Quantum Scattering Differential Probability

In summary: Therefore, the relation $\frac{d\sigma}{d\Omega} = |f(\theta)|^2$ is a quantum analog of the classical scenario, where a particle passing through an area $d\sigma$ is guaranteed to pass through the corresponding solid angle $d\Omega$ due to the symmetry of the target.
  • #1
CDL
20
1
I am reading Griffiths' Introduction to Quantum Mechanics, specifically the chapter on scattering. He is discussing the scenario where an incoming beam of particles scatter off an azimuthally symmetric target.

At large separation ##r## from the scattering centre, the wavefunction for incoming particles is $$\psi(r, \theta) \approx A \left[e^{ikz} + f(\theta) \frac{e^{ikr}}{r}\right] = \psi_\text{incident} + \psi_\text{scattered}$$ He says,

"The probability that the incident particle, traveling at speed v, passes through the infinitesimal area ##d \sigma##, in time ##dt## is equal to the probability that the particle scatters into the corresponding solid angle ##d \Omega##."

He equates ##dP =|\psi_{\text{incident}}|^2 dV = |A|^2 (v dt) d\sigma## to ##dP = |\psi_{\text{scattered}}|^2 dV = \frac{|A|^2 |f|^2}{r^2} (vdt) r^2 d \Omega##

This is used to derive the relation ##\frac{d \sigma}{ d \Omega} = | f (\theta) | ^2##.

Is this the quantum analog of the classical scenario, where when a particle passes through an area ##d \sigma## it is guaranteed to pass through the solid angle ##d \Omega##? Why should this be true? I can't wrap my head around why this claim would be made. What happens, say if we know that a particle passed through an area ##d \sigma##. Would that affect the probability of it passing through the corresponding solid angle ##d \Omega##? To me it seems like the claim is plausible, but a more in depth and hopefully intuitive explanation would be appreciated.
 
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  • #2
A:It's not true in general that the probability of passing through an area $d\sigma$ is equal to the probability of scattering into a solid angle $d\Omega$. That would only be true if the cross section for the scattering process was independent of the direction of the scattered particle, i.e. $\frac{d\sigma}{d\Omega}$ was independent of $\theta$. This is obviously not true in general, since even for a simple elastic scattering process, the magnitude of the scattering angle depends on the relative momentum of the incoming and outgoing particles.However, Griffiths' statement is true in the case of azimuthally symmetric targets. In this case, the scattering cross section does not depend on the direction of the scattered particle, and thus the probability of passing through an area $d\sigma$ is equal to the probability of scattering into a solid angle $d\Omega$. This can be seen by noting that the probability of scattering into a solid angle $d\Omega$ is proportional to the product $\frac{d\sigma}{d\Omega} \times d\Omega$, where $\frac{d\sigma}{d\Omega}$ is independent of $\theta$. Since the area element $d\sigma$ is related to the solid angle element $d\Omega$ by $d\sigma = r^2 d\Omega$, the two probabilities must be equal.
 

FAQ: Quantum Scattering Differential Probability

1. What is quantum scattering differential probability?

Quantum scattering differential probability refers to the likelihood of a quantum particle being scattered in a particular direction when it interacts with another particle or potential barrier. It is a fundamental concept in quantum mechanics that is used to calculate the scattering of particles in a quantum system.

2. How is quantum scattering differential probability calculated?

The calculation of quantum scattering differential probability involves solving the Schrödinger equation for the wave function of the particle in the presence of a potential barrier or scattering potential. The square of the wave function gives the probability of finding the particle at a particular position, which can then be used to calculate the differential probability of the particle being scattered in a particular direction.

3. What factors affect quantum scattering differential probability?

The quantum scattering differential probability is affected by the properties of the scattering potential, such as its shape, strength, and distance from the particle. It is also influenced by the properties of the incident particle, such as its energy, mass, and spin. Additionally, the angle at which the particle approaches the scattering potential can also affect the differential probability.

4. How is quantum scattering differential probability related to quantum tunneling?

Quantum scattering differential probability is closely related to quantum tunneling, which is the phenomenon of a particle passing through a potential barrier that it does not have enough energy to overcome in classical mechanics. The differential probability of a particle being scattered at a particular angle is affected by the probability of tunneling through the barrier at that angle.

5. How does quantum scattering differential probability impact real-world applications?

Quantum scattering differential probability plays a crucial role in understanding and predicting the behavior of particles in various physical systems, such as in nuclear reactions, semiconductors, and nanotechnology. It also has practical applications in fields such as quantum computing and quantum cryptography, where the manipulation of particles through scattering is used to perform quantum operations and secure communication, respectively.

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