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When discussing the Schroedinger equation in spherical coordinates, it is standard practice in QM handbooks to point out that the radial part of the 3-dimensional wave equation bears a strong analogy to the corresponding 1-dimensional case. This is because the Laplace operator in spherical coordinates can be written in the form $$L = \frac{1}{r}\,\frac{d^2}{dr^2}r.$$ Hence my making the substitution $\psi\to r\psi$, the 3D radial wave function $r\psi(r)$ satisfies exactly the same Schroedinger equation as the wave function $\psi(x)$ in the 1D case.

So far so good. However, at this stage some authors point out that this substitution is merely feasible, as long as the condition $r\psi\to0$ in the limit $r \to0$ is met. If not, then the wave function diverges at the origin, and this is unacceptable on physical grounds. [One may counter this argument, by pointing out that in the calculation of expectation values the square of the absolute value of the wave function is always multiplied by the spherical shell $4\pi r^2\,dr$. The factor $r^2$ neutralizes the above mentioned divergence.]

Indeed there appears to be a slight difference between the 3D and 1D case, when one observes the elementary case of a particle in a box. In both cases, outside of the box the wave function is exponentially decreasing. Inside, where the particle is "free", the solution is oscillatory. In the 1D case, the wave function can be written as $$\psi(x) = A\sin(kx) + B\cos(kx).$$ In the 3D case, the wave function is given by the zeroth order Bessel function $$\psi(r) = C\sin(kr)/r.$$ [Apparently the term $D\cos(kr)/r$ is omitted since it is considered unphysical, see previous paragraph.]

Now if the wave function inside the box has only one free parameter $C$, all one can really do is determine its value by demanding continuity of the wave function on the boundary of the box. On the other hand, QM textbooks claim that one should always seek a solution in which not only the wave function itself, but also its first derivative is continuous on a boundary!

All in all I find the omission of the $D\cos(kr)/r$ term on physical grounds confusing. First of all it creates a difference between the 3D and 1D case. This seems odd to me since the corresponding wave functions satisfy essentially the same Schroedinger equation. Secondly, by omitting this term one loses an adjustable parameter $D$, which plays an important role in meeting the full set of boundary conditions.

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The 3D case is different than the 1D case, so don't feel too committed to the analogy. However, in the 1D case, you get only sine functions as solutions if you place a "hard wall" (an infinite potential barrier) at $x=0$. That is basically what is happening here: the origin is a sort of "hard" boundary for the radial equation since "negative radius" makes no sense.

Furthermore, the cosine-like solution diverges as $\frac{1}{r}$ at $r=0$ which does not satisfy Schroedinger's Equation. More precisely, a $\frac{1}{r}$-type solution is normalizable, but $\nabla^2(\frac{1}{r})=-4\pi\delta(r)$ and to satisfy Schrödinger's Equation in this case we need $\nabla^2(\frac{1}{r})=\frac{E}{r}$.

Finally, there is a solution to the equation outside of the sphere, too. It is (as you point out) exponentially decaying. That solution has an undetermined constant as well, so you need two boundary conditions to find the two undetermined constants: the one inside and the one outside.

Edit:Since there is not enough room in the comments to adequately show my point about the two boundary conditions, I'm replying here.

Yes, you are correct that with just the 2 conditions (normalization and continuity) you can get a reasonable solution, but you cannot get a correct solution. This is because normalization is a requirement we impose, but continuity and smoothness are mathematical requirements of the system. I could just as easily require that $\psi(3)=16$ and it would have just as much mathematical relevance to the solution of the problem as normalization does.

However, it is true that normalization is a restriction on the solutions of the system; therefore, this system is actually over-specified and only admits a countable number of solutions (a hallmark of any Sturm-Liouville Theory). In other words, the normalization condition is important, but it does not negate the other, rigorous mathematical requirements of the problem (namely, smoothness). In what follows I try to explain why that is the case in a mathematically rigorous way.

