You are on the right track in calculating the uncertainty in momentum using the uncertainty principle.

The new position will be

$$x_2 = x_1 + \frac{p}{m} t$$

There is a well known technique of error propagation which works like

$\delta(f(x_1, x_2, x_i, ...) = \sqrt{\Sigma\left(\frac{\partial f }{\partial x_i} \delta x_i\right)^2 } $,

where $\delta x_i$ means the uncertainty in $x_i$, which is an independent coordinate (including momenta and times) of the motion. You sum over every measurement that has an uncertainty.

This comes from the Taylor series.

Applying this, you will get
$$\delta x_2 = \sqrt{\delta x_1^2 + \left(\frac{t}{m} \delta p\right)^2}$$

EDIT -

I thought about this a little more and I think that the addition in quadrature is not so appropriate here. Usually you use this for measurement uncertainties, where you look for one-sigma intervals, but for quantum mechanics, where you look for complete uncertainty, it might be more correct to add the components directly.

$$\delta x_2 = \delta x_1 + \left(\frac{t}{m} \delta p\right)$$

There is a very interesting story behind your question. In the early 1900's (after Special Relativity had been introduced) the solution to wave equation in vacuum:

$a\exp(i(kx-wt)) + b\exp(-i(kx-wt)) $

where $k$ is the wave vector and $w/k=v$ and v is the phase velocity.

De Broglie was the first to notice that the phase factors of the equation at every event remain invariant under Lorentz transformations if $(k,w)$ is considered a 4-vector. This meant invariant amplitude at every event. This is because the scalar product of the two 4-vectors $(k,w)$ and $(x,t)$ remains invariant under Lorentz transformations. From this remarkable piece of insight, he deduced that $(k,w)$ could represent the 4-momentum,
$(p,E)$, of a massive particle. This is how wave-particle duality was first discovered. Special Relativity gave birth to quantum mechanics in its proper form!

The commutation relation discovered subsequently

$[p,x]=-i\hbar$

on solving gives (setting the arbitrary phase factor in p to 1): $p=-i\hbar \frac{∂}{∂x} $. The uncertainty relation in momentum is derived from here.

The fundamental form of Schrodinger's equation

$i\hbar \frac{∂T}{∂t} = ET$

Where $E$ is the Hamiltonian and $T(t_0,t)$ is the Unitary time evolution operator has its origins in the relationship between $x$ and $p$ being extended to $t$ and $E$.

When introducing the Schrodinger equation in his book, Dirac points out that its derivation comes mainly from considerations of relativity. In fact Schrodinger's original equation was actually relativistic, where $E$ was the relativistic energy. Schrodinger wasn't sure whether to take the positive or negative root of $E^2$. So he discarded it in favor of its widely known non-relativistic form.

So in fact, other than the discovery of quantized energy levels in blackbody radiation and the photoelectric effect, every breakthrough in QM owes it's existence to special relativity.

**EDIT:** I earlier said w/k = c for a massive particle, this is incorrect. w/k is equal to the phase velocity, which is proportional to 1/u. u is the group velocity which is equal to the classical velocity of a massive particle. I have fixed the offending sentences.

## Best Answer

The uncertainty principle can be seen as a result of space $x$ and momentum $p$ being a Fourier transform pair. The free-particle wave function has, similarly to the exponential $e^{-\frac{i}{\hbar} px}$ an exponential $e^{-\frac{i}{\hbar} E t}$. Thus one could expect a similar uncertainty relation for the variable pair $(E, t)$. An immediate result is that the solutions with a perfectly defined energy, solutions of $\hat H \psi = E \psi$ are stationary, i.e. their physical content does not change with time.

This is unprecise, though. The (minimal) theoretical uncertainty of any two variables can be expressed through their commutator (see Wiki) $$ \sigma_A \sigma_B \geq \big \langle [ \hat A, \hat B ] \big\rangle $$ The hard thing to do is to find an operator that represents time, as non-relativistic quantum mechanics deals with time only as a parameter.

For unstable states, there is a way to derive the uncertainty you give where $t$ is not time in general but the lifetime of the state, thus giving an explanation for the natural width of spectral lines. The derivation (in brief) can be found on the above-linked wiki page.