# Double-Slit Wave Equation?

Gold Member
Hi,
I was wondering if anybody could help me understand a derivation connected to the double-slit experiment that I came across within an introduction to quantum theory paper. I was interested in understanding this approach because it seems to provide a useful correlation of the meaning of the square of the amplitude for an electromagnetic wave $$[Intensity= \Psi^2]$$ in comparison to a mechanical wave $$[Energy=A^2]$$. The derivation starts off explaining that it intends to use complex notation:

[1] $$\Psi = \Re (e^{i (kx-wt)})$$

As far as I can see, only the real part of this equation is being referenced throughout the derivation, which I assume corresponds to the following relationship?

[2] $$\Re (e^{i \theta}) = cos \theta = (e^{i \theta}+ e^{-i \theta})/2$$

I have attached a diagram showing all the basic parameters used:

[3] $$r_1 \approx r - d/2(sin \theta); \; r_2 \approx r + d/2(sin \theta)$$

This approximation is based on R>>d and the derivation continues based on Huygen’s theory, where the slits are 2 sources of light having equal amplitude and phase, which meet in superposition at a given point [P] defined by $$[ \theta]$$

[4] $$\Psi = A exp \; i \left[ k \left( r- \frac{d}{2}sin \theta\right) – wt \right] + A exp \; i \left[ k \left( r+ \frac{d}{2}sin \theta\right) – wt \right]$$

[5] $$\Psi = A exp \; i \left( kr-wt\right) \left( exp \left[ ik \left( \frac{d}{2} \right)sin \theta \right] + exp \left[ -ik \left( \frac{d}{2} \right) sin \theta \right] \right)$$

[6] $$\Psi = A exp \; i \left( kr-wt\right)*2 cos \left[ k \left( \frac{d}{2} \right)sin \theta \right]$$

As far as I can see, the step from [5] to [6] is based on [2], which seems to suggest that we are only dealing with the real part of the complex wave function? However, the derivation then goes on to point out that the square of this wave function, i.e. [6], corresponds to the intensity (I) of the resulting superposition wave, but then only equates (I) via proportionality to part of [6]:

[7] $$I \propto 4A^2 cos^2[k(d/2)sin \theta]$$

If we are only dealing with the real part of the complex notation, why can whole of [6] be converted by into a trigonometric form via applying [2] as follows?

[8] $$I = \left( Acos(kr-wt) * 2 cos \left[ k \left( \frac{d}{2} \right)sin \theta \right] \right)^2$$

On the basis that all the variable parameters are driven by the position [P], then the main parameter is $$[ \theta]$$ and we should be able to plot the intensity across the screen by setting the other parameters in [8] to relative values as follows:

$$A=1; R=1; r = Rsin \theta; d=0.001; k=2 \pi/500nm; t=0$$?

Therefore [8] becomes:

[9] $$I = \left( cos[(2 \pi/500*10^{-9})Rsin \theta] * 2 cos \left[(2 \pi/500*10^{-9})) (0.001/2)sin \theta \right] \right)^2$$

Unfortunately, plotting [9] against $$\theta$$ doesn’t seem to produce the expected alternating constructive/destructive interference intensity, so I must assume that I have made an error or wrong assumption. I realise that this post may be too involved to interest anybody, but any insights by somebody who understands both the maths and physics would be appreciated. Thanks

#### Attachments

• 5.5 KB Views: 303

jtbell
Mentor
As far as I can see, the step from [5] to [6] is based on [2], which seems to suggest that we are only dealing with the real part of the complex wave function?
No, it doesn't. $\cos \theta = (e^{i\theta} + e^{-i\theta})/2$ is simply being used as an identity here.

It may help to take your step [8] and rearrange it slightly:

$$\Psi = \left( 2 A \cos \left[ k \left( \frac{d}{2} \right) \sin \theta \right] \right) e^{ i ( kr - \omega t)}$$

The complex exponential at the end gives the (complex) oscillating aspect of $\Psi$. The big parentheses in the front enclose the (real) amplitude of that oscillation. To get the probability density at the screen we use the usual rule $P = \Psi^* \Psi$:

$$P = {\left( 2 A \cos \left[ k \left( \frac{d}{2} \right) \sin \theta \right] \right)}^2 e^{ -i ( kr - \omega t)} e^{ i ( kr - \omega t)}$$

The two exponentials multiply to give 1, and the rest gives you the probability which varies with position on the screen. So in this case, we don't use the "take the real part" trick at all. The relationship between the QM wave function $\Psi$ and the probability density P is guaranteed to give you a real P from a complex $\Psi$.