(Also, I want to point out that in the 1D case you don't just have continuity and smoothness at one boundary: you have them at 2 boundaries. You actually have 4 equations with 4 unknowns in that case - since there will be a decaying solution at the other side of the well, too. So don't confuse 1D and 3D, they only correspond here when there is a "hard wall" at $r=0$ (in which case you will only have 1 unknown inside the well since cosine solutions are not admitted).)

First, since you are happy to accept that the wave function must be continuous at $r=a$, I will prove that that implies that the wave function is also smooth at $r=a$ (even in the 3D case). Second, I will show that with the smoothness condition we recover a discrete spectrum of bound states (as expected). Third, I will prove that without the smoothness condition we recover a continuum of bound states. Therefore, by showing that smoothness is required and that it implies a discrete spectrum, I will show that solving the problem without considering smoothness is incorrect because it does not give the same results as the mathematically rigorous method.

First, since we believe that the function is continuous (but possibly not differentiable) at $r=a$, let's solve the following integral equation: $$ \lim_{\epsilon\rightarrow 0}\Big[\int_{a-\epsilon}^{a+\epsilon}\frac{-\hbar^2}{2m}\frac{d^2\psi}{dr^2}dr\Big]+\lim_{\epsilon\rightarrow 0}\Big[\int_{a-\epsilon}^{a+\epsilon}V(r)\psi(r)dr\Big]=\lim_{\epsilon\rightarrow 0}\Big[\int_{a-\epsilon}^{a+\epsilon}E\psi(r)dr\Big] $$ Basically, I'm just integrating the Schrödinger's Equation for this problem in a small region around the point where we expect the wave function to (possibly) have a cusp. Here, $V(r)=0$ (when $r<a$) or $V(r)=V_0$ (when $r>a$). That means that $$ \lim_{\epsilon\rightarrow 0}\Big[\int_{a-\epsilon}^{a+\epsilon}V(r)\psi(r)dr\Big]=\lim_{\epsilon\rightarrow 0}\Big[\int_{a}^{a+\epsilon}V_0\psi(r)dr\Big]=0 $$ Solving the equations above gives: $$ \frac{-\hbar^2}{2m}\Big(\frac{d\psi}{dr}\Big|_{r=a^+}-\frac{d\psi}{dr}\Big|_{r=a^-}\Big)=0 \\ \therefore \ \frac{d\psi}{dr}\Big|_{r=a^+}=\frac{d\psi}{dr}\Big|_{r=a^-} $$ And that final equation is exactly the definition of smoothness.

Now, let's take the 3D problem for $l=0$ (which is the case that you ask about in your question) and solve it: $$ \frac{-\hbar^2}{2m}\frac{d^2\psi_1}{dr^2}=E\psi_1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{for r<a)}\\ \frac{-\hbar^2}{2m}\frac{d^2\psi_2}{dr^2}+V_0\psi_2=E\psi_2 \ \ \ \text{(for r>a)} $$ Let's define $k_1=\frac{\sqrt{2mE}}{\hbar}$ and $k_2=\frac{\sqrt{2m(V_0-E)}}{\hbar}$. Therefore, $$ \frac{d^2\psi_1}{dr^2}=-k_1^2\psi_1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{for r<a)}\\ \frac{d^2\psi_2}{dr^2}=k_2^2\psi_2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(for r>a)} $$ Let's assume that $E<V_0$ so that we get exponential solutions when $r>a$ instead of oscillatory solutions. When $E>V_0$, our solutions are not bound states and do not decay exponentially as $r\rightarrow\infty$. Solving these equations gives $$ \psi_1(r)=A\sin(k_1r)+B\cos(k_1r) \\ \psi_2(r)=Ce^{k_2r}+De^{-k_2r} $$ Now we must apply our boundary conditions. First, since we have an infinite potential barrier at $r=0$, $\psi_1(0)=0$; therefore, $B=0$. Next, by normalizability,we must require that $\psi_2(\infty)=0$; therefore, $C=0$. This leaves just $$ \psi_1(r)=A\sin(k_1r) \\ \psi_2(r)=De^{-k_2r} $$ Now, we apply our continuity condition at $r=a$: $$ A\sin(k_1a)=De^{-k_2a} \\ \therefore \frac{A}{D}=\frac{e^{-k_2a}}{\sin(k_1a)} $$ Now, if you want to uniquely solve the problem, you must apply smoothness at $r=a$, which will give you $$ k_1A\cos(k_1a)=-k_2De^{-k_2a} \\ \therefore \tan(k_1a)=-\frac{k_1}{k_2} $$

