Position Probabilities · Quantum Mechanics (Phillips)

Born’s rule is bolted onto an equation that does not contain it. Nothing in the Schrödinger equation says |Ψ|² is a probability, and the identification is a separate postulate.

This is the page the book has been building toward since Chapter 1. It answers the question the chapter opened with — how something spread out can arrive as a lump — and the answer, due to Max Born in 1926, changed what physics claims to be about.

Phillips’ strategy is worth naming before the algebra starts, because it is what makes the argument convincing: he runs the two-slit experiment twice. Once with a classical wave, using real functions and ordinary trigonometry. Once with a quantum particle, using the complex wave function of Chapter 2. Then he puts the two answers side by side. They agree about everything you can see — and differ by exactly one factor, whose disappearance is the whole of quantum measurement.

The geometry

Fig. 3.1 — A wave-like entity passes through two slits $S_1$ and $S_2$ and is detected on a screen. Constructive interference occurs at $P$ when the path difference $R_1 - R_2$ is a whole number of wavelengths. §1.1–1.2 works out the geometry that converts a position on the screen into a path difference; here only $R_1 - R_2$ matters.

First run: a classical wave

A classical wave is a real function of space and time. Two waves emerge from the slits and add at $P$ :

\Psi = A_1\cos\,(kR_1 - \omega t) + A_2\cos\,(kR_2 - \omega t)\tag{3.9}

The amplitudes $A_1$ and $A_2$ fall off as $1/R_1$ and $1/R_2$ . When the screen is far away compared with the slit separation, the two distances are nearly equal and we may set $A_1 = A_2 = A$ . The energy density and intensity go as the square of the wave, so what a detector responds to is $\Psi^2$ .

step 1 of 3

Pure trigonometry — but it is the step that separates the pattern from the flicker, so it is worth watching.

1Set A₁ = A₂ = A and use the sum-to-product identity cos X + cos Y = 2 cos((X+Y)/2) cos((X−Y)/2).
$Ψ = 2 A cos (\frac{k ( R _{1} + R _{2} )}{2} - ω t) cos (\frac{k ( R _{1} - R _{2} )}{2})$
2Look at what each factor depends on. The second factor is a function of POSITION only. The first oscillates in time and contains the mean distance k(R₁+R₂)/2 — which barely changes across the pattern, since moving along the screen alters R₁ − R₂ by whole wavelengths while R₁ + R₂ stays almost fixed. Absorb it into the origin of t.
$Ψ ≃ 2 A cos (ω t) cos (\frac{k ( R _{1} - R _{2} )}{2})$
3Square it. This is Eq. (3.10) — with the coefficient the algebra actually gives; see the Caution below.
$Ψ^{2} = 4 A^{2} cos^{2} (\frac{k ( R _{1} - R _{2} )}{2}) cos^{2} ω t$
A fixed pattern in space, multiplied by a factor that flickers in time.

The book prints this as

\Psi^2 = 2A^2\cos^2\left(\frac{k(R_1 - R_2)}{2}\right)\,\cos^2\omega t\tag{3.10}

Maxima occur where $k(R_1-R_2)/2$ is an integer multiple of $\pi$ and minima where it is a half-integer multiple — that is, maxima where the path difference is a whole number of wavelengths, minima where it is a half-integer number. Rewriting $k = 2\pi/\lambda$ turns the condition into exactly the statement in Fig. 3.1’s caption.

Both Eq. (3.10) and Eq. (3.13) are printed with 2A²; the coefficient should be 4A²

This is a slip in the book, and it is worth catching because the check that exposes it is one you should be doing anyway.

The check. At the centre of the pattern $R_1 = R_2$ , so the two contributions arrive exactly in step. Two amplitudes $A$ in phase give a total amplitude $2A$ , hence an intensity $(2A)^2 = 4A^2$ . Equation (3.10) evaluated there gives $2A^2\cos^2 0 = 2A^2$ — half the right answer.

The same factor is missing from the quantum result, Eq. (3.13). The derivation on the book’s own page reaches $|\Psi|^2 = 2A^2(1 + \cos\theta)$ and then substitutes $1 + \cos\theta = 2\cos^2(\theta/2)$ , which gives $4A^2\cos^2(\theta/2)$ .

