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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

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  • $\begingroup$ $\psi_1$ and $\psi_3$ are real, this helps you simplify things. But in general it's just complex numbers, the dependence on $x$ doesn't change anything. $\endgroup$ – Javier Feb 10 '18 at 8:29
  • $\begingroup$ I am not quite sure what your question actually is, sorry. One "trick" is that which you are using yourself, to use $ \Psi$ as a placeholder for the actual superposition of possible states. It can't be further simplified when you need to calculate amplitudes, AFAIK. $\endgroup$ – user184116 Feb 10 '18 at 10:16
  • $\begingroup$ You need first to account for the fact that the wavefunction $\psi$ is in general complex so you need to compute $\psi\psi^*$. Then use Euler’s formula to convert the exponentials to trigonometric functions. $\endgroup$ – ZeroTheHero Feb 10 '18 at 14:37
  • $\begingroup$ Note also that you have some sign errors in your expressions, i.e. the term in $\psi_3\psi_3^*$ should be positive. $\endgroup$ – ZeroTheHero Feb 10 '18 at 14:42
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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.

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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.

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    $\begingroup$ The identification $\text{Re}[\psi_1^* \psi_3 e^{i(E_1-E_3)t/\hbar}] = \text{Re}[\psi_1^* \psi_3]\text{Re}[ e^{i(E_1-E_3)t/\hbar})]$ is incorrect. The real part of a product is not the product of the real parts - compare e.g. $\mathrm{Re}(i^2) \neq \mathrm{Re}(i)\mathrm{Re}(i)$. $\endgroup$ – Emilio Pisanty Feb 11 '18 at 0:09
  • $\begingroup$ @EmilioPisanty yes that was a mistake! thanks for reminding me of that. I can edit the solution to reflect that $\endgroup$ – bbachu Feb 11 '18 at 18:38
  • $\begingroup$ Note also that you require $\psi_3$ to be real for your final step to work. Otherwise you would have a complex final result. $\endgroup$ – Emilio Pisanty Feb 11 '18 at 22:46

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