That last equation is a transcendental equation which allows us to solve for the energy levels of the bound-states of the problem. Here is a WolframAlpha plot of the 2 sides of that equation when $a=10$ and $V_0=50$. (http://www.wolframalpha.com/input/?i=Plot[{Tan[Sqrt[x]*10]%2C+-Sqrt[x%2F%2850-x%29]}%2C{x%2C0%2C50}]) As you can see, there are only a finite number of intersections between the two graphs, and therefore, there are only a finite number of permissible energy values (i.e. the energy spectrum is discrete).

Finally, let's take a look at trying to solve the problem using only continuity and normalization. $$ A\sin(k_1a)=De^{-k_2a} \\ 1=\int^a_0A^2\sin^2(k_1r)dr+\int^\infty_aD^2e^{-2K_2r}dr \\ \therefore \ A=\frac{1}{\sqrt{\frac{a}{2}-\frac{\sin(2k_1a)}{4k_1}+\frac{e^{-k_2a}\sin(k_1a)}{2k_2}}} \\ \ \ \ D=\frac{e^{k_2a}\sin(k_1a)}{\sqrt{\frac{a}{2}-\frac{\sin(2k_1a)}{4k_1}+\frac{e^{-k_2a}\sin(k_1a)}{2k_2}}} $$ As you can see, there is no restriction on what values $k_1$ and $k_2$ can take on. Any value between $0$ and $V_0$ gives a sensible answer.

Therefore, we must conclude that ignoring the smoothness condition at the boundary in the 3D, $l=0$ case is unacceptable because doing so gives a fundamentally different answer than the one derived when smoothness was not ignored. Moreover, smoothness is required of the problem, and that fact is mathematically derivable (as shown above).

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  • $\begingroup$ The cosine-like solution appears normalizable. The divergence at the origin is neutralized by the spherical shell. Hence the radial probability density is analytic and smooth. $\endgroup$ – M. Wind Dec 21 '13 at 18:09
  • $\begingroup$ Yes, the exponentially decaying solution outside of the box (or sphere) has an undetermined constant as well. Its value is determined by the condition that the probability is normalized. $\endgroup$ – M. Wind Dec 21 '13 at 18:24
  • $\begingroup$ Thank you for pushing me on this point. I've edited my post to include a more accurate explanation for why the Spherical Neumann functions are excluded. To address your second point: no, the undetermined constant outside is not determined by normalization. The continuity condition only gives the ratio of the two constants; therefore, I could easily scale the outside solution up or down while maintaining continuity so long as I scale the inside solution by an appropriate amount. $\endgroup$ – Geoffrey Dec 21 '13 at 19:37
  • $\begingroup$ Thank you! The second point is trivial. I am sure that we actually agree (though it now seems as if we don't). Clearly there is one parameter describing the amplitude outside of the box. In the 1D case there are two parameters inside the box. Three parameters, three boundary conditions: continuity of psi, continuity of its derivative and normalization to unity. In the 3D case there is one parameter outside the sphere and one inside. One can match two boundary conditions: continuity of psi and normalization to unity. $\endgroup$ – M. Wind Dec 21 '13 at 20:12
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    $\begingroup$ I like your first comment. If one accepts the validity of the mathematical transformation from 3D to 1D, then ultimately only one key difference remains. Namely the fact that in 1D the domain for x is [-inf, +inf] while in 3D the domain for r is [0, +inf]. So in 1D there are anti-symmetric (sine-like) and symmetric (cosine-like) solutions. In 3D this distinction is meaningless. I will have to think hard whether this implies that a "hard wall" at r=0 should be imagined... $\endgroup$ – M. Wind Dec 21 '13 at 20:53
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First of all, normalization is not a condition to fix the constant. Schodinger equation is a linear equation because the solutions form a linear space - Hilbert space. A physical state is an equivalent class of the vectors in Hilbert space, where two vectors belong to the same class if they differ by only a nonzero constant complex number. And it's clear that normalized wave-functions don't form a linear space, they are in fact only a representative (class) of the equivalent class with an ambiguity of phase. In fact normalized state is not a necessary ingredient in quantum mechanics, because we can always define the expectable value to be \begin{equation} \langle O \rangle = \frac{\langle \psi| O |\psi \rangle }{\langle \psi |\psi \rangle} \end{equation} nevertheless normalization are convenient in the practical calculation.