Why nothing downstream breaks. The slip is the same in both equations, so the classical–quantum comparison — the entire point of the section — is untouched, as are the positions of every maximum and minimum. Only the absolute scale is off by two, and neither expression was normalized to begin with.

Where $2A^2$ does belong. It is the average intensity across the pattern, $\langle 4A^2\cos^2\rangle = 2A^2$ — which is also the correct energy bookkeeping, since two slits deliver twice the power of one. The printed equations put the average value where the peak value goes.

The <Compute> block at the foot of this page checks all of this against the definitions rather than against my arithmetic.

This is the first of four misprints in chapter 3, and it is worth knowing they are coming so you do not spend an evening assuming you have made the mistake. The others are Eq. (3.23)‘s reversed integration limits and a lost exponent on page 50 — both flagged in §3.5–3.6 — and a sign error in the probability current of problem 8. All four are typesetting slips; none changes a result, and each is caught by a check you can run yourself in under a minute.

Second run: quantum particles

Now the same experiment with a current of quantum particles. Chapter 2 leaves no choice about the form: a particle with definite momentum $p = \hbar k$ and energy $E = \hbar\omega$ is described by a complex wave function, and a particle that could have gone through either slit is a linear superposition of one wave from each:

\Psi = A_1\,e^{+i(kR_1 - \omega t)} + A_2\,e^{+i(kR_2 - \omega t)}\tag{3.11}

with $A_1$ and $A_2$ complex constants of approximately equal value.

What happens next at the screen is, in Phillips’ words, a very complicated process: a measuring device magnifies a microscopic event until there is a visible signal that a particle did or did not arrive. The book declines to explain it. Instead it makes one bold assumption — that the probability of detecting a particle somewhere is proportional to the effective intensity of the complex wave function there — and defines that intensity, in analogy with classical waves, as a real number:

|\Psi|^2 = \Psi^*\Psi\tag{3.12}

Setting $A_1 = A_2 = A$ and multiplying out:

|\Psi|^2 = \left(A^*e^{-i(kR_1-\omega t)} + A^*e^{-i(kR_2-\omega t)}\right)\left(Ae^{+i(kR_1-\omega t)} + Ae^{+i(kR_2-\omega t)}\right)

The four cross-multiplied terms give

|\Psi|^2 = |A|^2 + |A|^2 + |A|^2 e^{+ik(R_1-R_2)} + |A|^2 e^{-ik(R_1-R_2)}

Every $\omega t$ has vanished. The two “diagonal” terms lost it because $e^{-i\phi}e^{+i\phi} = 1$ ; the two cross terms lost it because both waves carry the same $\omega$ , so the time parts cancel between them as well. Using $\cos\theta = (e^{+i\theta} + e^{-i\theta})/2$ and then $\cos\theta = 2\cos^2(\theta/2) - 1$ :

|\Psi|^2 = 2A^2\cos^2\left(\frac{k(R_1 - R_2)}{2}\right)\tag{3.13}

The comparison

Set Eq. (3.10) beside Eq. (3.13) and one factor is missing:

\underbrace{4A^2\cos^2\!\left(\tfrac{k(R_1-R_2)}{2}\right)}_{\text{same fringes}}\times\underbrace{\cos^2\omega t}_{\text{classical only}}

Same maxima, same minima, same dependence on path difference — so a current of quantum particles builds an interference pattern that looks like a classical one. But the classical pattern blinks, twice per optical period, and the quantum one does not.

Classical wave — Eq. (3.10)

Ψ² = 4A² cos²(k(R₁−R₂)/2) · cos²ωt

Quantum particles — Eq. (3.13)

|Ψ|² = 4A² cos²(k(R₁−R₂)/2)

ωt0.00finite slit width (as in Fig. 3.2)

classical central peak now: 4.00 A²

quantum central peak now: 4.00 A² — and at every other instant

Run it. The left pattern collapses to nothing and returns, twice per optical period, because cos²ωt passes through zero twice per cycle — the faint dashed curve is its time average. The right pattern never changes: Eq. (3.13) contains no t. That is not a simplification, it is the result — the time factor e^−iωt has modulus one and cancels in Ψ*Ψ. What builds up over time on the right is not the pattern but the statistics: particles arriving one at a time, each landing somewhere at random, gradually filling in a curve that was already there.

