Suppose \(\psi(x)\) is a properly normalized wave function describing the state of an electron. Consider a second wave function, \(\psi_{\text {new }}(x)=e^{i \phi} \psi(x),\) for some real number \(\phi .\) How does the probability density associated with \(\psi_{\text {new }}\) compare to that associated with \(\psi ?\)

We are given that \(\psi(x)\) is a properly normalized wave function describing the state of an electron. To calculate the probability density associated with \(\psi(x)\), we need to find \(|\psi(x)|^2\). The probability density is given by: $$ P_{\psi}(x) = |\psi(x)|^2 $$

We are given that \(\psi_{\text{new}}(x) = e^{i\phi}\psi(x), \text{ where } \phi \text{ is a real number}\). We need to find \(|\psi_{\text{new}}(x)|^2\). The probability density is given by: $$ P_{\psi_{\text{new}}}(x) = |\psi_{\text{new}}(x)|^2 $$ Let's find \(|\psi_{\text{new}}(x)|\) first. We have \(\psi_{\text{new}}(x) = e^{i\phi}\psi(x)\); thus, $$ |\psi_{\text{new}}(x)| = |e^{i\phi}\psi(x)| $$ Now, let's use the property \(|ab| = |a||b|\) for complex numbers \(a\) and \(b\). We have: $$ |\psi_{\text{new}}(x)| = |e^{i\phi}||\psi(x)| $$ Since \(|e^{i\phi}| = 1\), we find: $$ |\psi_{\text{new}}(x)| = |\psi(x)| $$ Now we can calculate \(P_{\psi_{\text{new}}}(x)\): $$ P_{\psi_{\text{new}}}(x) = |\psi_{\text{new}}(x)|^2 = |\psi(x)|^2 $$

We have calculated the probability densities associated with the wave functions \(\psi(x)\) and \(\psi_{\text{new}}(x)\). We found that: $$ P_{\psi}(x) = |\psi(x)|^2 $$ and $$ P_{\psi_{\text{new}}}(x) = |\psi_{\text{new}}(x)|^2 = |\psi(x)|^2 $$ Since \(P_{\psi}(x) = P_{\psi_{\text{new}}}(x)\), we conclude that the probability density associated with \(\psi_{\text{new}}(x)\) is the same as that associated with \(\psi(x)\).