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In quantum mechanics, Hermitian operators play a very important role because they possess real eigenvalues.

Considering $-\frac{d^2}{dx^2}$, it is a Hermitian operator (Actually it's the simplest Hamiltonian) and its eigenfunction can be expressed by $e^{ikx}$.

However, what if $k$ is imaginary? if $k$ is imaginary, $-\frac{d^2}{dx^2}e^{ikx}=k^2e^{ikx}$ still holds, with its eigenvalue $k^2$ imaginary!

Can anyone tell me what's wrong here?

EDIT: As one comment states, the problem may lie in the restriction of $L^2$. Unfortunately, Most physics textbooks just ignore the restriction of $L^2$. Can anyone give me a rigorous proof that plain wave $e^{ikx}$(with $k$ real) is somehow appropriate in most situations?

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    $\begingroup$ Related: physics.stackexchange.com/q/81041/2451 and links therein. $\endgroup$ – Qmechanic Dec 12 '14 at 9:15
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    $\begingroup$ No, $e^{ikx}$ are no eigenfunctions, they don't lie in $L^2$. Since $d^2/dx^2$ is symmetric/Hermitian, the spectrum has to be real. $\endgroup$ – Martin Dec 12 '14 at 9:43
  • $\begingroup$ Yeah, I know the problem may lie in the restriction of $L^2$. Unfortunately, Most physics textbooks just ignore the restriction of $L^2$. Can anyone give me a rigorous proof that plain wave $e^{ikx}$ is somehow enough in most situations? $\endgroup$ – Chong Wang Dec 12 '14 at 10:32
  • $\begingroup$ @Martin: Careful with the terminology: Hermitian operators, for which the domains of the operator and their adjoint do not coincide, have real eigenvalues, but can have imaginary elements in their spectra. Only for self-adjoint operators the spectrum is also real, and only for them, we have a basis of real eigenvectors of the whole space. $\endgroup$ – ACuriousMind Dec 12 '14 at 13:05
  • $\begingroup$ @ACuriousMind: Right. I know that, but shouldn't have been lenient since this question arises from a too lenient use of the mathematics in the first place... $\endgroup$ – Martin Dec 12 '14 at 13:39
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Your "imaginary eigenvalues" don't work, because the eigenfunctions are no eigenfunctions. They do not lie in $L^2$, as you seem to be aware of.

So, let's deal with the Laplacian itself: $-\Delta=-\frac{d^2}{dx^2}$. What I want to do is, I want to calculate the Fourier transform of this operator, because the Fourier transform diagonalizes $-\Delta$, as we will see. From the diagonal form, we can read off the spectrum and thus conclude that the spectrum consists of $\mathbb{R}^+$. Of course, I don't answer the question "why $e^{ikx}$ are enough for real $k$", simply because it's a meaningless question in the Hilbert space setting.

So, let's do it. Take a function $\psi\in L^2(\mathbb{R})$ and calculate:

$$ \mathcal{F}(-\Delta \psi)(k)=\int e^{ikx}\left(-\frac{d^2}{dx^2}\right)\psi(x)\,dx = \int k^2e^{ikx}\psi(x)\,dx = k^2 \mathcal{F}(\psi)(k) $$

where we integrated by parts two times (using that $\psi$ necessarily vanishes at infinity) and then differentiated $e^{ikx}$. This formula means that $\mathcal{F}$ diagonalizes the Laplacian, because we have just seen that the Fourier transform of the Laplacian $\mathcal{F}(-\Delta)\mathcal{F}^*$ is a multiplication operator. The idea is now that from a multiplication operator, we can read off the spectrum: It's the essential range of the multiplication operator. Just from the definition, we can see that this will be $[0,\infty)$ in our case, hence $\sigma(-\Delta)=[0,\infty)$.

At no point did we even talk about $e^{ikx}$, so how to solve the $e^{ikx}$ with imaginary $k$ business? You don't have to - none of these functions is in $L^2$, but: if you now put in $\psi(x)=e^{ikx}$ with real $k$ (because $k$ is real in the Fourier transform!), then it looks like it is an eigenfunction of $-\Delta$. In other words: If you imagine a space with your basis being the functions $e^{ikx}$, then the Fourier transform of the Laplacian is just the infinite matrix with eigenvalues $k^2$ on the diagonal. However, this is not the rigorous picture!

Edit: A very good explanation with more maths can be found here: https://math.stackexchange.com/questions/766479/what-is-spectrum-for-laplacian-in-mathbbrn Note that this is not easy maths, but in order to understand the finer points of this business, you can't avoid it.

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  • $\begingroup$ That's a nice explanation, and a good complement to my attempt, I feel. $\endgroup$ – xebtl Dec 12 '14 at 11:32
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First off, if $k =: i\kappa$ is imaginary, the eigenvalue (“energy”) is $-\kappa^2$, i.e. real but negative: $$-\frac{d^2}{dx^2} e^{ikx} = -\frac{d^2}{dx^2} e^{-\kappa x} = -\kappa^2 e^{-\kappa x}.$$

Physically, that is an evanescent wave in one direction, but grows without bound in the other, so if your space is all of $x\in\mathbb R$, it is not a valid wavefunction for very tangible physical reasons: it is not normalizable. The underlying reason is that its energy is smaller than the minimum of the potential it lives in. Any introductory QM lecture or book should cover this, and also feature this type of wave function (probably in the context of finite potential barriers).

For the more formal, mathematical side of things this document has to be recommended: http://arxiv.org/abs/quant-ph/9907069.

As for the plane waves, they are not really valid states either (physically, a particle will never have a perfectly well-defined momentum), but they are convenient to work with and often adequate (prototypical situation: a particle beam incident on a target being described by a plane wave). When a single plane wave is not good enough (to be more realistic, a particle might be described as a wave packet), you can still do a decomposition in terms of plane waves , i.e. a Fourier transform. Your actual (well-behaved, mathematically valid) state can be described as a sum over plane waves.

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