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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)?

Qmechanic
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Rajesh D
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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
  • @Noldorin: I mean both being same...i did'nt know how to type Psi hence i typed S. – Rajesh D Nov 18 '10 at 15:39
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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 01 '11 at 09:09

2 Answers2

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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.

Gordon N. Fleming
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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.

r_31415
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  • @Robert Smith: "some of the conditions", do you there are more ? – Rajesh D Nov 18 '10 at 16:25
  • @Rajesh: Well, I take for granted that each statement can be reformulated in other way. Furthermore, I can't claim that such a list is comprehensive. I covers a lot, though. – r_31415 Nov 18 '10 at 16:34
  • what about in QED ? – Rajesh D Nov 18 '10 at 16:37
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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. – r_31415 Nov 18 '10 at 16:55
  • It violates the condition in 0 of course. – Raskolnikov Nov 18 '10 at 17:12
  • There is no wavefunction in 0! – r_31415 Nov 18 '10 at 17:18
  • Which proves Keenan's point. – Raskolnikov Nov 18 '10 at 17:21
  • Wow. These conditions are statements about the wavefunctions, not about particular systems obeying those wavefunctions. To see if a wavefunction follows the rules I referenced, you need first and foremost, a wavefunction. Unless you can find a postulate saying that a system needs to be described in terms of a single wavefunction continuous at all points, you're misreading the argument. – r_31415 Nov 18 '10 at 17:30
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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
  • @Robert: the moral of all this being that you should add some clarification of where does the $x$ in your story come from. – Marek Nov 18 '10 at 18:16
  • @Marek: Again, that statement refers to a wavefunction (in your case, a generalization), which is perfectly fine because each of the two wavefunctions for the Delta potential, are continuous at every point (that is, they comply with the condition 3). What it doesn't say is that a system needs to have a single wavefunction with the properties aforementioned. I don't think there is any need to blame the artificial attributes of the Delta potential since it needs to be mathematical consistent, the same way that these conditions on the wavefunction maintain such consistency. – r_31415 Nov 18 '10 at 18:21
  • @Robert: all right then. But in that case you should specify that you are working on the subset of $\mathbb{R}^d$ in stark contrast with what people usually expect from the usual quantum particle located in the usual world around us. You should just state that given the set of possible positions $S \subset \mathbb{R}^d$, the Hilbert space is defined as $L^2(S)$. – Marek Nov 18 '10 at 18:32
  • @Robert,@Marek,@Raskolnikov : what about mechanics (obeying special and general relativity) ? Is the state function well behaved everywhere ? – Rajesh D Nov 18 '10 at 19:40
  • @Marek: Good enough. Just a thing: Don't you think is a bit redundant to say that for all positions $x$, the wavefunctions are defined on $L^2(S)$ when we already said that wavefunctions must be square-integrable. Then, I would only need to add that $x \in \mathbb{R}$ (as in the hyperphysics link) or $x \in \mathbb{R^{d}}$ in general. – r_31415 Nov 18 '10 at 20:45
  • @Rajesh: What do you mean by 'state function' in mechanics? – r_31415 Nov 18 '10 at 20:45
  • @Robert Smith: I mean position of a particle or its momentum – Rajesh D Nov 18 '10 at 21:13
  • @Rajesh: good question. As for momentum, yes it is well defined. Momentum is much more fundamental than position because it corresponds to a translational symmetry of the space-time (and the equations), e.g. invariance of physical laws in time gives you the law of conservation of the energy. But it turns out that position is not really a good observable. Already when working with Dirac equations there are big problems with interpretation and consistency and when one moves to Quantum Field Theory one has to give up the concept of position as a well-defined operator completely. – Marek Nov 18 '10 at 21:32
  • @Robert: Right. I think the most concise way of writing all the conditions 2., 3. and 4. is $\psi \in L^2(S)$ and $\psi \in C^1(S)$. But maybe you should just leave it as it is (because probably not everyone here understands math notation) and just add the quantificators $x \in S$. – Marek Nov 18 '10 at 21:39
  • @Robert: seems to me the bound state of the Dirac potential doesn't fulfill condition 4. Keenan is still correct. Anyway, wether we agree to include 0 or not, I don't think condition 4 is necessary. – Raskolnikov Nov 18 '10 at 22:22
  • @Marek: I agree. I will add the change in a couple of hours I guess. – r_31415 Nov 18 '10 at 23:23
  • @Raskolnikov: You meant this http://en.wikipedia.org/wiki/Delta_potential#Bound_state_.28negative_energy.29, right? In that case, the plot of the graph certainly is not differentiable at $0$ (that is, a function which depicts such a graph is not differentiable at $0$), however, there is not wavefunction with that behavior. Instead, we have two wavefunctions and both follow condition 4. – r_31415 Nov 18 '10 at 23:31
  • @Raskolnikov: it depends on whether you allow infinite potentials. But they are not physical. For finite physical potential, Robert's conditions are fine. – Marek Nov 19 '10 at 00:25
  • @Marek: I think that every wavefunction (physical or not) need to satisfy the condition 4. Otherwise, how could it satisfy condition 1? Granted, you don't start knowing the wavefunction, but as I said, this whole thing needs to be consistent. – r_31415 Nov 19 '10 at 02:18
  • @Robert: it depends in how you treat the equation. If you indeed require that it holds everywhere then surely the function itself must be twice differentiable. But because we are working on a Lebesgue space, the inner product (which is the main structure of the theory) doesn't "see" sets of measure zero and so function diverging at a point (or not even being defined there) need not really be a problem from a mathematical point of view. Also, you can work with distributions. But all of this is getting rather technical. I think your answer is fine without getting into these problematic areas. – Marek Nov 19 '10 at 02:26
  • Take the Bloch electrons in Noldorin's post. Some of the models describing the wave functions of these electrons satisfy neither 1 nor 4. As a result, the wave functions can be fractal. – Raskolnikov Nov 19 '10 at 08:27
  • @Raskolnikov: I haven't studied Bloch electrons, however, from what I've seen, the wavefunction is (obviously) derived from the Schrödinger equation (then, it satisfies condition 1) for a periodic potential. If satisfies Schrödinger without problems, then satisfies condition 4. – r_31415 Nov 19 '10 at 15:34
  • @Marek: Oh, I didn't understand the Lebesgue measure part quite well. As for the distributions, sure (even if I haven't seen a working example trying to obtain more leeway when dealing with ill-behaved functions in QM). – r_31415 Nov 19 '10 at 15:43
  • @Raskolnikov, Robert: I think that you are both correct but you are each obviously talking about different things. AFAICS Robert talks about physical examples and for those his definition is fine while Raskolnikov talks about various mathematical idealizations of potentials and solutions (like infinite walls, delta functions, etc.) which are surely very useful but not really physical. – Marek Nov 19 '10 at 15:50
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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
  • @Raskolnikov: very nice point. After thinking about it for a little bit I guess you are right. – Marek Nov 21 '10 at 22:11