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I don't really know much about Quantum mechanics, but would like to know one simple fact.

The state function $\Psi(r, t)|$ whose magnitude gives the probability density of the position of the particle and the magnitude of its ($\Psi(r, t)$) fourier transform gives probability density of its momentum. Is there any rule that these state functions are smooth (possess infinite order derivatives everywhere) (derivatives of all orders exist)?

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I'm not sure your statement about the Fourier transform is quite correct. Foruier-transforming the wavefunction in terms of position will indeed give the momentum wavefunction, but whether this can be done on the probability distribution ($|\psi|^2$), I do not know. Hopefully someone more mathematically adept can enlighten me. –  Noldorin Nov 18 '10 at 15:35
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@Noldorin: I meant it on the wave function itself, not on the magnitude/probability distribution. Thanks for the clarification in the question. –  Rajesh D Nov 18 '10 at 15:36
    
Ok, sure. That makes more sense now. :) (And in your question, I'm also presuming you define $S(r, t) = |\Psi(r, t)|^2|$.) –  Noldorin Nov 18 '10 at 15:37
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Can you change title to something meaningful like "Is it guaranteed that wavefunction is well behaved everywhere?"? –  Pratik Deoghare Nov 18 '10 at 16:11
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related: Is the world $C^\infty$? –  Tobias Kienzler Mar 1 '11 at 9:09
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The only general requirement on the state function for a single, spinless, quantum particle (quanton) in a physically realistic state is that the state function be square integrable, i.e., the integral of its absolute value squared over all space be finite. Non-square integrable state functions are used for many purposes, but they are all idealizations that do not, individually, represent realistic states. If the state function is also to belong to the domain of definition of the Hamiltonian, then, in non-relativistic QM, the state function must be spatially differentiable to second order as well. State functions which are square integrable but not second order differentiable do not satisfy the Schroedinger equation. But their time evolution is still determined by continuity considerations since the second order differentiable state functions are everywhere dense in the state space, i.e., Hilbert space.

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Is there any derivative operator in the QM ? –  Rajesh D Nov 18 '10 at 16:19
    
Momentum is represented by the derivative operator, up to a factor. –  Raskolnikov Nov 18 '10 at 17:20
    
derivative of what ? –  Rajesh D Nov 18 '10 at 21:19
    
This is a slightly convoluted answer. I'm actually not sure what point you're trying to get across, I'm afraid. –  Noldorin Nov 19 '10 at 19:40
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Some of the conditions for wavefunctions $\Psi(x)$, for all elements $x$ of a subset of $\mathbb{R^{d}}$ (in the hyperphysics link, they use $x \in \mathbb{R}$).

1.- Must be a solution of the Schrodinger equation.

2.- Must be normalizable.

3.- Must be a continuous function of $x$.

4.- The slope of the function in $x$ must be continuous, that is, $\displaystyle \frac{\partial \Psi(x)}{\partial x}$ must be continuous.

The property of being square-integrable is included in the condition 2.

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@Robert Smith: "some of the conditions", do you there are more ? –  Rajesh D Nov 18 '10 at 16:25
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The solution to the Dirac delta potential is not continuously differentiable, so it violates condition 4. –  Keenan Pepper Nov 18 '10 at 16:40
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The assumption for the Delta potential is separate the space for $x<0$ and $x>0$. Therefore, the solution is one wavefunction for $x<0$ and other for $x>0$. Is that what you're saying? I don't see how that violates the condition 4. –  Robert Smith Nov 18 '10 at 16:55
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@Robert: your point 3. says precisely that it has to be continuous at every point $x$. What you forgot to include (in this formulation) is that a particle in QM lives in Hilbert space $H = L^2(\mathbb{R}^d)$ so that indeed it needs to be defined (and continuous) for every $x \in \mathbb{R}^d$. The problem with Delta potential arises only because it's not quite physical to assume infinite jump in potential. You do this to make things simple, e.g. to disallow movement through walls. But in reality walls are made of atoms so the potential is smooth (just very fast growing). –  Marek Nov 18 '10 at 18:13
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For sure Marek, but I don't see why the Schrödinger equation is not considered a mathematical idealization then? After all, it's only a non-relativistic approximation. And if we're gonna start like this, everything that has ever been conceived of in physics is an idealization. Your decision to consider one more physically relevant than the other is arbitrary if you don't specify the bounds within which the approximation is valid or not. So, without doubt, the Schrödinger equation can do more than models for Anderson localization which are not unphysical, only less broadly applicable. –  Raskolnikov Nov 19 '10 at 15:57
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