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If we have a free particle of energy E incident on a potential $V$ $$V(x) = \begin{cases}0 & x \leq 0 \\ V_0 & 0 < x < L \\ 0 & x \geq L\end{cases}$$ We find that the wave function $\phi$ $$\phi(x) = \begin{cases} e^{ikx}+re^{-ikx} & x < 0 \\ A + Bx & 0 < x < L \\ te^{ikx} & x > L \end{cases}$$ Is a Hamiltonian eigenfunction with energy $E = V_0$. I can show for $x < 0$ and $x > L$, that $\phi$ satisfies $\hat{H}\phi = V_0\phi$ but for $0 < x < L$ we have $\hat{H}\phi = 0$.

I am trying to understand this as a result of $E - V_0 = 0$, thus in this region $\phi$ has an eigenvalue of $0$? So in this region the particle has zero energy? But I cannot find a rigorous explanation of this online anywhere and was hoping someone here could clear it up for me.

Additionally: Wikipedia states that a complete solution can be found by using constraints on $A$ & $B$ that can be found by matching the wave function and its derivatives at $0$ & $L$. Could somebody please explain why the wave function and its derivative must be continuous everywhere.

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  • $\begingroup$ Under which hypothesis you find that wavefunction in the region $0<x<L$? $\endgroup$
    – GiorgioP
    Mar 13 at 15:37
  • $\begingroup$ This Wikipedia article If you look at the section titled $E = V_0$ $\endgroup$ Mar 13 at 15:41
  • $\begingroup$ Ok, I see. I did not notice the condition in the title. $\endgroup$
    – GiorgioP
    Mar 13 at 15:43
  • $\begingroup$ I think you should improve the formulation of the question by starting with "If we have a particle of energy $E$ incident ..." and after formula for $\phi$, "is a solution with energy ...". $\endgroup$
    – GiorgioP
    Mar 13 at 16:39
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Let me start removing a possible misunderstanding. Scattering solutions in terms of plane waves asymptotic states (as the solutions for $x<0$ and $x>L$ ) imply that the global wavefunction is not normalizable. Therefore, it is not a true eigenfunction, and the associated value for $E$ is not the eigenvalue. Of course, once one has solutions for different values of $E$, it is always possible to find linear combinations corresponding to normalized wavepackets representing physical states.

The kind of problem solved by the cited solution is the following: "assuming that there is an incoming state corresponding to a plane wave of energy $E$ and wavevector $k$ such that $E=\frac{\hbar^2 k^2}{2 m}$ traveling towards the barrier, what is the solution corresponding to stationary fluxes?". Put this way, it is clear that $E$ is the energy, and it is a constant independent of the position.

Now let me touch on your final question. There is widespread confusion about the reasons why solutions of the Schrödinger equation should be continuous. Even the links proposed by @AFG are not satisfactory on this point, and some of the answers are definitely wrong.

Any acceptable wavefunction has the only constraint of being square-integrable. Even though the plane waves are not, the above-mentioned idea that they can be used to build a square-integrable function resolves this problem. The constraint of continuity comes into play once we require that the wavefunction is also in the domain of the Hamiltonian operator. To be more precise, since most of the Hamiltonians are unbounded operators, they belong to the domain, making the Hamiltonian a self-adjoint (or essentially self-adjoint ) operator. Such a requirement is a key ingredient to ensure the Hamiltonian's correct properties and the unitary evolution operator.

It is a simple exercise to check that in the case of a piece-wise continuous potential, the continuity of the wavefunction (and its first derivative) is a condition for a self-adjoint (or essentially self-adjoint ) Hamiltonian. More details can be found in this Wikipedia page or good Quantum Mechanics textbooks. I remember a related discussion about the boundary conditions for the 3D Coulomb potential in Messiah's textbook.

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The eigenvalue of the Hamiltonian is $E=V_0$ in the three regions

$$\hat{H}\phi(x)=\Big(\hat{T\,}+\hat{\,V}(x)\Big)\phi(x)=E\phi(x)=V_0\phi(x).$$

For $0<x<L$, what you have is

$$\Big(\hat{T\,}+V_0\Big)\phi(x)=V_0\phi(x)\,\,\rightarrow\,\,\hat{T\,}\phi(x)=0,$$

so $0$ is the eigenvalue of the kinetic energy.

With regard to your last question, more info here and here.

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  • $\begingroup$ Being an eigenvalue is a global property,. There is nothing like a piece-wise eigenvalue. $\endgroup$
    – GiorgioP
    Mar 13 at 16:23
  • $\begingroup$ Yes, I just wanted to make clear that it is not different in the regions as OP suggests. $\endgroup$
    – AFG
    Mar 13 at 16:30
  • $\begingroup$ Yes, but $\hat T \phi \neq 0$ out of the $0<x<L$ interval. Therefore it is not possible to speak about an eigenvalue of the kinetic energy $\endgroup$
    – GiorgioP
    Mar 13 at 16:34
  • $\begingroup$ Oh you are right! I'll delete my answer then, I think yours is perfect. @redpanda2236, please accept GiorgioP's answer. $\endgroup$
    – AFG
    Mar 13 at 16:38
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I don't understand how you arrive at:

$\phi(x) = \begin{cases} e^{ikx}+re^{-ikx} & x < 0 \\ A + Bx & 0 < x < L \\ te^{ikx} & x > L \end{cases}$

The TISE for your system is: $$\Big[-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}+V_0\Big]\psi=E\psi$$ $$-\frac{\hbar^2}{2m}\psi''+V_0\psi=E\psi$$ $$\psi''=-\frac{2m}{\hbar^2}(E-V_0)\psi$$ Make: $$k^2=\frac{2m}{\hbar^2}(E-V_0)$$ $$\psi''+k^2\psi=0$$ $$\psi(x)=c_1\sin kx+c_2\cos kx$$ Boundary conditions: $$\psi(0)=0\text{ and }\psi(L)=0$$ $$\Rightarrow c_2=0\text{ and }\sin kL=0$$ $$\sin kL=0 \Rightarrow kL=2\pi n$$ with $n=1,2,3,...$ $$k=\frac{2\pi n}{L}$$ $$\frac{2m}{\hbar^2}(E-V_0)=\frac{4\pi^2 n^2}{L^2}$$ From which $E$ can then be extricated.

So the wavefunctions for $0<x<L$ are given by:

$$\psi_n(x)=c_n\sin\Big( \frac{2\pi n}{L}\Big)$$

where the coefficients are to be determined from normalisation (not shown here)

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  • $\begingroup$ It was also my misunderstanding. It is in the title. The question ids for an energy exactly equal to the height of the barrier. In that case $k=0$ and the solution in terms of $sin$ and $cos$ cannot directly used. $\endgroup$
    – GiorgioP
    Mar 13 at 16:32
  • $\begingroup$ @GiorgioP Poorly phrased question, IMHO. $\endgroup$
    – Gert
    Mar 13 at 16:34

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