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Is there anything in the physics that enforces the wave function to be $C^2$? Are weak solutions to the Schroedinger equation physical? I am reading the beginning chapters of Griffiths and he doesn't mention anything.

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Related: physics.stackexchange.com/q/1067/2451 –  Qmechanic Jan 17 '12 at 23:57
    
Thanks, but I don't think a good answer was given there. –  StuartHa Jan 18 '12 at 0:05

3 Answers 3

Some of this was discussed elsewhere. See « significance of unbounded operators » http://physics.stackexchange.com/a/19569/6432 .

It is not true the wave function has to be continuous, it just has to be measurable (i.e., a limit of step functions almost everywhere). Naturally you might wonder what sense Schroedinger's equation makes if you apply it to a step function...but the answer is easier than worrying about distributional weak solutions. The point is that you can solve the time-dependent Schroedinger equation with the exponential $$e^{itH},$$ which is a family of unitary operators, and which is better behaved than the $H$ you have to use in Schroedinger's equation. The $H$ you have to use, for example $$-{\partial ^2\over\partial x^2} + \mathrm{other\ stuff}, $$ is unbounded. And non-differentiable functions are not in its domain. But plugging it in to the power series for exponential converges in norm anyway, and so the resulting operator, being bounded and even unitary on a dense domain of the Hilbert space, can be extended painlessly to the entire space, even step functions. So it makes more sense to say that the solution to Schroedinger's equation with a given initial condition $\psi_o$ is $$\psi_t (x) = e^{itH}\cdot \psi_o (x)$$ and there is no need to bring in distributional weak solutions. These considerations are called the Stone--von Neumann theorem.

But such functions are not very important and indeed it is possible to do all of Quantum Mechanics with smooth functions, especially if you take the attitude that, for example, a square well potential would also be unphysical and is really just a simplified approximation of a physical potential which smoothed off those square corners but had a formula that was unmanageable.... See Anthony Sudbery, Quantum Mechanics and the Particles of Nature, which since it is written by a mathematician, is careful about unimportant issues like this.

That family of operators I wrote down is called the time-evolution operators, and they are an example of a unitary group with one-parameter, time. It is easy to see that if $\psi_o$, the initial condition, the state of the quantum system at time $t=0$, is nice and smooth, then all the future states will be nice and smooth too. Furthermore, all the usual quantum observables have eigenstates which are nice and smooth, so if you perform a future measurement, you will get a function which is nice and smooth and its future time evolution will remain that way, until the next measurement, etc. until Doomsday.

That said, for all practical purposes you may assume all wave functions are smooth and that the only reason you study discontinuous ones is as convenient approximations.

The comment one sometimes hears is that a wave function that was not in the domain of the Hamiltonian would « have infinite energy » but this is nonsense. In Quantum Mechanics, you are not allowed to talk about a quantum system as having a definite value of an observable unless it is in an eigenstate of that observable. What you can ask is, what would be the expectation of that observable. If the wave function $\psi$ is discontinuous and not in the domain of the Hamiltonian, it cannot be an eigenstate, but if its energy is measured, the answer will always be finite. Yet, the expectation of its energy does not exist, or you could say, the expectation « is infinite ». Not the energy, its expectation. There is nothing very unphysical about this because expectation itself is not very directly physical: you cannot measure the expectation unless you make infinitely many measurements, and your estimated answer, even for this discontinuous function, will always be a finite expectation. It's just that those estimates are way inaccurate, the expectation really is infinite (like the Cauchy distribution in statistics).

But even for such a « bad » wavefunction, all the axioms of Quantum Mechanics apply: the probability that the energy, if measured, will be 7 erg, is calculated the usual way. But these bad wave functions never arise in elementary systems or exercises so most people think they are « unphysical ». And, as I said, if the initial condition is a « good » wave function, the system will never evolve outside of that. This, I think, is connected with the fact that in QM, all systems have a finite number of degrees of freedom: this would no longer be true for quantum systems with infinitely many degrees of freedom such as are studied in Statistical Mechanics.

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Right, there's nothing wrong about step functions, delta-functions (the derivatives of the former), and others, and that's why physicists freely work with them and never mention artificial mathematical constraints. Still, some discontinuities may make the kinetic energy infinity, so they don't exist in the finite-energy spectrum. I would add that the most natural space of functions to consider is $L^2$, all square-integrable functions. They may be Fourier-transformed or converted to other (discrete...) bases. A subset also has a finite (expectation value of) energy. –  Luboš Motl Jan 18 '12 at 7:22

The time-independent Schroedinger equation for the position-space wavefunction has the form $$\left(\frac{-\hbar^2}{2m}\nabla^2 +(V-E) \right)\Psi=0$$

Where $E$ is the energy of that particular eigenstate, and $V$ in general depends on the position. All physical wavefunctions must be in some superposition of states that satisfy this equation.

At least in nonrelativistic QM, the wavefunction is not allowed to have infinte energy. If the second derivative of the wavefunction does not exist or is infinite, it implies that either $V$ has some property that "cancels out" the discontinutiy (as in the infinite square well), or that the wavefunction is continuous and differentiable everywhere.

Generally, $\Psi$ must always be continuous, and any spatial derivative of $\Psi$ must exist unless $V$ is infinite at that point.

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Here we want to show that there is an easy mathematical bootstrap argument why solutions to the time independent 1D Schrödinger equation

$$-\frac{\hbar^2}{2m} \psi^{\prime\prime}(x) + V(x) \psi(x) ~=~ E \psi(x) \qquad\qquad (1)$$

tend to be rather nice. First rewrite eq. (1) in integral form

$$ \psi(x)~=~ \frac{2m}{\hbar^2} \int^{x}\mathrm{d}y \int^{y}\mathrm{d}z\ (V(z)-E)\psi(z) .\qquad\qquad (2)$$

There are various cases.

  1. Case $V \in {\cal L}^2_{\rm loc}(\mathbb{R})$ is a locally square integrable function. Assume the wavefunction $\psi \in {\cal L}^2_{\rm loc}(\mathbb{R})$ as well. Then the product $(V-E)\psi\in {\cal L}^1_{\rm loc}(\mathbb{R})$ due to Cauchy–Schwarz inequality. Then the integral $y\mapsto \int^{y}\mathrm{d}z\ (V(z)-E)\psi(z)$ is continuous, and hence the wavefunction $\psi$ on the lhs. of eq. (2) is smooth $\psi\in C^{1}(\mathbb{R}).$

  2. Case $V \in C^{p}(\mathbb{R})$ for a non-negative integer $p\in\mathbb{N}_0$. Similar bootstrap argument shows that $\psi\in C^{p+2}(\mathbb{R}).$

The above two cases do not cover a couple of often-used mathematically idealized potentials $V(x)$, e.g.,

  1. the infinite wall $V(x)=\infty$ in some region. (The wavefunction must vanish $\psi(x)=0$ in this region.)

  2. or a Dirac delta distribution $V(x)=V_0\delta(x)$. See also here.

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