Where the flicker went

It is tempting to think the time dependence was averaged away, or was too fast to see. Neither is true — it was never there.

Every term of Eq. (3.11) carries $e^{-i\omega t}$ , a complex number of modulus exactly one. Forming $\Psi^*\Psi$ pairs it with $e^{+i\omega t}$ and the product is 1. Nothing was approximated and nothing was averaged; the time factor cancelled identically.

This is not a quirk of the two-slit problem. Any wave function of the form $\psi(x)e^{-i\omega t}$ — and §2.2 showed these are the states of definite energy — has a probability density that is completely static, while the wave function itself keeps rotating in the complex plane. Chapter 4 is built on this, under the name stationary state.

A real classical wave cannot do it. $\cos(\theta - \omega t)$ has no partner to cancel against, so squaring it leaves $\cos^2$ behind, and the intensity oscillates. That footnote of Phillips’ — that he deliberately avoided complex exponentials for the classical wave — is doing real work: it keeps the distinction visible instead of hiding it in a convention.

Amplitudes are arrows

The algebra above is arrow addition in disguise. Each slit contributes a complex number — a length and a direction — and the two add tip to tail. The path difference sets the angle between them; $|\Psi|^2$ is the squared length of the resultant.

from S₁from S₂

Re Ψ: 2.000
Im Ψ: 0.000
|Ψ|²: 4.000

Engineer's bridge — coherent versus incoherent summation

$|A_1 + A_2|^2 \ne |A_1|^2 + |A_2|^2$ . The difference is the cross term $2\,\mathrm{Re}(A_1^*A_2)$ , and it is the entire interference pattern.

You have met this. In an antenna array or a phased-array radar, element contributions are summed as complex amplitudes before taking the magnitude, which is what produces a beam pattern with nulls; sum the powers instead and you get a smooth blob with no nulls at all. Same in optics (coherent versus incoherent illumination), in beamforming, and in the difference between abs(sum(x))**2 and sum(abs(x)**2) in any DSP routine you have written. In signal-processing language: quantum mechanics superposes amplitudes, and the detector measures power.

Where the analogy breaks — and it matters here. In an antenna array, every element genuinely radiates at once, and the field at a point is a real physical sum of real physical waves. In the two-slit experiment the electrons go through one at a time, and each one lands as a single dot. There is no physical field being summed over, and no second electron for the first to interfere with.

The superposition is of possibilities available to one particle, and the cross term is a statement about the probabilities of a single detection event. That difference is precisely what §3.6 will refuse to explain, and it is why this is a new kind of theory rather than classical wave optics with new symbols.

What builds up

Each dot is one particle arriving at one place. No individual dot shows any sign of interference; the pattern is a property of the ensemble, and it emerges only as a statistical statement about many identically prepared particles. This is what Fig. 3.2’s caption means when it distinguishes a classical pattern that oscillates in time from a quantum one that builds up gradually.

Switch on the which-path detector and the fringes vanish. The book’s assumption is stated plainly: identifying which slit the particle went through changes the wave function, collapsing it to a single wave from one slit. With one term instead of two there is no cross term and no pattern. §1.4 argues the same point from the momentum recoil of the screen. As Phillips notes, it is standard practice to assume a measurement can affect a wave function — and equally standard not to look too closely at how.

The Born interpretation

Generalising away from slits and screens: the wave function $\Psi(\mathbf r, t)$ is a complex function of position whose modulus squared measures the probability of finding the particle at $\mathbf r$ . The particle can be found anywhere, but is more likely where $|\Psi|^2$ is large. This is the Born interpretation , proposed in 1926:

|\Psi(\mathbf r, t)|^2\,\mathrm d^3\mathbf r = \left\{\begin{array}{l}\text{the probability of finding the particle}\\\text{at time } t \text{ in the volume element } \mathrm d^3\mathbf r\end{array}\right.\tag{3.14}

So $|\Psi|^2$ is a probability density for position , and $\Psi$ itself is called a probability amplitude for position — the thing you superpose, whose square gives the thing you measure.

