Math Problem Statement

Solve

A spin-1/2 particle is in a state described by the spinor χ = A [(1+i), 2], where A is a constant. (a) Determine the normalization constant A. (b) What is the probability of finding the particle with spin projection Sz = +ħ/2?

Solution

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Problem Overview:

The spinor state of a spin-12\frac{1}{2} particle is given by: χ=A(1+i2),\chi = A \begin{pmatrix} 1 + i \\ 2 \end{pmatrix}, where AA is a normalization constant. The questions are:

  1. (a) Determine the normalization constant AA.
  2. (b) Calculate the probability of finding the particle with spin projection Sz=+2S_z = +\frac{\hbar}{2}.

Part (a): Normalization Constant AA

The spinor must satisfy the normalization condition: χχ=1,\chi^\dagger \chi = 1, where χ\chi^\dagger is the Hermitian conjugate (complex conjugate transpose) of χ\chi.

Step 1: Write out the normalization condition.

χχ=A2[(1+i)(1+i)+22]=1,\chi^\dagger \chi = |A|^2 \left[ (1 + i)^* (1 + i) + 2^* 2 \right] = 1, where (1+i)(1 + i)^* is the complex conjugate of 1+i1 + i.

Step 2: Calculate the terms.

  1. (1+i)(1+i)=(1i)(1+i)=1+1=2(1 + i)^* (1 + i) = (1 - i)(1 + i) = 1 + 1 = 2,
  2. 22=2×2=42^* 2 = 2 \times 2 = 4.

So: χχ=A2(2+4)=A26.\chi^\dagger \chi = |A|^2 (2 + 4) = |A|^2 \cdot 6.

Step 3: Solve for A2|A|^2.

A26=1    A2=16.|A|^2 \cdot 6 = 1 \implies |A|^2 = \frac{1}{6}.

Thus, the normalization constant is: A=16.A = \frac{1}{\sqrt{6}}.


Part (b): Probability for Sz=+2S_z = +\frac{\hbar}{2}

The spin projection operator SzS_z has eigenstates: Sz=+2    corresponding eigenvector: (10).S_z = +\frac{\hbar}{2} \implies \text{corresponding eigenvector: } \begin{pmatrix} 1 \\ 0 \end{pmatrix}.

The probability of finding the particle in the Sz=+2S_z = +\frac{\hbar}{2} state is given by: P=+χ2,P = |\langle + | \chi \rangle|^2, where +=(10)|+\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} and χ=A(1+i2)\chi = A \begin{pmatrix} 1 + i \\ 2 \end{pmatrix}.

Step 1: Compute the inner product +χ\langle + | \chi \rangle.

+χ=(10)A(1+i2)=A(1+i).\langle + | \chi \rangle = \begin{pmatrix} 1 & 0 \end{pmatrix} A \begin{pmatrix} 1 + i \\ 2 \end{pmatrix} = A (1 + i).

Step 2: Calculate the modulus squared.

+χ2=A21+i2.|\langle + | \chi \rangle|^2 = |A|^2 |1 + i|^2.

The modulus squared of 1+i1 + i is: 1+i2=(1+i)(1+i)=(1i)(1+i)=1+1=2.|1 + i|^2 = (1 + i)^* (1 + i) = (1 - i)(1 + i) = 1 + 1 = 2.

Thus: +χ2=A22=162=26=13.|\langle + | \chi \rangle|^2 = |A|^2 \cdot 2 = \frac{1}{6} \cdot 2 = \frac{2}{6} = \frac{1}{3}.


Final Answers:

(a) The normalization constant is: A=16.A = \frac{1}{\sqrt{6}}.

(b) The probability of finding the particle with spin projection Sz=+2S_z = +\frac{\hbar}{2} is: P=13.P = \frac{1}{3}.


Would you like further clarification or more examples on this topic? Here are some related questions:

  1. What are eigenstates and eigenvalues of the spin operator SzS_z?
  2. How does the probability change if the spinor is rotated?
  3. What is the physical meaning of the normalization constant?
  4. Can we calculate probabilities for Sz=2S_z = -\frac{\hbar}{2} with the same method?
  5. How does this formalism generalize to systems with higher spin values?

Tip: Always ensure spinor wavefunctions are normalized before calculating physical quantities like probabilities.

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Math Problem Analysis

Mathematical Concepts

Quantum Mechanics
Spinors
Probability
Normalization

Formulas

Normalization condition: χ†χ = 1
Inner product: ⟨ψ|χ⟩
Probability: P = |⟨ψ|χ⟩|^2

Theorems

Hermitian Conjugate Properties
Spin Eigenstate Projections

Suitable Grade Level

Undergraduate Physics