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Selection of an interval affects the energy eigenstates - impossible

  1. Aug 3, 2013 #1
    I haven't found a comparison like this in any book that I have been reading so let me explain.

    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?
     
  2. jcsd
  3. Aug 3, 2013 #2

    Bill_K

    User Avatar
    Science Advisor

    Doing it this way (assuming B ≠ 0) you've gotten half the solutions. To get the other half, subtract one equation from the other and assume A ≠ 0.
     
  4. Aug 3, 2013 #3
    Thank you. If i substract these two equations i get:

    \begin{align}
    0&=-2A\sin\left(\sqrt{\tfrac{2mE}{\hbar^2}}\tfrac{d}{2}\right)
    \end{align}

    I can solve this with ##A=0## or ##d=0## but the later means that width of the well is 0 and can't be true which means ##A=0## is an only possible solution. Is this assumption ok? If i continue in this maner i get better values for the wavefunctions which must be

    \begin{align}
    \psi = B\cos\left(\sqrt{\tfrac{2mE}{\hbar^2}} x\right)
    \end{align}

    If i try to sum up all this in my head now. By summing the equations i got a ##B=0## solution, which now must not be used because if it is, both ##A=0## and ##B=0## which means ##\psi=0## and the particle wasn't supposed to be in the well. But it is! So i am left with only 3 possible equations which are ##A=0## the ##\psi = B\cos\left(\sqrt{\frac{2mE}{\hbar^2}} x\right)## and the one for energies - i mean the second boxed equation.

    What now? Am i ready to normalize the wavefunction and Find ##B##?

    PS: I can't forget the energy equation which is still not the same as for interval ##0<x<d##.
     
  5. Aug 3, 2013 #4

    Bill_K

    User Avatar
    Science Advisor

    The solutions are √(2mE/ħ2) d/2 = Nπ.
     
  6. Aug 3, 2013 #5
    Oh right :) my solution was a bit trivial.

    So if i use this i get another equation for energies which looks like:

    $$\boxed{W=\frac{2N^2 \pi^2 \hbar^2}{md^2}}$$

    I have now 2 equations for the energy levels on an interval ##-d/2<x<d/2## will the sum of those two return the same energy equation as i got for interval ##0<x<d##?
     
  7. Aug 3, 2013 #6

    jtbell

    User Avatar

    Staff: Mentor

    [By the way, it would make things easier if each equation were in a separate block of LaTeX, enclosed in its own pair of tags. Then people could easily quote one equation at a time to make a point. I can't figure out how to do that, with your huge blocks of LaTeX.]

    Here's another approach. Consider the general solution to the DE, with the coefficients A and B. Apply one of the boundary conditions, say psi(+d/2) = 0. The resulting equation (which you have a little further down the page) can be true if either

    Case 1: A = 0 and cos(...) = 0. This gives you solutions of the form psi = B cos(kx), with the values of k (and E) determined by the condition cos(...) = 0.

    or

    Case 2: B = 0 and sin(...) = 0. This gives you solutions of the form psi = A cos (kx), with the values of k (and E) determined by the condition sin(...) = 0.

    If you now start over again with the the general solution, and apply the other boundary condition (psi(-d/2) = 0), you get the same final result. The two boundary conditions are redundant because of the symmetry of the potential function.

    Taken together, the allowed values of E for cases 1 and 2 are the same ones as for the other method (interval from 0 to d). The psi functions of course have somewhat different forms because of the shifted origin.
     
  8. Aug 3, 2013 #7

    jtbell

    User Avatar

    Staff: Mentor

    There are also the following cases which don't enter into the final solution:

    Case 3: A = 0 and B = 0, which is not useful, for obvious reasons!

    Case 4: cos(...) = 0 and sin(...) = 0, which is impossible.
     
  9. Aug 3, 2013 #8
    I don't think you understood what i wanted to ask. I derived these two equations for an interval ##-d/2>x>d/2##:

    $$W=\frac{2N^2 \pi^2 \hbar^2}{md^2}$$
    $$W=\frac{(2N-1)^2 \pi^2 \hbar^2}{2md^2}$$

    and this one using interval ##0>x>d##:

    $$W=\frac{N^2 \pi^2 \hbar^2}{2md^2}$$

    Energy levels should be equal but i still can't see the equality. What should i do with the first two to see that they are in fact the same as the later one?
     
  10. Aug 3, 2013 #9

    Bill_K

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    Science Advisor

    Write the first line as W = (2N)2π2ħ2/2md2. Then in the third line, N is either odd or even. If it's odd, that's the second line. If it's even, that's the first line.
     
  11. Aug 3, 2013 #10

    jtbell

    User Avatar

    Staff: Mentor

    Re-write the first equation as Bill indicated. Then write out a list of the values of 2N2, and of (2N-1)2, for N = 1, 2, 3...

    See how they "fit together" with each other, and with the third equation?
     
  12. Aug 3, 2013 #11
    Thanks to both of you. I understand now.
     
  13. Aug 3, 2013 #12
    One more question. If i now want to derive wavefunctions for an interval ##-d/2 >x>d/2## i am left with two options. Either i take ##\psi = A\sin \left(\sqrt{\tfrac{2mW}{\hbar^2}} x\right)##, insert ##W=\frac{(2N-1)^2\pi^2\hbar^2}{2md^2}## and then normalise it OR i take ##\psi = B\cos \left(\sqrt{\tfrac{2mW}{\hbar^2}} x\right)##, and insert ##W=\frac{(2N)^2\pi^2\hbar^2}{2md^2}## and then normalise it.

    I believe that in both cases i should get the same wavefunction? Please correct me if i am wrong.
     
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