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Double slit Experiment & The Aharnov-Bohm
effect
Manoj Krishna
May 2025
1 The double slit experiment
The below schematic describes the setup for the two-slit experiment.
Observations from the two-slit experiment performed with electrons are as fol- lows:
- The plot of probability density of an electron being found at a distance ‘x’ from the centre of the screen vs ‘x’ is similar to that of when the plot of intensity (I) vs ‘x’ when the experiment is performed with waves such as water waves.
- Each time the detector detects an electron, it is a complete electron.
From the above observations, quoting Feynman, ”the electrons arrive at the de- tector as a whole, like particles, and the probability of arrival of these electrons
is distributed like the distribution of intensity of a wave. It is in this sense that an electron behaves sometimes like a particle and sometimes like a wave”.
A remarkable feature of the two-slit experiment: It is reasonable to say that any electron that reaches the detector must have passed through either of the two holes. Any attempt to determine the hole through which an electron reaching the screen had passed will interfere with the experiment in such a way that the interference pattern (P 12 in Fig3-1) will be lost and the probability density will be as shown below (P 12 in Fig1-1)
Hence, the probability when we look at the electrons is different from when we do not look at the electrons!! General principles of Quantum mechanics in the given context (as mentioned in the third chapter of Feynman lectures Vol.III which follow from the First principles mentioned in the first chapter of Feynman lectures Vol.III).
- The probability that a particle will arrive at x, when let out at the source s, can be represented quantitatively by the absolute square of a complex number called a probability amplitude.
- When a particle can reach a given state by two possible routes, the total amplitude for the process is the sum of the amplitudes for the two routes considered separately.
- When a particle goes by some particular route the amplitude for that route can be written as the product of the amplitude to go part way with the amplitude to go the rest of the way.
Following the first and second principles mentioned above one can say < x|s >bothholesopen = < x|s >through 1 + < x|s >through 2 Let us denote < x|s >both holes open by ψ and < x|s >through 1 by ψ 1 and < x|s >through 2 by ψ 2. Then the above expression becomes ψ = ψ 1 + ψ 2
Note λ′^ is a function of position and time i.e. λ′(r, t)
2.2.2 In Quantum dynamics
We note that the Schr¨odinger’s equation is Gauge invariant.
i¯h
∂ψ′ ∂t
= H′ψ (5)
i¯h
∂ψ ∂t = Hψ (6)
Where H is the hamitonian operator of the form for a particle subjected to aan electromagnetic field which is given by H = (^21) m ( ¯hi ∇ − qc A)^2 + qϕ where A is the vector potential and ϕ is the scalar potential
Claim: The Schr¨odinger’s equation is Gauge invariant i.e. equation(5) ⇔ equation(6), if the wave function transforms according to ψ′^ = U ψ where U = exp(i (^) hc ¯q λ) ; where = λ(r, t) Proof: Observe that ( ¯hi ∇ − qc A’)U ψ = U ( ¯hi ∇ − qc Aψ) substituting the above equation into the equation (5) and using the fact that U = e(^ iqλ ¯hc ) the equation simplifies into equation (6).
Supplementry calculation: Expand the Hamiltonian as follows, (^21) m ( ¯hi ∇ − qc A′).( ¯hi ∇ − qc A′) + qϕ′
H = (^21) m (p.p − qc (p.A) − qc (A.p) + q
2 c^2 A.A) +^ pϕ
By evaluating ¯hi ∇.(Aψ) we can say p.A = ¯hi (∇.A) + A.p Hence the Hamiltonian equation becomes H = (p.p + 2 iqmc¯h (∇.A) − (^) mcq (A.p) + q
2 2 mc^2 A.A) +^ qϕ.
Conclusion: Hence we have showed that a gauge transformation results in a phase shift of the wave function by a factor U.
2.3 The double-slit set up revisited
An infinite current carrying solenoid is introduced between the slits as shown below (we are still performing the double slit experiment using electrons)
Let us compare what is difference when the solenoid is present as to when the solenoid is not present. Case 1: The solenoid is absent i.e.B=0,A=0 Let ψ be the wave function of the electrons satisfying the Schr¨odinger’s equation. (Recall the equation ψ = ψ 1 + ψ 2 from Section 1 and the fact that this equation suggests that the wave function ψ is a result of interference of ψ 1 and ψ 2 .) Case 2: B=0,A′^ ̸= 0 B=0 since the trajectories of the electrons with with we are concerned are out- side the solenoid since the solenoid is impenetrable. Since B=0 in both cases but A is not the same we can use Gauge transforma- tions. Outside the solenoid the magnetic vector potential is given by (in the cylindrical co-ordinates) A’(r) = 2 ϕπr θ; where ϕ is the magnetic flux density of the solenoid. Now let’s find the suitable λ such that A′^ = A + ∇λ Since A=0 we have A’(r) = 2 ϕπr θ = ∇λ λ = 2 ϕπ is an appropriate choice since using ∇λ = ∂λ∂r r + (^1) r∂λ∂θ + ∂λ∂z give us A Let us denote the phase difference between ψ 1 and ψ 2 by △Φ then, △Φ = (^) ¯hce △λ = (^) ¯hce ϕ; since U = exp(i (^) ¯hcq λ) and △λ = ϕ This phase difference can be more appropriately calculated by using the path integral (^) ¯hq
H
aba A dl^ followed by the use of Stokes theorem.
The interference of the waves will depend on this phase difference, and hence we have a measurable effect of the potentials even in the absence of the fields. Varying the potential will result in the shifting of the interference pattern. The potential can be varied by varying the solenoid current and in turn varying the magnetic field within the solenoid.
