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I am confused. I have to show that $f(x)$ that satisfies the ODE below "has definite parity or can be chosen to have def parity".

$$\frac{d^2}{dx^2}f(x) = (h(x)+c)f(x)$$ where $h(x)$ has even parity and $c$ is a constant.

Operator $\frac{d^2}{dx^2}$ has even parity $\implies$ LHS has even parity $\implies$ RHS has even parity.

Also $h(x)+c$ has even parity by the even parity of $h(x)$.

Does this naturally imply that $f(x)$ has even parity? But if so, what is the thing about choosing? I am also convinced that my argument is wrong since I am later told to consider the odd-parity solutions.

Thanks.

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3 Answers 3

up vote 4 down vote accepted

Let $g$ denote any solution of the ODE, then the even function $f$ defined by $f(x)=g(x)+g(-x)$ is a solution as well. Furthermore $f$ is nonzero as soon as $g$ is not odd. But the ODE cannot have only odd solutions (consider the initial condition $g(0)=1$), hence you are done.

(Likewise, starting from any solution $g$ and considering $h(x)=g(x)-g(-x)$ yields an odd solution $h$ of the ODE. And one can manage that $h$ is nonzero by considering the initial condition $g'(0)=1$ which ensures that $g$ is not even.)

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The ODE being second order, it admits two solutions. The parity symmetry of the DE maps the space of solutions into itself, but not necessarily each individual solution. One can find a particular solution that is indeed symmetric. Consider $y''(x) - y(x)=0$. Its two solutions can be $y_1(x) = \mathrm{e}^x$ and $y_2(x) = \mathrm{e}^{-x}$ which are not themselves symmetric. Yet there symmetric linear combination is, i.e. $\cosh(x)$. The odd parity solution would be $\sinh(x)$.

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More generally, if $G$ is a symmetry group that leaves a linear equation invariant, so that if $f$ is a solution, $Sf$ is also a solution for all symmetry operations $S\in G$, then you can extract from any solution of the equation the component that transforms according to some irreducible representation of $G$ by forming the linear combination

$$\sum_{S\in G}\chi(S)Sf\;,$$

where $\chi$ is the character of the irreducible representation. In the present case, $G$ is $\mathbb Z_2$, with the two irreducible characters $(1,1)$ and $(1,-1)$, and Didier's prescription for forming an even and an odd solution from a given solution corresponds to extracting the components of the solution that transform according to these representations.

[Edit in response to the comments]

Here's another example. This approach is quite useful in solid state physics and quantum chemistry, where one often solves (some approximation to) the Schrödinger equation under various symmetries. Suppose you've calculated some wavefunctions or orbitals for an isolated atom and you now want to use these as a starting point for a calculation involving that atom in another situation, say, in a molecule or a crystal or an external field. For instance, you might want to do a perturbation expansion starting from the solution for the atom.

The field of the isolated nucleus has spherical symmetry, so your atomic functions transform according to irreducible representations of $O(3)$. For instance, you might have some valence orbitals for a carbon atom that transform under the irreducible representations for $l=0$ and $l=1$ (which are labelled as $s$ and $p$, respectively), and perhaps also some unoccupied orbitals with $l=2$ (labelled as $d$) that you want to include in the calculation to improve the accuracy. The new environment will typically not have full $O(3)$ symmetry, but will often have some remaining symmetry. For instance, in a methane molecule, there is full tetrahedral symmetry (symmetry group $T_d$), in a simple cubic lattice there might be full octahedral symmetry (symmetry group $O$), and in an external field there might be $O(2)$ symmetry left. (There isn't much to do in that last case, since the functions for a particular $l$ are typically already set up to correspond to a particular representation of $O(2)$ about the $z$ axis, so you just have to orient the $z$ axis along the field.)

Now the question is how to find combinations of your $O(3)$-symmetric functions that transform under irreducible representations of the reduced symmetry group. You can deduce from the character tables which functions have components belonging to which representations, but you need the projection method above to actually find those components.