(There is little originality in the following statement, I just elaborate section 8.4 of the reference[1] for the context of this problem. Thanks OP, you drive me learn something.)

What's important however, is normalizablity(at least for the bound states). Physically the total probability for this particle to present in the whole space for in bound state is $1$ by definition. Mathematically, we require physical observables to be self-adjoint operators with respect to some (weighted) inner product (the weight for example is $r^2 \sin \theta$ in spherical polar coordinates). So for Laplace equation in 3d, when eigenvalue is $0$ \begin{equation} -\frac{1}{r^2} \frac{d }{d r} (r^2 \frac{d\psi}{dr}) + \frac{l(l+1)}{r^2}\psi = 0 \end{equation} for $l \ne 0 $, the solution $\psi = r^{-(l+1)}$ has to be excluded, because it's not normalizable in the vicinity of $r=0$. And in this case, no boundary condition at $r=0$ is needed.

Things become interesting when $l=0$. The two solutions $1$ and $\frac{1}{r}$ are all normalizable near $r=0$. Their behavior around the $r=0$ is irrelevant about eigenvalue. For a general case, the two independent normalizable solutions are $j_0(kr)$ and $n_0(kr)$ have the same expansions as $k=0$, \begin{equation} j_0(kr) \sim 1 \quad -k n_0(kr) \sim \frac{1}{r} \end{equation}

The problem arises here is due to the fact that $r=0$ is a singular point of this differential equation, where the coefficients of ODE are ill-defined. My personal point of view is that transformation from Cartesian to spherical polar coordinate is singular at $r =0$ and we somehow lose the information there! The way to restore the corrupted piece is to provide a boundary condition. Yet this boundary condition is not completely arbitrary: we require this boundary condition associated with inner product makes the Hamiltonian a self-adjoint(mathematical alias for Hermitian) operator. Weyl theorem systematically deal with this issue: the solution to this case(limit circle) is that we specify the boundary condition at $r=0$ to be \begin{equation} \psi \sim A(1 + \frac{a_s}{r} ) \end{equation} where $a_s$ is called the scattering length and could be determined experimentally. As pointed out by Geoffrey, the physical meaning of this parameter is that it mimics a delta potential(see psedo-potential ) $\nabla \frac{1}{r}= -4\pi \delta(r)$ at the center where our differential equation can't reach. The delta function, like the boundary condition, in a sense restore the information that is missing in the transformation. So we if ignore the $n_0$ branch by specifying $a_s =0$, that's simply because we believe in reality there is no potential in the center, a evident fact before the coordinate transformation.

