Machine-translated by Google. Questions are authored in English.
Top

Discussion

The normalized wave function of a particle can be written as

\[ \begin{aligned} \psi(x) = N\sum_{n=0}^{\infty}&\left(\frac{1}{\sqrt{7}}\right)^{n} \\ &\phi_{n}(x) \end{aligned} \]

where \(\phi_{n}(x)\) are the normalized energy eigenfunctions of a given Hamiltonian. The value of N is :

  • A.\(\sqrt{\frac{6}{7}}\)
  • B.\(\sqrt{\frac{1}{7}}\)
  • C.\(\sqrt{\frac{3}{7}}\)
  • D.\(\sqrt{\frac{13}{7}}\)

Answer: A

The normalisation constant is \(N = \sqrt{6/7}\).

The eigenfunctions are orthonormal, so the norm of \(\psi\) is the sum of the squared moduli of its coefficients:

\[ \begin{aligned} 1 &= N^{2}\sum_{n=0}^{\infty}\left(\frac{1}{7}\right)^{n} \\ &= N^{2}\cdot\frac{1}{1 - 1/7} = N^{2}\cdot\frac{7}{6} \end{aligned} \]

so \(N^{2} = 6/7\) and \(N = \sqrt{6/7}\).

The coefficients \((1/\sqrt{7})^{n}\) must be squared before summing, giving a geometric series of ratio 1/7. Forgetting to square them, or stopping after the first term (which gives \(N = 1\)), leads to the wrong values; \(\sqrt{1/7}\), \(\sqrt{3/7}\) and \(\sqrt{13/7}\) do not make the sum equal to 1.

No comment is present. Be the first to comment.
Loading…

Post your comment

Please log in or register to post a comment.