I haven't found a comparison like this in any book that I have been reading so let me explain.(adsbygoogle = window.adsbygoogle || []).push({});

I decided that I will derive an equation for an energy levels of a particle in an infinite potential well in two ways. 1st I tried to derive it using the interval ##0<x<d## where ##d## is a width of a well. What I got was an equation wich matches perfectly with any of the books out there:

\begin{align}

&\boxed{E=\frac{N^2\pi^2\hbar^2}{2md^2}}\longrightarrow N=1,2,3\dots\\

&E_1=\pi^2\hbar^2/2md^2\\

&E_2=2^2E_1\\

&E_3=3^2E_1\\

&\dots

\end{align}

So far so good. Now I decided to use the interval ##-d/2<x<d/2## and what I got was a different equation for energy levels. This is impossible because energies can't depend on our selection of an interval. I included my derivation:

\begin{align}

\frac{d^2\psi}{dx^2} &= - \left(\frac{2mE}{\hbar^2}\right)\psi \longleftarrow \substack{\text{Diferential eq. (DE) I get if I}\\\text{set $V=0$ in a Schrodinger. eq. }}\\

&\downarrow\\

\psi &= A\sin\left(\sqrt{\tfrac{2mE}{\hbar^2}}x\right)+B \cos\left(\sqrt{\tfrac{2mE}{\hbar^2}}x\right)\longleftarrow \substack{\text{A general solution}\\\text{to the above DE}}

\end{align}

Let us not forget ##\psi## is in fact ##\psi(x)##. At this point I used the border equations ##\psi(-d/2)=0## and ##\psi(d/2)=0## and what I got was a system of two equations:

\begin{align}

0 &= A\sin\left(\sqrt{\tfrac{2mE}{\hbar^2}} \left(-\tfrac{d}{2}\right)\right)+B \cos\left(\sqrt{\tfrac{2mE}{\hbar^2}}\left(-\tfrac{d}{2}\right)\right)\\

\substack{\text{What I get if I insert $x=-\frac{d}{2}$}}\longrightarrow 0 &= -A\sin\left(\sqrt{\tfrac{2mE}{\hbar^2}} \tfrac{d}{2}\right)+B \cos\left(\sqrt{\tfrac{2mE}{\hbar^2}}\tfrac{d}{2}\right)\\

\phantom{d}\\

\substack{\text{What I get if I insert $x=\frac{d}{2}$}}\longrightarrow 0 &=A\sin\left(\sqrt{\tfrac{2mE}{\hbar^2}} \tfrac{d}{2}\right)+B \cos\left(\sqrt{\tfrac{2mE}{\hbar^2}}\tfrac{d}{2}\right)\\

\end{align}

I sum these two equations and I get:

\begin{align}

0&=2B\cos\left(\sqrt{\tfrac{2mE}{\hbar^2}}\tfrac{d}{2}\right)

\end{align}

And the possible solutions of this eq. are ##B=0## (this will help me find ##\psi##) but if ##B\neq 0## than even ##d\neq 0## or else we would get ##0=2B##. So if ##d\neq 0## than it must be true that:

\begin{align}

\sqrt{ \tfrac{2mE}{\hbar^2}} \tfrac{d}{2} &= (2N-1) \tfrac{\pi}{2} \longleftarrow N=1,2,3\dots\\

\sqrt{ \tfrac{2mE}{\hbar^2}} d &= (2N-1) \pi\\

&\!\!\!\!\!\!\boxed{E=\tfrac{(2N-1)^2\pi^2\hbar^2}{2md^2}}\\

E_1&=\pi^2\hbar^2/2md^2\\

E_2&=3^2E_1\\

E_3&=5^2E_1\\

&\dots

\end{align}

Now if I check two boxed equations I see that they are different! Even ##N## is defined the same for those two. Why did I get different energy equation by simply choosing a different interval (only the ground state ##E_1## is the same in both cases while ##E_2, E_3, E_4\dots## differ)? This means that my selection of an interval changes the particle? Please explain. I couldn't spot any mistake in my derivation. Can you?

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# Selection of an interval affects the energy eigenstates - impossible

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