The pattern on the screen is easier to "see" if you call the position on the screen y, instead of $R \tan \theta$, and use the small-angle approximation:

$$\sin \theta \approx \tan \theta = \frac{y}{R}$$

Last edited:
Gold Member
It may help to take your step [8] and rearrange it slightly:

$$\Psi = \left( 2 A \cos \left[ k \left( \frac{d}{2} \right) \sin \theta \right] \right) e^{ i ( kr - \omega t)}$$

The complex exponential at the end gives the (complex) oscillating aspect of $\Psi$. The big parentheses in the front enclose the (real) amplitude of that oscillation. To get the probability density at the screen we use the usual rule $P = \Psi^* \Psi$
jtbell:
Thanks very much; your comments were really helpful. I guessed that I was missing some key issues regarding the complex component of the wave function. However, having only just started to look quantum theory, I wasn’t sure of the physical interpretation of the complex component. However, could I just clarify a couple of the points you have highlighted?

$$\Psi$$= probability amplitude
P = probability density = $$\Psi \Psi^{*}$$

Are you multiplying the complex wave function $$\Psi$$ by its complex conjugate $$\Psi^*$$ to get [P]?

[1] $$\Psi = \left( 2 A \cos \left[ k \left( \frac{d}{2} \right) \sin \theta \right] \right) e^{ i ( kr - \omega t)}$$

[2] $$\Psi^* = \left( 2 A \cos \left[ k \left( \frac{d}{2} \right) \sin \theta \right] \right) e^{ -i ( kr - \omega t)}$$

Therefore, based on $$e^{-ix}*e^{ix} = 1$$, the equation becomes?

[3] $$P = {\left( 2 A \cos \left[ k \left( \frac{d}{2} \right) \sin \theta \right] \right)}^2 * 1$$

This aligns to equation [7] in post-1 and the derivation. So, in this case, is the probability density [P] equal the intensity at any given offset on the screen, i.e. [y]?

The reason for asking is that I am still having problems using this equation to draw a graph of [P] against $$\theta$$ that aligns to the constructive and destructive interference patterns normally associated with Young’s experiment, i.e. I was expecting the amplitude of the intensity to be brightness in the centre of the screen?

I need to check the figures tomorrow, but thanks again for the help.

jtbell
Mentor
Are you multiplying the complex wave function $$\Psi$$ by its complex conjugate $$\Psi^*$$ to get [P]?
Yes.

So, in this case, is the probability density [P] equal the intensity at any given offset on the screen, i.e. [y]?
Yes, more or less. Actually, I call it "intensity" only when I'm dealing with classical electromagnetic waves, because you get patterns of light and dark on the screen. The visual "brightness" depends on the calculated intensity.

In the QM case, when you send one particle through the apparatus, it arrives at a single point on the screen, and it doesn't make much sense (to me) to talk about "brightness" or "intensity" in this case, just the probability density which describes where it is more likely or less likely to arrive.

The reason for asking is that I am still having problems using this equation to draw a graph of [P] against $$\theta$$ that aligns to the constructive and destructive interference patterns normally associated with Young’s experiment, i.e. I was expecting the amplitude of the intensity to be brightness in the centre of the screen?
Set $\theta = 0$ and you get $P = 4A^2$. This corresponds to what you see at the center of the screen in the classical Young's experiment: the intensity with both slits open is four times what you get with only one slit open. In the QM case, the particle is four times as likely to arrive at the center of the screen, with both slits open, versus with only one slit open.

Gold Member
In the QM case, when you send one particle through the apparatus, it arrives at a single point on the screen, and it doesn't make much sense (to me) to talk about "brightness" or "intensity" in this case, just the probability density which describes where it is more likely or less likely to arrive.
I understand your comments concerning the classical versus quantum view. However, would it still be correct to equate intensity (I) with the probability density [P] as the number of photons of a given energy (hf) per square metre per second?

$$P = I = n * hf /unit-area/unit-time$$
Set $\theta = 0$ and you get $P = 4A^2$. This corresponds to what you see at the center of the screen in the classical Young's experiment: the intensity with both slits open is four times what you get with only one slit open. In the QM case, the particle is four times as likely to arrive at the center of the screen, with both slits open, versus with only one slit open.
That makes sense, however I would expect the probability of individual photons hitting the screen to continue fall as the angle $\theta$ or [y] is increased. As such, successive bands of constructive interference have less intensity due to the lower probability density as [y] is increased. This is the bit that I having problems with as far as the following equation is concerned:

