$\Delta x_j \Delta p_k \geq \frac{1}{2} \delta_{jk}$$.

The limit is reached for a Gaussian wave packet (it's of course Gaussian in both momentum and position since one wave function is the Fourier transform of the other).

 

[dm4b]

I think maybe the OP was also asking if an electron could come to a stop, or have zero momentum, even if we did not measure it.

In other words does the HUP mean an electron can have no rest?

But, maybe I misunderstood the question ..

 

[LostConjugate]

Thank you for all your answers.

I think maybe the OP was also asking if an electron could come to a stop, or have zero momentum, even if we did not measure it.

In other words does the HUP mean an electron can have no rest?

I am curious, if you place an electron in a vacuum with no initial momentum, would the electron move?

It seems as if though the uncertainty in momentum could mean that the electron may pick up momentum and start to move. Though the accuracy of our knowledge of the momentum gives such a great spread in the position that it appears the electron would need to gain an immense amount of momentum to quickly reach some of the distances it has a probability of being found.

How could such a small uncertainty in momentum lead to such high momentum values?

 

[Feldoh]

You're still thinking in terms of "if I measure the momentum the particle must be at the place where I took the momentum measurement". It isn't.

If you measure the momentum exactly you have not idea where the electron is. It's not like you're measuring the momentum and then the electron goes flying off to some random position.

 

[LostConjugate]

You're still thinking in terms of "if I measure the momentum the particle must be at the place where I took the momentum measurement". It isn't.

If you measure the momentum exactly you have not idea where the electron is. It's not like you're measuring the momentum and then the electron goes flying off to some random position.

The electron must be in the lab though in order to have made a measurement. I find that it has close to 0 momentum. How then is it just as likely for it to then be in another galaxy as it is to be still in the lab?

In fact the closer to 0 my measurement is the more likely the electron is to be found further and further away. The more certainty I have that the electron is not moving very fast, the higher the probability I find it further away in the next moment of time.