# Connection between a simple matter wave and Heisenberg's uncertainty relation

When looking at the wave function of a particle, I usually prefer to write

$$\Psi(x,t) = A \exp(i(kx - \omega t))$$

since it reminds me of classical waves for which I have an intuition ($k$ tells me how it moves through space ($x$) and $\omega$ tells me how it moves through time ($t$), roughly speaking). However, I noticed that you can translate this from the $(k,\omega)$ into the $(p,E)$ space by extracting $\hbar$:

$$\Psi(x,t) = A \exp(i(px - Et)/\hbar)$$

When looking at this representation, I couldn't help but be remembered of the uncertainty relations

$$\Delta p \Delta x \geq \hbar\\ \Delta E \Delta t \geq \hbar$$

This cannot be a coincidence, but the derivation of these relations (as described e.g. on Wikipedia) are simply over my head and my textbook merely motivates them with an argument based on wave packets.

What is the connection here? Is there an intuitive explanation for the recurrence of both uncertainty relations in the wave function of a matter particle?

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You are absolutely right. In fact, the uncertainty relations are a direct consequence of this wave nature and the "conjugated" variables being $p \leftrightarrow x$ and $E \leftrightarrow t$.

I will attempt a simple explanation. Remember that the wavefunction gives you a probability amplitude for finding the particle at some location $x$ at some time $t$? The probability to find the particle at some point $x$ at time $t$ is given by $|\Psi(x,t)|^2$.

Now here's the catch: If you put your plane wave into that equation, you'll find that $|\Psi(x,t)|^2 = A$, independent of $x$ and $t$. So for your plane wave, the particle has equal probability to be anywhere. In terms of uncertainty, then, we could say that the uncertainty in position is at a maximum, and so is the uncertainty in time.

What about momentum and energy? Well, your plane wave has a definite momentum of $p = \hbar k$ and a definite energy of $E = \hbar \omega$, so their uncertainty is zero. This is an extreme case of the uncertainty relation: Certainty in one variable means maximum uncertainty in the other variable.

Any other possible form of the wave function can always be written as a sum of plane waves (that's the Fourier theorem):

$$\Psi(x,t) = \int dE \int dk A(k,E) \exp(i(kx-E/\hbar t)$$

You can think of $\Psi(x,t)$ as the wavefunction in position and time domain and $A(k,E)$ as the wavefunction in momentum and frequency domain, and the conversion between those two is achieved via the Fourier transform.

You can now mathematically show how if $\Psi(x,t)$ is "narrow" (i.e. position has low uncertainty) then $A(k,E)$ must be broad and vice versa.

The intuition behind that: You need to add up a lot of plain waves with lots of different momenta to go from the broad plain wave distribution to a narrow wave packet, but this then implies that you have high uncertainty in momentum.

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Awesome answer. Thank you so very much! –  bitmask May 3 '13 at 2:48