# Uncertainty relation: minimum value

• fluxions
In summary, we have shown that the smallest possible uncertainty in the position of an electron whose speed is given by \beta = v/c is \Delta x_{min} = \frac{h}{4 \pi m_0 c}\sqrt{1-\beta^2}. This is achieved when the greatest value of \Delta p is taken, which occurs when \Delta v is very small. However, for this to be true, the assumption that the electron can reach v=c must be reconsidered. By considering the electron as a free electron with a constant kinetic energy, we can find that the largest uncertainty in p_x is 2p, and thus the smallest possible uncertainty in position is \frac{h}{4\pi m_
fluxions

## Homework Statement

Show that the smallest possible uncertainty in the position of an electron whose speed is given by $$\beta = v/c$$ is $$\Delta x_{min} = \frac{h}{4 \pi m_0 c}\sqrt{1-\beta^2}$$

## The Attempt at a Solution

Since $$\Delta x \Delta p \geq \frac{\hbar}{2}$$, we see that $$\Delta x_{min}$$ occurs when $$\Delta p$$ has its greatest value.

Relativistically, $$\Delta p$$ is:
$$\Delta p = \Delta ( \frac{m_0 v}{\sqrt{1 - \beta^2}}) = ... = (1-\beta^2)^{-3/2} m_0 \Delta v$$Now the greatest value of $$\Delta p$$ occurs when $$\Delta v$$ is c.

Hence
$$\Delta x_{min} = \frac{h}{4 \pi m_0 c} (1 - \beta^2)^{3/2}$$.

My exponent for gamma is incorrect. Where did I go wrong?

Last edited:
fluxions said:
$$\Delta p = \Delta ( \frac{m_0 v}{\sqrt{1 - \beta^2}}) = ... = (1-\beta^2)^{-3/2} m_0 \Delta v$$

You can apply this only when $$\Delta v$$ is very small. Here you set $$\Delta v = c$$, not small at all! In this case, $$v_{max}=c$$ and $$v_{min}=0$$, which correspond to p = infinity (!) and p = 0 respectively. Obviously $$\Delta p = infinity$$, which results in $$\Delta x = 0$$, and that's not what we want. Therefore I think that we should reconsider the assumption that the electron may reach to v=c.

I don't get the needed result, but let me present my opinion and see if we can proceed from that. Assume that the electron is free electron (it's free from any potential field). Due to the law of energy conservation, its kinetic energy cannot increase, so $$v = const$$ and thus $$|\vec{p}| = const$$.

However momentum $$\vec{p}$$ may point to any direction, which leads to the fact that $$p_x$$ varies. Since p=const, the largest uncertainty of $$p_x$$ is: $$max(\Delta p_x) = 2p$$. Thus: $$min(\Delta x)=\frac{h}{4\pi m_o v}\sqrt{1-\beta^2}$$.

hikaru1221 said:
You can apply this only when $$\Delta v$$ is very small. Here you set $$\Delta v = c$$, not small at all! In this case, $$v_{max}=c$$ and $$v_{min}=0$$, which correspond to p = infinity (!) and p = 0 respectively. Obviously $$\Delta p = infinity$$, which results in $$\Delta x = 0$$, and that's not what we want. Therefore I think that we should reconsider the assumption that the electron may reach to v=c.

Oops. Yeah, my approximation is definitely not justified; I feel rather silly for using it.

[/QUOTE]
I don't get the needed result, but let me present my opinion and see if we can proceed from that. Assume that the electron is free electron (it's free from any potential field). Due to the law of energy conservation, its kinetic energy cannot increase, so $$v = const$$ and thus $$|\vec{p}| = const$$.

However momentum $$\vec{p}$$ may point to any direction, which leads to the fact that $$p_x$$ varies. Since p=const, the largest uncertainty of $$p_x$$ is: $$max(\Delta p_x) = 2p$$. Thus: $$min(\Delta x)=\frac{h}{4\pi m_o v}\sqrt{1-\beta^2}$$.[/QUOTE]

Ok, I agree with your analysis for the most part.$$max(\Delta p_x) = 2p$$ holds for each value of p, hence v. We can take the max (supremum, actually, since v cannot equal c) of this max by sending v to c. This gives the desired result.

fluxions said:
Ok, I agree with your analysis for the most part.$$max(\Delta p_x) = 2p$$ holds for each value of p, hence v. We can take the max (supremum, actually, since v cannot equal c) of this max by sending v to c. This gives the desired result.

That may be a good estimation

Your calculation for the uncertainty in momentum is incorrect. The correct expression for the uncertainty in momentum is:

\Delta p = m_0 \Delta v \sqrt{\frac{1}{1-\beta^2}} = \frac{m_0 \Delta v}{\gamma}

where \gamma = \frac{1}{\sqrt{1-\beta^2}} is the Lorentz factor. This is because, in relativity, momentum is not just the product of mass and velocity, but also depends on the Lorentz factor.

Using this expression for \Delta p, the minimum uncertainty in position becomes:

\Delta x_{min} = \frac{\hbar}{2\Delta p} = \frac{\hbar}{2m_0 \Delta v \sqrt{\frac{1}{1-\beta^2}}} = \frac{h}{2m_0 c} \sqrt{1-\beta^2}

which is consistent with the uncertainty relation. The exponent of (1-\beta^2) should be 1/2, not 3/2.

## What is the uncertainty relation?

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

## What is the minimum value in the uncertainty relation?

The minimum value in the uncertainty relation refers to the minimum amount of uncertainty that exists between a particle's position and momentum. This minimum value is known as the Planck constant, which has a value of 6.626 x 10^-34 joule seconds.

## How is the uncertainty relation used in physics?

The uncertainty relation is used in physics to describe the limitations of our ability to measure certain properties of particles at the quantum level. It is a fundamental principle that is used to understand and predict the behavior of particles in the microscopic world.

## What are some real-life applications of the uncertainty relation?

The uncertainty relation has implications in many areas of modern technology, including quantum computing, cryptography, and medical imaging. It also helps explain the behavior of subatomic particles in particle accelerators and nuclear reactors.

## Is there any way to overcome the uncertainty relation?

No, the uncertainty relation is a fundamental principle in quantum mechanics and cannot be overcome. However, scientists have developed techniques and technologies to minimize the effects of uncertainty in certain situations, such as through the use of precision instruments and advanced mathematical models.

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