Question:

Consider the wavefunction πœ“(π‘₯) = 𝑁[exp(π‘–π‘˜π‘₯) + exp(βˆ’π‘–π‘˜π‘₯)]. The complex conjugate πœ“ βˆ— (π‘₯) is [Given: 𝑁 is the normalization constant; 𝑖=\(\sqrt-1\)]

Updated On: Jan 16, 2025
  • 𝑁[exp(βˆ’π‘–π‘˜π‘₯) βˆ’ exp(π‘–π‘˜π‘₯)]
  • π‘βˆ—[exp(βˆ’π‘–π‘˜π‘₯) βˆ’ exp(π‘–π‘˜π‘₯)]
  • π‘βˆ—[exp(π‘–π‘˜π‘₯) + exp(βˆ’π‘–π‘˜π‘₯)]
  • 2𝑁[sin(π‘˜π‘₯)]
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The Correct Option is C

Solution and Explanation

  • The complex conjugate Οˆβˆ—(x) is obtained by replacing i with βˆ’i and taking the conjugate of the normalization constant N, resulting in Nβˆ—.
  • Applying this to ψ(x) = N[exp(ikx) +exp(βˆ’ikx)]:

    \( ψ^*(x) = N^*[\exp(-ikx) + \exp(ikx)] \)

    \( = N^*[\exp(ikx) + \exp(-ikx)]. \)

Thus, the correct expression for the complex conjugate is \(N^*[\exp(ikx) + \exp(-ikx)].\)

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