Uncertainty relation: minimum value

[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?

 

[hikaru1221]

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

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: $$min(\Delta x)=\frac{h}{4\pi m_o v}\sqrt{1-\beta^2}$$.

 

[fluxions]

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.

 

[hikaru1221]

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