Let's take $T_d$ as the symmetry group (think of CH$_4$). Here's the character table for that (reproduced from here):

$$ \begin{array}{|l|r|r|r|r|r|} &E&8C_3&3C_2&6S_4&6\sigma_d\\ \hline A_1&1&1&1&1&1\\ \hline A_2&1&1&1&-1&-1\\ \hline E&2&-1&2&0&0\\ \hline T_1&3&0&-1&1&-1\\ \hline T_2&3&0&-1&-1&1 \end{array} $$

The rows correspond to the irreducible representations; their labels are arbitrary conventions. The columns correspond to the conjugacy classes of $T_d$; their labels consist of their sizes and abbreviations for the corresponding symmetry operations: identity, rotation about a three-fold axis, rotation about a a two-fold axis, inversion and rotation about a four-fold inversion axis, and reflection in a plane (i. e. inversion and rotation about $\pi$), respectively. With respect to the induced permutations of the corners of a tetrahedron, these correspond to the cycle types $()$, $(123)$, $(12)(34)$, $(1234)$, $(12)$, respectively.

The irreducible representations of $O(3)$ have character

$$P^l\sum_{m=-l}^l\mathrm e^{\mathrm im\phi}=P^l\left(1+2\sum_{m=1}^l\cos(m\phi)\right)\;,$$

where $\phi$ is the rotation angle and $P=\pm1$ according to whether the operation involves an inversion. Here are the characters for $l=0,1,2$ for the operations in $T_d$:

$$ \begin{array}{|l|l|r|r|r|r|r|} &&E&8C_3&3C_2&6S_4&6\sigma_d\\ \hline l=0&1& 1&1&1&1&1 \\ \hline l=1&P(1+2\cos\phi)& 3&0&-1&-1&1 \\ \hline l=2&1+2\cos\phi+2\cos(2\phi)& 5&-1&1&-1&1 \end{array} $$

Let's start with the easiest case, an $s$ orbital, that is, a function proportional to the spherical harmonic $Y_{00}$. This transforms according to the $l=0$ representation of $O(3)$. If you didn't already know or guess, you can see from the character table that this transforms under the one-dimensional $A_1$ representation of $T_d$. So there's nothing to do here; you can directly use your $s$ orbital for the methane calculation.

So let's move on to the $p$ orbitals, with $l=1$. This was perhaps less obvious, but from the character table you can see that these transform under the three-dimensional $T_2$ representation of $T_d$. So here again there's nothing to do and you can use them as is for the methane calculation. Since this might be less expected, we can check this result explicitly with the projection method. Take the $l=1$ function for $m=0$, which is proportional to $Y_{10}$, and thus to $\cos\theta$. The operations in $T_d$ all transform this function either into itself or into the corresponding function oriented towards one of the other three corners of the tetrahedron. Since those three corners are all equivalent, we only have to count how often each conjugacy class transforms the function into itself and to any other corner. Here are the counts, which you can verify by going through the possible ways of orienting the respective axes of the transformations, or somewhat more easily using the corner permutation cycle types above:

$$ \begin{array}{|l|c|c|c|c|c|c|} &E&8C_3&3C_2&6S_4&6\sigma_d\\ \hline \text{same corner}&1&2&0&0&3\\ \hline \text{other corner}&0&6&3&6&3 \end{array} $$

Let's project this $p$ function onto the $A_1$ representation. We have to add up all $24$ transformed versions of the function, using the $A_1$ characters as coefficients. These are all $1$, so this adds up to $6$ functions on the same corner and $18$ on the others, or $6$ functions per corner, which is perhaps not surprising. Since the tetrahedral angle is $\arccos(-\frac13)$, pointing the function $\cos\theta$ towards a tetrahedral corner in the $x$-$z$-plane yields $-\frac13\cos\theta+\sqrt{8/9}\sin\theta\cos\phi$, and then rotating that about the $z$ axis through $2\pi/3$ and $4\pi/3$ yields $-\frac13\cos\theta+\sqrt{8/9}\sin\theta(-\frac12\cos\phi\pm\frac{\sqrt3}2\sin\phi)$. So the projection of our $l=1$, $m=0$ function onto the $A_1$ representation of $T_d$ is

$$ \begin{eqnarray} 6\left(\cos\theta-\frac13\cos\theta+\sqrt{8/9}\sin\theta\cos\phi\right. &-&\frac13\cos\theta+\sqrt{8/9}\sin\theta\left(-\frac12\cos\phi+\frac{\sqrt3}2\sin\phi\right) \\ &-& \left.\frac13\cos\theta+\sqrt{8/9}\sin\theta\left(-\frac12\cos\phi-\frac{\sqrt3}2\sin\phi\right) \right)\;, \end{eqnarray} $$

which is zero as predicted from the character table. For the remaining three representations, $A_2$, $E$ and $T_1$, the projection immediately yields zero; we can deduce this from the symmetries of the function even without using its concrete form, since the scalar product of both the "same corner" and the "other corner" row with the characters of these representations vanishes. (Here you need to take the scalar product without multiplying by the conjugacy class sizes as usual, since the number of operations is already included in the table of the counts.) You can perform the projection for $T_2$, too, but you don't need to since we already know now that there are no other components in the $l=1$ functions.