Because the particle is certain to be somewhere, $\rho = |\Psi|^2$ must satisfy the normalization condition of §3.1, which is Eq. (3.5) with $\rho$ named:

\int |\Psi(\mathbf r, t)|^2\,\mathrm d^3\mathbf r = 1\tag{3.15}

In one dimension both statements shed their vector notation:

|\Psi(x, t)|^2\,\mathrm dx = \left\{\begin{array}{l}\text{the probability of finding the particle}\\\text{at time } t \text{ between } x \text{ and } x + \mathrm dx\end{array}\right.\tag{3.16}

\int_{-\infty}^{+\infty} |\Psi(x, t)|^2\,\mathrm dx = 1\tag{3.17}

Normalization is a choice you get to make, not a constraint you must check

The Schrödinger equation is linear: if $\Psi$ solves it, so does $c\Psi$ for any constant $c$ . It therefore cannot fix the overall size of a wave function, and any solution you derive arrives with an arbitrary multiplier attached.

Equation (3.17) is what pins it down. In practice you solve the equation, then multiply by whatever constant makes the integral equal 1. This is why the book says a wave function “is normalized by multiplying a solution by an appropriate constant” — normalization is a final tidying step, not a property you have to engineer into the solution.

Two consequences worth carrying forward. First, a wave function whose integral diverges — the plane wave $Ae^{i(kx-\omega t)}$ , for instance, whose $|\Psi|^2$ is the constant $|A|^2$ over an infinite line — cannot be normalized at all, and is a mathematical idealisation rather than a state a particle can be in. Second, normalizing at $t = 0$ is enough: the Schrödinger equation preserves the total probability for all later times, which problem 8 establishes via the probability current.

The footnote about h/mc — a limit that does not apply here

Phillips attaches a caveat: the Born interpretation lets a particle’s position be pinned down arbitrarily precisely, at the cost of total uncertainty in its momentum — yet §1.4 showed that a particle of mass $m$ cannot be localized better than about $h/mc$ , its Compton wavelength.

The two are not in conflict, because they belong to different theories. The $h/mc$ limit comes from relativistic reasoning: squeeze a particle harder than that and the momentum uncertainty implies an energy uncertainty exceeding $mc^2$ , enough to create new particles, and a single-particle description stops making sense. Non-relativistic quantum mechanics — the whole of this book — never goes near that regime, so within it Eq. (3.14) is unrestricted.

Worth remembering as a boundary marker: it is where this book’s framework ends and quantum field theory begins.

Put a number on it — where does an electron land?

Take a two-slit experiment with 50 keV electrons, slit separation $d = 1\ \mu\mathrm m$ , screen $D = 1\ \mathrm m$ away.

The de Broglie wavelength from §1.2 is $\lambda = h/p \approx 5.4\ \mathrm{pm}$ — about 100 000 times shorter than visible light. The fringe spacing is $\lambda D/d \approx 5.4\ \mu\mathrm m$ : small, but resolvable, and this is essentially why electron interference took until 1927 to see.

Now the point of the section. At a bright fringe $|\Psi|^2$ is at its maximum; at a dark fringe it is exactly zero. Neither number tells you where the next electron will land. What they give you is the probability of each landing site, and the pattern is a prediction about ten thousand electrons, not about one. Fire a single electron and it makes a single dot — very probably in a bright region, never in a dark one, and beyond that unpredictable in principle.

Reproduce it in NumPy

Everything the section claims, evaluated from Eqs. (3.9)–(3.13) rather than asserted — including the factor of two flagged in the Caution above.

