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In time independent perturbation theory we can calculate the first and second energy corrections resulted by a potential V in the Hamiltonian $ H=H_o + λV , $ , λ<<, by the expressions:

$$ε_1 = \langle φ_n| V | φ_n \rangle ,$$

$$ε_2 = \sum{ {| \langle φ_m | φ_n \rangle |^2 } \over {E_m - E_n} } ,$$ where ε are the corrections, E the unpertubed energy states and the sum takes all the possible values besides n.

The current problem for which I would like to ask is this:

For an infinite potential well we have the following wavefunctions given different boundary conditions. In case 1 we take the limits of the well to be $(0,L)$, where in case 2 we take the limits to be $(-L/2 , L/2)$. Then:

Case 1:

$ψ(x) = \sqrt {{2} \over {L} } sin (nπx/L) $ and

Case 2:

$ψ_{even}(x) = \sqrt {{{2} \over {L}} }cos(nπx/L) , $ n=1,3,5... $ ψ_{odd} (x) = \sqrt {{{2} \over {L}} }sin(nπx/L), $ n= 2,4,6...

We add a perturbation of the form $V=-qAx $, and we seek the corrections for the ground state n=1.

We see that the first energy correction for the first case is $ ε_1 = \langle φ_n| V | φ_n \rangle ={{2} \over {L}} \int{_0}^L {sin^{2}(πx/L) x dx} = { {-qAL} \over {2}} $ and thus non-zero, but for the second case is $ε_1 = {{2} \over {L}} \int_{-L/2}^{L/2} {x cos^{2}(πx/L) } = 0 ,$ it' s zero from symmetry considerations.

But the length of the area without potential, the legth of the well is the same in both cases. Can the energy corrections in the two cases be indeed different or have I made some mistake I don't notice. If they are different how can this be? Does the system react differently under the perturbation and if so why? How, in the end, can the two cases not produce the same result?

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    $\begingroup$ The difference between the two systems is a constant addition to $V$. There will be no observable consequences, because this is just the same as adding a constant number to $H$. All of the energy levels will get bumped up by the same amount, so their differences will be the same. $\endgroup$ – knzhou Feb 24 '16 at 0:58
  • $\begingroup$ @knzhou Thanks. Do you mean that despite the difference between the corrections of the two cases, in the end, the sum of the unperturbed energies with the corrections will be equal amongst the two considerations here? $\endgroup$ – Constantine Black Feb 24 '16 at 8:45
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    $\begingroup$ Note that \sin and \cos will give you $\sin$ and $\cos$, rather than $sin$ and $cos$. $\endgroup$ – Kyle Kanos Feb 25 '16 at 0:57
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Consider the average energy shift of the bottom of the well for your two cases. In case 1, the well bottom is shifted on average net downwards since it exists at positive z where your defined potential is negative. In case 2, the well bottom has zero average shift since one side sits at more positive potential and the other at more negative. So, it's just that you've defined the problem with different locations of $V=0$ with respect to the well's location. If for case one you choose $V=-qA(x-L/2)$, this makes the two equivalent.

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  • $\begingroup$ Thanks. So the main problem here is the fact that by changing the system' s boundaries I didn't changed the function of the potential. Is my understanding of your answer correct? Even so, the particle in the well moves in a space of same length. The problem is just how we define the potential? $\endgroup$ – Constantine Black Feb 24 '16 at 8:42
  • $\begingroup$ Yes, all that's different is the definition of where V=0, which can be chosen as whatever you like. This is fine because nothing is physically different about the two situations, since absolute energy is not a real observable. Instead, consider any energy differences, for example energy w/respect to the well bottom or energy difference between two states - these differences will be equivalent for both of your cases. $\endgroup$ – dbq Feb 24 '16 at 17:58

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