And yes, dimensionality is definitely crucial. For a $n$-dimensional radial Laplace operator \begin{equation} \frac{1}{\sqrt{g}} ( \sqrt{g} \psi^{i} )_{,i} = \frac{1}{r^{n-1}} \frac{\partial}{\partial r}(r^{n-1} \frac{\partial}{\partial r}\psi) \end{equation} even if $l=0$, one of the solutions $\psi = r^{-(n-2)}$ become non-normalizable when $n\ge 4$. So when the dimension is greater than $4$, you can completely ignore the point interaction in the center without disturbing the solution, whatever the strength it has. It is said to be irrelevant in the theory.

[1]: Mathematics for Physics: A Guided Tour for Graduate Students, Michael Stone, Paul Goldbart.

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  • $\begingroup$ Thank you very much for your response. For me the most interesting aspect is your claim that the transformation of Cartesian to spherical coordinates leads to loss of information in the point r=0. Therefore one has to be careful not to introduce singular solutions. In practice this means a careful check on the solutions of the SE to see which solutions are physically acceptable and which ones must be discarded. I will try to keep this in mind! $\endgroup$ – M. Wind Dec 24 '13 at 6:13
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You have received other answers, so I would like to focus on the overall problem of necessary regularity of the solutions of Schroedinger equation. More precisely: why one should require the regularity conditions on $\psi$ you have read in the other answers?

The point can be traced back to one of the most important axioms of QM: observables are self-adjoint operators. The reason for this requirement is that self-adjoint observables admits a spectral decomposition in terms of orthogonal projectors labeled by (Borelian) subsets of $R$ interpreted as the set of results of the measurement of the observable. (It is possible to make weaker the requirement dealing with decompositions of bounded positive operators, but I only stick to the elementary case here.)

In our case the relevant observable is the Hamiltonian one. But for the sake of simplicity I intend to start focusing on the momentum observable along the $k$th axis: $M_k$. One usually assumes that:

$$M_k := -i\hbar \frac{\partial}{\partial x_k}\:,\qquad (0)$$

where, for instance the domain ${\cal D}(M_k)$ is $C_0^\infty(R^3)$ of $\cal S(R^3)$ (the Schwartz'space), what follows is independent on this choice.

It is true that, if $\psi,\phi \in {\cal D}(M_k)$ then:

$$\langle \psi| M_k \phi \rangle = \langle M_k\psi | \phi \rangle\:.$$

Indeed, that identity only says that $M_k$ is symmetric (a densely defined operator is symmetric if, on its domain, coincides to the adjoint operator). However, it does not say that $M_k$ is self-adjoint. The self-adjoint condition (that implying the existence of the spectral decomposition) is instead:

$$M_k^\dagger = M_k\:. \qquad (1)$$

Above $M^\dagger$ is defined as, follows. First one defines its domain:

$${\cal D}(M^\dagger_k) := \left\{\psi \left.\in L^2(R^3) \:\right|\: \exists \psi' \in L^2(R^3)\qquad \mbox{with}\quad\langle \psi' | M_k \phi\rangle = \langle \psi |\phi\rangle\quad \forall \phi \in {\cal D}(M_k)\right\}$$

Since ${\cal D}(M_k)$ is dense, $\psi'$ is uniquely determined by $\psi$ and thus the map:

$${\cal D}(M_k^\dagger) \ni \psi \mapsto \psi' =: M_k^\dagger \psi$$

is well-defined. One easily sees that ${\cal D}(M_k^\dagger)$ is a subspace of $L^2(R^3)$ with ${\cal D}(M_k^\dagger) \supset {\cal D}(M_k)$ and that $M_k^\dagger$ is linear.

In this case ${\cal D}(M_k^\dagger)$ turns out to be considerably greater than ${\cal D}(M_k)$, so that (1) fails and $M_k$ is not self-adjoint with the given (standard) definition. What it is true is that $M_k^\dagger$, defined as above, is self-adjoint and that it is the only self-adjoint extension of $M_k$. Mathematically one says that $A$ is essentially self-adjoint when it is symmetric its adjoint operator $A^\dagger$ is self-adjoint $A^\dagger = (A^\dagger)^\dagger$. Therefore $M_k$ is essentially self-adjoint.

