Double-slit experiment

Click here to read the Wikipedia article on path integral formulation.



The path integral formulation of quantum mechanics developed by Richard Feynman uses the so-called time-slicing technique. It involves evolving the system one step at a time with each step covering a very small slice of the entire time interval. To do this, Feynman in his book Quantum Mechanics and Path Integrals first described the double-slit experiment for photons or light particles. He then patterned similar reasoning to come up with the so-called probability amplitude K(b,a) to go from position x_a at the time t_a to position x_b at t_b. Then the probability to go from x_a to x_b is given by

P(b,a) = |K(b,a)|².

We only consider one-dimensional motion x(t) here to illustrate the idea.

Double-slit experiment
K(b,a) is the sum of all amplitude φ[x(t)] for each path x(t). Let us consider first the double-slit experiment.

Figure 1
Figure 2
Figure 3
Figure 4

In the four figures above the red curve gives the intensity of light on the screen when light is considered as a wave. If light is a particle, then the curve is interpreted as proportional to the probability that the light particle (or photon) will hit at a certain location on the screen. Actually, the curves in Figures 1, 3, and 4 are the same except that they are translated at different positions depending on where the single slit is located.

Now, if light is a particle and using the probabilistic interpretation of the curves, we would say that the probability that the particle will pass through one slit (P₁) is equal to the probability that the particle will pass through the other slit (P₂). Since the particle will pass through one slit or through the other, the properties of probability tells us that the curve should show P₁ + P₂ as shown below (Figure 5).

Figure 5

But this is not the case in actual double-slit experiment. Thus, there must be other way of interpreting the probability curve.

Classical Action

One-dimensional system

Action
S[x(t)] = ∫(L[x,x^*,t]*dt) t_a to t_b

Lagrangian
L[x,x^*,t] = T - V = mv²/2 - V(x)

In classsical mechanics, the solution x(t) is the path x^-(t) that will give the minimum action. This condition gives the Euler-Lagrange equation:

d/dt{∂L/∂x^*)} - ∂L/∂x = 0

Example: Free particle
Calculate the action of the path of a free particle in one dimension.
L = m*x^*2/2

Solution
1. ∂L/∂x = 0
2. ∂L/∂x^*) = m*x^*
3. d/dt{result in 2} = m*x^**
4. Euler-Lagrange equation is m*x^**= 0
5. E-L equation gives x(t) = A + B*t
6. Let x_a = x(t_a) and x_b = x(t_b)
7. x^* = B
8. From #5 and #6, B = (x_b-x_a)/(t_b-t_a)
9. L = m*(B)²/2 = m*B²/2
10. S = ∫(m*B^2/2 * dt) = m*B^2/2*∫(dt) from t_a to t_b
Finally,
S = m*(x_b-x_a)^2/(t_b-t_a)

That's all!

Schrodinger equation (1D)

discussed last saturday july 9 for maricris.

Time-independent potential
The SE is just an eigenvalue problem for the hamiltonian H = T+V. That is,

Hφ(x) = Eφ(x)

where
T = kinetic energy = (1/2)mv² = p²/2m
V = V(x) <--- position representation

In the position representation, the momentum p is given by

p^→ = (hbar/i)Del^→

or in 1D

p_x = (hbar/i)d/dx

Thus, 1D SE in position representation becomes

-(hbar²/2m)d²φ/dx² + V(x)φ(x) = Eφ(x)

Complete SE (time-dependent)

HΨ(x,t) = i*hbar*∂Ψ(x,t)/∂t

To arrive at the time-independent case, just remember the following position-time-separated form of Ψ(x,t):

Ψ(x,t) = e^-i*E*t/hbarφ(x)

That's all.

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