[1] $$P = {\left( 2 A \cos \left[ k \left( \frac{d}{2} \right) \sin \theta \right] \right)}^2$$

If I normalise the terms $4A^2=1$, then the maximum intensity at the centre will be unity and we are left with just the $$(cos(x))^2$$ function, where we might approximate [x] for small angles as follows:

[2] $$x = \frac{kd}{2} * \frac {y}{R} = \frac {\pi d}{\lambda} * \frac {y}{R}$$

I am assuming the cos(x) function oscillates between [+1] and [-1], which when squared is normalised to [0] and [+1]. Therefore, although the cosine function would suggest alternating bands of probability density [P], the intensity doesn’t seem to fall as [y] is increased, only cycles between the max and min value?

As a reference point, I considered the difference in the paths associated with each slit as a function of the angle, i.e.

[3] $$r_1-r_2 = d sin(y/R)[/itex] A value of [d=3cms] and [R=100cms], the difference in the path length would be 1.5mm at an offset of [y=5cms]. Of course, I am assuming this difference has to be put into the perspective of the wavelength, i.e. red light=650nm. As such, the granularity of incrementing [y] affects any graph plotted, which I believe was the root cause of some of the strange results, but I have attached one that appears to reflect that while [P] oscillates it doesn’t fall off as a function of [y], which I still don’t understand. Sorry, to belabour the point, but it seems important to any subsequent understanding. Thanks #### Attachments • 29 KB Views: 266 jtbell Mentor I understand your comments concerning the classical versus quantum view. However, would it still be correct to equate intensity (I) with the probability density [P] as the number of photons of a given energy (hf) per square metre per second? Yes, although I would say "proportional to" rather than "equal" because the units are different. Intensity has units of energy / (area . time) whereas probability density has units of 1 / (area). I would expect the probability of individual photons hitting the screen to continue fall as the angle $\theta$ or [y] is increased. As such, successive bands of constructive interference have less intensity due to the lower probability density as [y] is increased. Yes, successive interference maxima should really become smaller as you move away from the center of the interference pattern. Our equation for P doesn't show this because of the approximation in step [3] of your original post, which is valid only for smallish $\theta$. [added] There's another approximation which is more important: the original assumption that the individual waves are of the form [tex]\Psi = A e^{i(kr - \omega t)}$$

This is a plane wave with flat wavefronts, whose amplitude remains constant. Actually, the waves from a finite source diverge, with curved wavefronts, and their amplitude decreases as they move away from the source. From a point source you get a spherical wave that goes like

$$\Psi = \frac {A_0} {r} e^{i(kr - \omega t)}$$

So if you have two small pointlike holes instead of two slits, you have to find

$$\Psi = \frac {A_0} {r_1} e^{i(kr_1 - \omega t)} + \frac {A_0} {r_2} e^{i(kr_2 - \omega t)}$$

If the distance between the holes (d) is much smaller than the distance from the holes to the screen, you can still use your equation [3] as a good approximation for $r_1$ and $r_2$, but the sum is still a lot messier than with plane waves.

For long narrow slits you have to use cylindrical waves, which I think go like

$$\Psi = \frac {A_0} {\sqrt{r}} e^{i(kr - \omega t)}$$

Last edited:
Gold Member
I really appreciate all the help you have extended. It is often very easy to get some very wrong ideas when you are starting out on a subject and your clarifications have save me a lot of time.
Yes, successive interference maxima should really become smaller as you move away from the center of the interference pattern. Our equation for P doesn't show this because of the approximation in step [3] of your original post, which is valid only for smallish $\theta$.
OK. I can go back to this point, at a later date, now that I understand the basic problem. I think I also understand the implications of the points you have made about plane and curved waves, i.e. if we apply Huygen’s principle, then the actual distribution of photons/waves from the source depends on the shape of the source, e.g.
http://en.wikipedia.org/wiki/Huygens'_principle
The complex exponential at the end gives the (complex) oscillating aspect of [itex]\Psi[/itex
Based on this comment, I recognise the need to now take a closer look at the wider implications of Schroedinger's wave equation and get a better understanding of the meaning of the complex component. Again, many thanks.