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The Schrödinger equation in Physics is solved by different wave functions. The solution depends on the environment and the starting values. In one dimension and without regarding the time, the equation looks like $$-\frac{\hbar}{2m}*\partial_x^2 \Psi(x)+V(x)*\Psi(x)=E*\Psi(x).$$ $V(x)$ is an initial energy. In this simple case, we start with $V(x)=0$ on an interval $(0,L)$. Outside, it shall go against $\infty$ to both sides along $x$. Furthermore, the solution $\Psi$ shall have the value $$\Psi(0)=0\quad\land\quad \Psi(L)=0.$$

Then, the solution may follow two solution terms: $$\Psi=A*\sin(kx)+B*\cos(kx)\quad\lor\quad \Psi=A*e^{ikx}+B*e^{-ikx}$$ and both terms in the Schrödinger equation lead to the solution of the equation. $A,B$ are going to differ in both cases but solutions are going to be very similar: $$\Psi=\sqrt{\frac{2}{L}}*\sin\left ( \frac{\pi*x}{L} \right ) \qquad\lor \qquad\Psi=2*\sqrt{\frac{1}{2L}}*\sin\left ( \frac{\pi*x}{L} \right ) .$$

To find the coefficient in the solution, one must solve the condition that $$\int_{-\infty}^{\infty} |\Psi |^2 \,\mathrm{d}x = 1.$$

That does not make a difference since both solution terms only differ in the coefficient $i$ (or not $i$). As you can see, it does not help that we have the values $0$ as starting values on certain points, because $$0=0*i.$$

How to find a unique solution for this problem? What is needed? Where is the mistake?

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  • $\begingroup$ Hi, please have a look at my edit and confirm that it is correct - you had an extra "}". $\endgroup$ – Arnaud Mortier Feb 5 '18 at 10:21
  • $\begingroup$ The normalization condition should be $\int_{-\infty}^\infty |\Psi|^2 dx \color{red}{= 1}$. $\endgroup$ – achille hui Feb 5 '18 at 10:43
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It seems that you made an error solving for the constant factor. In the first case $\Psi_1(x) = A_1\sin(kx)+B_1\cos(kx)$, the boundary values for $\Psi$ tell us that $B_1=0$ and $k= \frac{n\pi}{L}$ for $n\in \mathbb{Z}$. So our solution using this form of the solution is: $$\Psi_1(x) = A_1\sin\left(\frac{n\pi x}{L}\right)$$ Using the other form of the solution, $\Psi(0) = 0$ tells us that $$\Psi_2(x) = A_2(e^{ikx}-e^{-ikx})=2A_2i\sin(kx)$$ The other boundary term gives, again, $k=\frac{n\pi}{L}$ for $n\in \mathbb{Z}$. In both situations, to obtain the value of $A$, you must normalize the wavefunction. Since you seem to be interested in the case $n=1$, I will set $n=1$. Notice the value of the following integral: $$\int_{[0,L]} \sin^2\left(\frac{\pi x}{L}\right)\;dx = \int_{[0,L]}\frac{1-\cos(2\pi x/L)}{2}\;dx = \frac{L}{2}$$ So for the first form of your wavefunction, we have: $$|A_1|^2\frac{L}{2}=1\Longrightarrow |A_1| = \sqrt{\frac{2}{L}}$$ For the second form, we have: $$4|A_2|^2\frac{L}{2} = 1\Longrightarrow |A_2| = \sqrt{\frac{1}{2L}}$$ You typically want the constant in front of $\sin(\pi x/L)$ to be real, so in the second case, let $A_2= \frac{1}{i}\sqrt{\frac{1}{2L}}$. Then we have in the first case: $$\Psi_1(x) = A_1\sin\left(\frac{\pi x}{L}\right) = \sqrt{\frac{2}{L}}\sin\left(\frac{\pi x}{L}\right)$$ In the second case, we have: $$\Psi_2(x) = 2A_2i\sin\left(\frac{\pi x}{L}\right)=2\frac{1}{i}\sqrt{\frac{1}{2L}}i\sin\left(\frac{\pi x}{L}\right)=\sqrt{\frac{2}{L}}\sin\left(\frac{\pi x}{L}\right)$$ The solutions are identical.

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  • $\begingroup$ Hm, your solution is $|A_2|=\sqrt{\frac{1}{2L}}$ and you decide that you want to at the factor $\frac{1}{i}$ because it is better to have a real coefficient, right? Because that does not affect the result in $|\cdot |$. But would the same term with a factor $1$ instead of $\frac{1}{i}$ be a solution as well? And that exactly is my question. There are multiple solutions for $$|A_2|\;\Rightarrow\;A_2=\,?$$ and I do not see how to no which to choose by the starting values given. $\endgroup$ – Kutsubato Feb 5 '18 at 12:00
  • $\begingroup$ Yes, it is also a solution to the Schrödinger equation. The phase of a wavefunction is meaningless, so you can freely assign any value to $A_2$ such that $|A_2| = \sqrt{\frac{1}{2L}}$. It does not matter if $A_2$ is real or complex; you will get the same results for $\Bbb E[x]$, $\Bbb E[E]$, etc. So the solutions are all the same in that sense. The only time that phase is important is in superpositions of wavefunctions. In this case, relative phase between wavefunctions is important. But this can be taken care of using the complex superposition coefficients anyway. $\endgroup$ – bames Feb 5 '18 at 21:07
  • $\begingroup$ What does that mean? There is no unique solution? And I can freely choose one? $\endgroup$ – Kutsubato Feb 6 '18 at 8:33
  • $\begingroup$ Yes, you can choose any constant $A_2$ with magnitude $\sqrt{\frac{1}{2L}}$. All physical quantities calculated from $\Psi_2$ will be the same as long as $|A_2| = \sqrt{\frac{1}{2L}}$. You just need to be careful when you start making superpositions of wavefunctions including $\Psi_2$, since relative phase of wavefunctions in a superposition is important. $\endgroup$ – bames Feb 6 '18 at 8:44
  • $\begingroup$ How can I understand that physically? Why is it not important what to choose? Because I am only regarding the real part for real physical effects? $\endgroup$ – Kutsubato Feb 6 '18 at 9:31

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