Question about calculating |Wavefunction|^2

Question

In one of my homework in Quantum mechanics, I was asked to find $|Ψ(x,t)|^2$, where \begin{align} Ψ(x,t)&=1/\sqrt{10}[3ψ_1(x)e^{-iE_1t/ħ}-ψ_3(x)e^{-iE_3t/ħ}]\, ,\\ &=1/\sqrt{10}[3\sqrt{2/a} \sin(\pi x/a)e^{-iE_1t/ħ}- \sqrt{2/a}\sin(3\pi x/a)e^{-iE_3t/ħ}] \end{align} So, from my calculation steps: \begin{align}|Ψ(x,t)|^2&=Ψ^*(x,t)Ψ(x,t)\, ,\\ &=\frac{1}{10}[3ψ_1(x)e^{-iE_1t/ħ}-ψ_3(x)e^{-iE_3t/ħ}]^* [3ψ_1(x)e^{-iE_1t/ħ}-ψ_3(x)e^{-iE_3t/ħ}]\, ,\\ &=\frac{1}{10}[3ψ_1^*(x)e^{iE_1t/ħ}-ψ_3^*(x)e^{iE_3t/ħ}]^* [3ψ_1(x)e^{-iE_1t/ħ}-ψ_3(x)e^{-iE_3t/ħ}]\, ,\\ &=\frac{1}{10}[9ψ_1^*(x)ψ_1(x)-3ψ_3^*(x)ψ_1(x)e^{iE_3t/ħ}e^{iE_1t/ħ}-3ψ_1^*(x)ψ_3(x)e^{iE_1t/ħ}e^{iE_3t/ħ}-ψ_3^*(x)ψ_3(x)] \end{align} However the answer from the internet shows that the result after tedious simplifications is: $$\frac{1}{10}[9ψ_1^2(x)+ψ_3^2(x)-6ψ_1(x)ψ_3(x)\cos((E_3-E_1)/ħ)t] $$ So, the question is that how does $9ψ_1^*(x)ψ_1(x)$ becomes $9ψ_1^2(x)$, $ψ_3^*(x)ψ_3(x)$ becomes $ψ_3^2(x)$ and $-3ψ_3^*(x)ψ_1(x)e^{iE_3t/ħ}e^{iE_1t/ħ}-3ψ_1^*(x)ψ_3(x)e^{iE_1t/ħ}e^{iE_3t/ħ}$ becomes $-6ψ_1(x)ψ_3(x)\cos((E_3-E_1)/ħ)t$?

Because the professor or tutor doesn't go very deeply in calculating wavefunction conjugation and the rules of conjugation, I have no idea how to further simplify the problem in to the form in the answer found in the internet. It would be great if someone can explain to me the rules or tricks in dealing with $ψ^*(x)$. Thanks

Answer 1

For a huge set of physically relevant hamiltonians, the energy eigenfunctions can be set as completely real-valued in the position representation if so desired. The solution you quote has made use of this possibility; this should normally be explicitly marked at some point but it's not an Earth-shattering omission if it's not.

If you don't want to make that assumption, then the simplest you can get your expression is $$ \frac{1}{10}\left[9|ψ_1(x)|^2+|ψ_3(x)|^2-6\,\operatorname{Re}\left(ψ_1(x)^*ψ_3(x)e^{i(E_3-E_1)t/ħ}\right)\right]. $$ Here the appearance of the product $ψ_1(x)^*ψ_3(x)$ is important, as it marks the times at which the interference cosine is at a maximum. This can be rephrased explicitly as $$ \frac{1}{10}\left[9|ψ_1(x)|^2+|ψ_3(x)|^2-6|ψ_1(x)^*ψ_3(x)|\cos\left((E_3-E_1)t/ħ+\arg(ψ_1(x)^*ψ_3(x))\right)\right], $$ where the cosine has an added phase given by the complex argument of the wavefunction product, $\arg(ψ_1(x)^*ψ_3(x))$, but that's not normally considered to be an improvement on the previous phrasing.

Answer 2

First note that in general the wavefunction is complex, not real. This answers the first two questions you asked, since for any complex number $z = a + ib$, $|z|^2 = z^*z$. Then, \begin{equation} \psi_1^*(x) \psi_1(x) = |\psi_1(x)|^2 \quad \text{and} \quad \psi_3^*(x) \psi_3(x) = |\psi_3(x)|^2. \end{equation} You can save time in the next part by recognizing that one term is simply the complex conjugate of the other. To illustrate, define $A$ as, \begin{equation} A = 3\psi_1^* \psi_3 e^{i(E_1-E_3)t/\hbar}, \end{equation} then you are trying to do the following sum, \begin{equation} -A - A^*. \end{equation} Since $A$ is complex, we can say $A = a+ib$, and then the sum looks like \begin{align} -A - A^* &= -(a+ib) - (a-ib) \\ &= -2a\\ &= -2 \text{Re}[A]\\ &= - 2 \text{Re}[3\psi_1^* \psi_3 e^{i(E_1-E_3)t/\hbar}] \\ &=-6 \psi_1 \psi_3 \cos[(E_1-E_2)t/\hbar] \end{align} Where I went from the second to last line, to the last line, by assuming $\psi_1$ was real, so $\psi_1^* = \psi_1$.

Also note that in your calculations you have dropped signs in a few places when expanding the products of the complex conjugates, notably, in the exponential factors.