Derivation: Uncertainty Relation Using a Single Slit

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In the following, we derive the Heisenberg uncertainty relation using a single slit.

Consider a single slit with width $ \Delta x $. A straight electron beam is sent onto this single slit. Each electron in the beam has a momentum $ p_{\text y} $ in the $y$ direction, which is directed exactly to the single slit. In this way, the electrons can cross the slit.

We choose the slit width $ \Delta x $ so that there is electron diffraction at the single slit. As a result, we can observe an interference pattern on the detector screen behind the single slit. It has a main maximum in the center and smaller secondary maxima around it. Obviously, after passing through the slit, the electrons have received a momentum in the $x$ direction. If this were not the case, we would only observe one fringe exactly in the center of the detector screen.

We do not know how large the momentum component of the electron is in the $x$ direction, because we do not know where exactly the electron will land on the screen once it has crossed the single slit.

Momentum $p$ of the electron after diffraction and its momentum components in a right triangle.

We can look at the detector screen and see that the farther the electron fringes are on the screen, the larger the momentum deviation $ \Delta p_{\text x} $ must be. This is the $x$-component of the momentum $p$, which was not present BEFORE crossing the slit. We refer to the deviation from $ p_{\text y} $ as $ \Delta p_{\text x} $ (momentum uncertainty).

With the help of illustration 1 we can establish the first of the two equations that will lead us to the Heisenberg uncertainty relation.

Condition #1

Let's look at the deviation $ \Delta p_{\text x} $ up to the 1st minimum. Then we can construct the following right triangle using the illustration:

The hypotenuse $p$ - as the total momentum of the electron after crossing the single slit.
The opposing cathetus $ \Delta p_{\text x} $ - as a deviation from the momentum component $ p_{\text y} $.
The adjacent cathetus $ p_{\text y} $ - as the initial total momentum of the electron before crossing the single slit.

The angle enclosed between $p$ and $ p_{\text y} $ is $ \theta $. The trigonometric relation holds: sine of the angle is opposing cathetus divided by hypotenuse:

$$ \begin{align} \sin(\theta) ~=~ \frac{ \Delta p_{\text x} }{ p } \end{align} $$

Condition #2

Let's still eliminate the angle $ \theta $ in 1 and bring the slit width $ \Delta x $ into play. The slit width $ \Delta x $ restricts the location of the electrons to the region $\Delta x$ and is important for the uncertainty relation.

For this we use the right triangle in illustration 2.

Here, $n\lambda$ is the phase difference of two electron waves at the single slit, which in our case lead to the 1st minimum. Note that for the single slit, the condition for destructive interference (i.e., a minimum on the screen) is a multiple of the wavelength $n \,\lambda$. At the double slit $n\,\lambda$ was a condition for constructive interference. Here, $n$ takes the values 1, 2, 3, ... and so on and stands for the 1st minimum, 2nd minimum and so on.

We read off the following relation from illustration 2 (again: sine is opposing cathetus divided by hypotenuse):

$$ \begin{align} \sin(\theta) ~=~ \frac{ n\, \lambda }{ \Delta x } \end{align} $$

Let's link the two conditions by substituting the sine 1 into equation 2:

$$ \begin{align} \frac{ \Delta p_{\text x} }{ p } ~=~ \frac{ n\, \lambda }{ \Delta x } \end{align} $$

Now we have to somehow bring in the Planck's constant $h$, because it occurs in the uncertainty relation. We do this by using the matter wavelength $\lambda$ (de Broglie wavelength) of electrons:

$$ \begin{align} \lambda ~=~ \frac{h}{p} \end{align} $$

Using the de Broglie relation 4, let us substitute the wavelength $\lambda$ into Eq. 3:

$$ \begin{align} \frac{ \Delta p_{\text x} }{ p } ~=~ \frac{ n\, h }{ \Delta x \, p} \end{align} $$

Thus we have eliminated the angle $\theta$. The momentum $p$ also cancels out:

$$ \begin{align} \Delta p_{\text x} ~=~ \frac{ n\, h }{ \Delta x } \end{align} $$

We are almost at our final result. Bring $\Delta x $ and $n$ to the other side:

$$ \begin{align} \frac{ \Delta p_{\text x} \, \Delta x }{ n } ~=~ h \end{align} $$

To get an inequality, we compare Eq. 7 with the following term for the 1st minimum ($n=1$): $\Delta p_{\text x} \, \Delta x$. Obviously, the left-hand side of Eq. 7 is smaller than $\Delta p_{\text x} \, \Delta x$ (or equal if $n=1$) (because we divide by a number $n$ that is greater than or equal to 1). So we can write:

$$ \begin{align} \Delta p_{\text x} \, \Delta x ~\geq~ \frac{ \Delta p_{\text x} \, \Delta x }{ n } ~=~ h \end{align} $$

If we omit the middle term, we get the following inequality:

Heisenberg's uncertainty relation

$$ \begin{align} \Delta p_{\text x} \, \Delta x ~\geq~ h \end{align} $$

For example, from Eq. 9 we can directly read off what happens when we decrease the gap width $\Delta x$. If the slit width is made smaller, the momentum deviation $ \Delta p_{\text x} $ must increase so that the right side in 9 always remains larger than $h$. This in turn means that the interference pattern becomes more extended. The electron momentum after crossing the slit deviates more strongly from the initial momentum $ p_{\text y} $.
If, on the other hand, you make the slit wider, the position of the electrons is less determined, and the momentum uncertainty $\Delta p_{\text x}$ decreases. The interference pattern becomes narrower. If the slit width $ \Delta x$ is sufficiently large, the interference pattern disappears and all electrons end up in the center of the screen, like classical particles.

Note also that the derived Heisenberg uncertainty relation 9 can occur with a slightly different factor on the right-hand side than with the usual factor $h/4\pi$. Important here is the order of magnitude! While $ h $ is in the order of magnitude $ 10^{-34}$, $h/4\pi$ is in the order of magnitude $ 10^{-35}$. So the factor does not make a big difference!