Question

\(\psi(x,y,z)\) describes the wavefunction of a particle. The probability of finding the particle between \(x\) and \(x+dx\), \(y\) and \(y+dy\), and \(z\) and \(z+dz\), can be expressed as

• A. \( \psi^*(x,y,z) \psi(x,y,z) \)
• B. \( |\psi(x,y,z)|^2 \, dx \, dy \, dz \)
• C. \( \psi^*(x,y,z) \psi(x,y,z) \, dx \, dy \, dz \)
• D. \( \int_{-\infty}^{\infty} \, dx \int_{-\infty}^{\infty} \, dy \int_{-\infty}^{\infty} \, dz \, \psi^*(x,y,z) \psi(x,y,z) \)

Hint:

The probability of finding a particle in a given region is the square of the wavefunction's magnitude, \( |\psi(x, y, z)|^2 \).

Answer 1

The Correct Option is B, C

Step 1: Analysis:

The probability density is |ψ(x,y,z)|², and the probability of finding the particle in a small volume element dx dy dz is:

P = |ψ(x,y,z)|² dx dy dz

Step 2: Evaluating each option:

*(A) ψ(x,y,z) ψ(x,y,z)**: ✗

• Missing the volume element dx dy dz
• This is just the probability density, not the probability

(B) |ψ(x,y,z)|² dx dy dz: ✓

• This is the standard form
• |ψ|² = ψ*ψ is the probability density
• Multiplied by the volume element gives probability

(C) ψ(x,y,z) ψ(x,y,z) dx dy dz*: ✓

• Since |ψ|² = ψ*ψ, this is equivalent to option (B)
• This explicitly shows the probability density as ψ*ψ
• Includes the volume element

*(D) ∫∫∫ dx dy dz ψ(x,y,z) ψ(x,y,z)**: ✗

• This integral over all space gives the normalization condition = 1
• It gives the total probability (which must equal 1 for a normalized wavefunction)
• Not the probability in a specific small volume element

Answer: (B) and (C) are correct.