This discussion leads to conclude that the true definition of the momentum operator is not (0) but is:

$$P_k = M^\dagger_k\:.$$

It is however important to stress that the naive, technically wrong, definition (0) uniquely defines $P_k$, since it is the only self-adjoint extension of $M_k$. This latter is very simple to handle, because is a differential operator. Conversely $P_k$ has a domain much more difficult to characterize (without using the Fourier transform). The domain of $P_k$ is made of the functions $\psi$ in $L^2(R^3)$ that admit weak $k$-derivative which, in turn, is a function in $L^2(R^3)$. One says that $\psi \in L^2(R^3)$ admits a weak $k$-derivative $\phi_k : R^3 \to C$, if there is a function - the mentioned $\phi_k$ - such that

$$\int_{R^3} \frac{\partial f}{\partial x_k} \psi \:d^3x = - \int_{R^3} f \phi_k \:d^3x \quad \forall f \in C_0^\infty(R^3)\:.$$

You see that if $\psi$ admits the standard $k$-derivative it coincides with the weak one (that therefore exists in this case). However, there are many functions admitting weak derivatives that are not differentiable anywhere!

Let us come to the problem of the Hamiltonian operator. The Hamiltonian operator, in the mathematically naive version for the non relativistic theory, always includes an added part proportional to the Laplacian operator $\Delta$. As a matter of fact, for $a := -\hbar^2/(2m)$ and for some function $V: R^3 \to R$ the naive Hamiltonian operator is:

$$A := a \Delta + V\:,$$

with domain ${\cal D}(A)$ made of sufficiently differentiable functions.

Again one would be sure that $A$ is self-adjoint to exploit all spectral technology, but, as before $A$ is not. At most, with a careful choice of the domain ${\cal D}(A)$, the operator $A$ turns out to be essentially self adjoint. Namely, $A^\dagger$ is self-adjoint and the true Hamiltonian observable can safely be defined as: $$H := A^\dagger\:.$$ As before, the correct domain (and one could have many choices!) involves weak derivatives: $\Delta$ has to be interpreted using (second) weak derivatives instead of standard derivatives. So the class of functions one should consider in solving problems like that of finding the eigenvalues of $H$ (the energies of stationary states) or other problems, like determining the scattering states, are a wide class of generally non differentiable functions.

This is not the whole story, because, differently from the case of the momentum operator, the presence of $\Delta$ in $A$ makes easier the problem in view of known results on elliptic regularity. The basic results (due to Weyl, Friedrichs and Sobolev) under suitable hypotheses establish that if a function (actually a distribution) in $R^n$ verifies an equation like $$\Delta f = g\:,$$ where $\Delta$ is interpreted in weak sense, then the degree of weak differentiable regularity of $f$ is that of $g$ plus $2$. Moreover, if $f$ (assumed to be locally $L^2$) has a certain degree $k$ of weak regularity, it also has another degree $k' = k-p$ of standard regularity, where $p>0$ is a number depending on $n$. (To write a rigorous statement I should introduce several mathematical notions and I will not do for the sake of simplicity, since I just wish to give an idea about the basic argument).

Taking this result into account, it turns out that, for instance, if $\psi \in {\cal D}(H)= {\cal D}(A^\dagger)$ is an eigenvector of $H$, so that

$$-a\Delta_w \psi = (E-V) \psi \quad \mbox{where $\Delta_w$ is the weak Laplacian}\:,$$

then $\psi \in C^\infty$ where $V$ is such.

These procedures and results lead to a precise theorem concerning the mathematical requirements for a function $\psi$ that stays in the domain of $H$ and, if it the case, solves the proper or generalized eigenvalue equation. The theorem takes into account the fact that the true self-adjoint Hamiltonian operator is not the differential operator $A$, but is its unique self-adjoint extension $A^\dagger$.