import numpy as np

A, k, w = 1.0, 2*np.pi, 3.0
d  = np.linspace(-2, 2, 400001)          # path difference R1 - R2
R1 = np.full_like(d, 5.0); R2 = R1 - d
theta = k*(R1 - R2)

print("two-slit intensity, checked against the definitions")
print("="*60)
Psi = A*np.exp(1j*k*R1) + A*np.exp(1j*k*R2)      # Eq (3.11) at t = 0
I   = np.abs(Psi)**2                             # Eq (3.12)
print(f"peak of |Psi|^2 computed directly from Eq (3.11) : {I.max():.4f} A^2")
print(f"  Eq (3.13) as printed, 2A^2 cos^2(...)          : "
      f"{(2*A**2*np.cos(theta/2)**2).max():.4f} A^2   <- half")
print(f"  with the coefficient 4A^2                      : "
      f"{(4*A**2*np.cos(theta/2)**2).max():.4f} A^2   <- matches")
print(f"mean of |Psi|^2 across the pattern               : {I.mean():.4f} A^2")
print("  (so 2A^2 is the correct AVERAGE, not the peak)")

t = np.linspace(0, 2*np.pi/w, 5001)
cl = (A*np.cos(-w*t) + A*np.cos(-w*t))**2                       # Eq (3.9), R1 = R2
qu = np.abs(A*np.exp(-1j*w*t) + A*np.exp(-1j*w*t))**2           # Eq (3.11), R1 = R2
print("\ntime dependence")
print(f"  classical Psi^2 at the centre, over one period : "
      f"min {cl.min():.4f}  max {cl.max():.4f}")
print(f"  quantum  |Psi|^2 at the centre, over one period: "
      f"min {qu.min():.4f}  max {qu.max():.4f}")
print("  -> the classical pattern vanishes twice per period; "
      "the quantum one is constant")

print("\namplitudes superpose, probabilities do not")
print(f"  |A1|^2 + |A2|^2          = {2*A**2:.4f} A^2   "
      f"(no interference: a smooth blob)")
print(f"  |A1 + A2|^2 at a maximum = {I.max():.4f} A^2")
print(f"  |A1 + A2|^2 at a minimum = {I.min():.4f} A^2")
print("  the cross term integrates to zero: total probability is only redistributed")

printstwo-slit intensity, checked against the definitions ============================================================ peak of |Psi|^2 computed directly from Eq (3.11) : 4.0000 A^2 Eq (3.13) as printed, 2A^2 cos^2(...) : 2.0000 A^2 <- half with the coefficient 4A^2 : 4.0000 A^2 <- matches mean of |Psi|^2 across the pattern : 2.0000 A^2 (so 2A^2 is the correct AVERAGE, not the peak) time dependence classical Psi^2 at the centre, over one period : min 0.0000 max 4.0000 quantum |Psi|^2 at the centre, over one period: min 4.0000 max 4.0000 -> the classical pattern vanishes twice per period; the quantum one is constant amplitudes superpose, probabilities do not |A1|^2 + |A2|^2 = 2.0000 A^2 (no interference: a smooth blob) |A1 + A2|^2 at a maximum = 4.0000 A^2 |A1 + A2|^2 at a minimum = 0.0000 A^2 the cross term integrates to zero: total probability is only redistributed

Where this leaves us

Two ideas carry the whole argument, and Phillips lists them explicitly: the wave function at the screen is a linear superposition of a wave from each slit, and the probability of detection is proportional to $|\Psi|^2$ at that point. Everything else on this page follows.

What has not been explained is how a spread-out wave function turns into one dot in one place. The book is candid that it is not going to try, and makes a striking claim about why that is acceptable: quantum mechanics succeeds because it avoids explaining how events happen. It predicts the probabilities and stops.

Next: §3.3 does for momentum exactly what this section did for position — and finds that the same wave function already contains the answer.

Check yourself

0 / 7 answered

1.Comparing Eq. (3.10) with Eq. (3.13), what is the one structural difference?
2.Why does the $e^{- iω t}$ in Eq. (3.11) leave no trace in $∣Ψ ∣^{2}$ ?
3.Two amplitudes $A$ arrive exactly in phase at the centre of the pattern. What is $∣Ψ ∣^{2}$ there?
4.At a dark fringe the probability of detection is zero. Where did that probability go?
5.The two-slit sum resembles coherent summation in a phased-array antenna. Where does that analogy break down?
6.A solution of the Schrödinger equation comes with an arbitrary overall constant. Why, and what fixes it?
7.Footnote 2 notes that §1.4 limited a particle's localization to about $h / m c$ , while Eq. (3.14) appears to allow unlimited precision. How is that resolved?