So let's move on to the $d$ functions. Here we get the first non-trivial result. From the character table, we can directly see that the $d$ functions don't transform according to an irreducible representation of $T_d$. But since they transform according to an irreducible representation of $O(3)$, we know they transform according to a representation of $T_d$ (which is a subgroup of $O(3)$), so this must be a reducible representation, which we can decompose using the character tables. Forming the scalar product of the $l=2$ row with the rows of the character table for $T_d$, we find that the five-dimensional $l=2$ representation decomposes into a two-dimensional $E$ representation and a three-dimensional $T_2$ representation. (You can check this by adding up those two rows to obtain the $l=2$ row.)

So now we can't simply take the $d$ functions as is. They split into different irreducible representations of $T_d$, and we need to find the components by projecting with the characters of those representations. Let's again use the simplest $d$ function we have, corresponding to $Y_{20}$, which is proportional to $3\cos^2\theta-1$. Again, all the operations in $T_d$ map this function to either itself or a rotated version pointing towards a different corner. From the above calculations of the scalar products with the "same corner" and "other corner" rows, we already know that such a function can only have $A_1$ and/or $T_2$ components. But from the character of $l=2$ we know that there's no $A_1$ present, so it follows without further calculation that the $l=2$, $m=0$ function belongs to $T_2$.

Things get more complicated for the other $m$ values, since in that case the operations of $T_d$ produce more different version of the functions. They must, since we know from above there's an $E$ component in $l=2$, and there can't be an $E$ component in a function that only has one version for each corner as the $m=0$ functions do. The actual projection of the remaining $d$ functions onto $E$ and $T_2$ is rather cumbersome to carry out by hand, but I hope it's become clear by now how one would have to go about.

On Didier's concern about ensuring a non-zero result of the projection: That depends on the setting and on how you choose which functions to project. As above, if we project a function that belongs to a different representation, we get $0$ (as we should). If we're solving an eigenvalue problem $H\psi=E\psi$ of a diagonalizable linear operator $H$, as we usually are in approximating a solution to the Schrödinger equation, then we can decompose the entire space according to the irreducible representations of the symmetry group, and we can choose the eigenfunctions to correspond to an irreducible representation, so in that case we can always find suitable functions. But if we're just solving a problem like $Lf=g$ for some linear operator $L$ and a fixed function $g$, then there may or may not be solutions that transform according to any particular irreducible representation of the symmetry group. You can see this directly in the case of $H\psi=0$, where $H$ is again a diagonalizable linear operator like the Schrödinger operator: This only has solutions if $0$ is in the spectrum of $H$, and if it is, there's no general way of telling which irreducible representations may be present in the decomposition of the corresponding eigenspace. In fact, by subtracting any eigenvalue $E$ from $H$, you can get an equation $(H-E)\psi=0$ (where $E$ is now a fixed part of the operator, not an eigenvalue to be determined) whose solutions are the eigenfunctions of $H$ corresponding to $E$, and these will typically only have components according to one or, if the operator is degenerate, perhaps a few irreducible representations.

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joriki: Yep, obviously, symmetry groups are the proper general setting. Which makes me wonder if there is anything else than ad hoc procedures (like the two I used) to make sure the resulting linear combination is not zero. –  Did Aug 25 '11 at 14:09
    
@joriki: Thanks! This is a very nice way of looking at the problem. Could you possibly give me another example using this method? (Maybe one with a more complicated sym group). It's really quite cool! Thanks again. –  littleme Aug 25 '11 at 17:34
    
@Didier: Do you mean if we weren't given the $g(0)$ and $g'(0)$? Thanks also for your answer! I am interested in the ans to your concern too. –  littleme Aug 25 '11 at 17:36
    
I have to finish something else; I'll hopefully be responding to these comments later in the evening. –  joriki Aug 25 '11 at 17:42
    
@littleme: Apologies for taking longer than expected to respond. I hope the example is of interest. –  joriki Aug 29 '11 at 7:44

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