The theorem considers an operator of the form:

$$A = a \Delta + V$$

where $V: R^3 \to R$ has the form for $N$ real constants $g_k$ and corresponding isolated points ${\bf x}_k$:

$$V({\bf x}) = \sum_{j=1}^N \frac{g_k}{|{\bf x}-{\bf x_j}|} + V_0({\bf r})\:,$$

$V_0$ is bounded below, diverges at most polynomially for $|{\bf x}|\to +\infty$ and it is a continuous function except for a finite number of 2-surfaces $\Sigma_i$ where the discontinuities are finite. With these hypotheses it is possible to establish that $A$ is essentially self adjoint on ${\cal D}(A)= C_0^\infty(R^3)$ or ${\cal D}(A)={\cal S}(R^3)$ with the same unique self-adjoint extension $H= A^\dagger$ in both cases. The domain of $H$ is much larger than these spaces and includes functions that so not admit proper second derivatives in the whole $R^3$.

It turns out that the functions (distributions in the generalize case) $\psi : R^3 \to C$ that can be used to solve the proper or generalized eigenvalue problem for the true Hamiltonian $H := A^\dagger$ must verify the following requirements (in addition to the eigenvalue problem):

1) away from the singularities of $V$, $\psi$ is $C^2$ and solves the eigenvalue equation interpreting $\Delta$as a proper differential operator;

2) crossing a singular surfaces $\Sigma_i$, if ${\bf y}\in \Sigma_i$, the function $\psi$ satisfies $$\lim_{{\bf x} \to {\bf y}^+}\psi({\bf x})= \lim_{{\bf x} \to {\bf y}^-}\psi({\bf x})$$ where the two limits are computed from the two semi-spaces separated by $\Sigma$ around ${\bf y}$ and, similarly: $$\lim_{{\bf x} \to {\bf y}^+}{\bf n}\cdot \nabla\psi({\bf x})= \lim_{{\bf x} \to {\bf y}^-}{\bf n}\cdot \nabla \psi({\bf x})$$ where ${\bf n}$ is the unit vector normal to $\Sigma_i$ at ${\bf y}$;

3) If ${\bf x}_k$ is an isolated singular point for $V$, the limit of $\psi$ for ${\bf x} \to {\bf x}_k$ exists and is finite.

Dealing with genuine $1D$ systems one finds similar requirements on the allowed wavefunctions.

Taking the above results into account, you can understand why, for instance a wavefunction in $R^3$ must not diverge at the origin: it is nothing but the requirement (3) above or even a consequence of requirement (1) when $V$ is regular at ${\bf r}=0$. For this reason $\psi(r) \sim r^{-1}\cos (kr)$ for $r\to 0$ cannot be accepted even if the corresponding $1D$ wavefunction $r\psi(r)$ is, in principle, allowed. The comparison from the $1D$ and the $3D$ case, based on the replacement $\psi(r) \to r\psi(r)$ is only formal and can only be used outside the singularities, while what is permitted or forbidden crossing a singularity must be discussed separately, remembering the true nature of the problem: $3D$ or $1D$. Notice also that, for the free particle, the accepted solution for $\ell=0$: $$\psi({\bf x})= A\frac{\sin (kr)}{r}$$ in accordance with (1) is $C^2$ (more strongly it is real analytic) and not only bounded in a neighborhood of ${\bf x}=0$, including that point. In fact there is no singularity in ${\bf x}=0$ for the free particle since $V\equiv 0$ in that case, and the apparent singularity is only due to the use of polar coordinates.

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  • $\begingroup$ Thank you very much for your lengthy reply. I am afraid that most of is too QM-technical for me, and I feel a bit lost... However the last paragraph is both clear and insightful. I will try to keep this in mind in my future calculations. $\endgroup$ – M. Wind Dec 24 '13 at 6:07
  • $\begingroup$ Do not worry, indeed these things are a bit technical if one wants to understand them deeply. $\endgroup$ – Valter Moretti Dec 25 '13 at 12:53

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