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The time-independent Schrödinger equation in one dimension for a free particle,

$$\frac{-\hbar^{2}}{2 m} \frac{\partial^{2} \Psi(x)}{\partial x^{2}}=\varepsilon\Psi(x)$$

can be solved as a homogeneous second-order linear ODE with constant coefficients, so the complex solutions are

$$\Psi(x)=Ae^{ikx}+Be^{-ikx} \tag{1}$$

with $A$ and $B$ two constants and $k=2m\varepsilon/\hbar$. However, when referring to the plane-wave solution, many sources omit the term with the negative exponential (for example, here), regardless of the boundary conditions,

$$\Psi(x)=Ae^{ikx}\tag{2}$$

While $(1)$ is a combination of a right- and a left- traveling waves, $(2)$ includes only a right- or a left- travelling wave, depending on the sign of $k$. So, why is the negative exponential omitted? Wouldn't this imply some loss of generality?

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  • $\begingroup$ Yes. Picking only positive exponent is lossing some generality. The text you linked is giving an example of travleling wave. If you wane to show a standing wave, then the negative exponent has to participate. It is nothing but a preferential choice of the author, $\endgroup$
    – ytlu
    Commented Aug 17, 2022 at 6:57

2 Answers 2

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I hope by telling you this, things will become clearer. If not, you can always comment.

So, to start, first I shall say that both $$\Psi_k(x)=A_ke^{ikx}+B_ke^{-ikx} \ \text{and}\ \Psi_k=A_ke^{ikx}$$ do not comprise the most general solution to the time-independent Schroedinger equation. To obtain the most general solution, one must sum over all possible (positive) values of $k$ in Eq. (1), or one must sum over all possible (positive and negative) values of $k$ in Eq. (2) (and the spectrum of $k$ will depend on the boundary conditions).

So, what exactly do you mean when you are referring to loss of generality? There is no generality in (1) or in (2), since both describe stationary solutions. The latter is, as you correctly noted, a right(left) travelling wave, whereas the former is simply the linear combination of a right-travelling wave and of a left-travelling one. No one claims that either Eq. (1) or Eq. (2) describe general solutions to the time-independent Schroedinger equation.

To obtain, though, the aforementioned solution, one needs to

  1. Take Eq. (1) and sum with respect to positive values of $k$: $$\Psi(x)=\sum_{n=0}^{\infty}\Big(A_ne^{2in\pi x/L}+B_ne^{-2in\pi x/L}\Big)$$
  2. Take Eq. (2) and sum with respect to negative values of $k$: $$\Psi(x)=\sum_{n=-\infty}^{\infty}A_ne^{2in\pi x/L}$$

where I have used the example of periodic boundary conditions, such that we have a concrete case of spectrum in our hands. So, to sum up, your equations (1) and (2) describe solutions to the time-independent Schroedinger equation, yes! The one is associated with a plane wave travelling to one direction, whereas the other is a superposition of two single-direction travelling ones, yes! But none of them is the most general solution and hence no loss of generality occurs when referring to one instead of the other.

I hope I understood well what the confusion is. If not, you are free to comment.

P.S.: The problem at hand was chosen in order to demonstrate the equivalence of summing with respect to all the possible eigenvalues of the wave-vector in the first case, with the case in which we sum with respect to both negative and positive eigenvalues of the wave-vector!

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    $\begingroup$ In your expressions with sums the $A$ and $B$ should instead be $A_n$ and $B_n$, since components of different frequencies in general have different amplitudes. $\endgroup$
    – Ruslan
    Commented Aug 16, 2022 at 22:27
  • $\begingroup$ Yes, that is correct @Ruslan, thanks for noticing. I will edit in a few moments $\endgroup$
    – schris38
    Commented Aug 16, 2022 at 23:00
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As Schris38 pointed out $\Psi_k$ is not the most general solution either way.

But even so, \begin{equation} A e^{ikx}+B e^{-ikx}=(A+iB) \cos(kx)+(A-iB)\sin(kx)\equiv C_1\cos(kx)+C_2\sin(kx) \tag{1} \end{equation}

and $$A e^{ikx}= \alpha \cos(kx)+\beta\sin(kx) \tag{2}$$ where $\alpha=Re A$, $\beta=Im A$. So it's just a matter of correctly choosing the coefficients. Both things work.

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    $\begingroup$ But $A e^{ikx}+B e^{-ikx}$ is $(A+B)\cos(kx)+i(A-B)\sin(kx)$, not $(A+iB) \cos(kx)+(A-iB)\sin(kx)$. $\endgroup$
    – Invenietis
    Commented Sep 1, 2022 at 11:35

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