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the general solution for the wave function of Helium is a linear combination of the two possible states:

$$ \text{state 1:} \quad \psi_a(\vec{r}_{1}) \, \psi_b(\vec{r}_{2}) \\ \text{state 2:} \quad \psi_a(\vec{r}_{2}) \, \psi_b(\vec{r}_{1}) $$

(where $\vec{r}_{1}/\vec{r}_{2}$ denotes electron 1/2.)

But aren't the products of these wavefunction the same? Why then make a linear combination; it would be just like same function +/- same function?

$$ \psi_a(\vec{r}_{1}) \, \psi_b(\vec{r}_{2}) = \psi_a(\vec{r}_{2}) \, \psi_b(\vec{r}_{1}) \, ? $$

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  • $\begingroup$ "the general solution for the wave function". It is not correct. The wavefunction is the solution of the Schrödinger equation. Also, that combination is not the general solution of the T.I.S.E. $\endgroup$ – user1420303 Sep 28 '16 at 17:16
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First of all, note that $$ \psi_a(\vec{r}_{1}) \, \psi_b(\vec{r}_{2}) \neq \psi_a(\vec{r}_{2}) \, \psi_b(\vec{r}_{1}) \, , $$ since same functions in these products ($\psi_a$ and $\psi_b$) have different arguments ($\vec{r}_{1}$ and $\vec{r}_{2}$) to the left and to the right of the inequality sign.


Now, to the question, why do we have a linear combination. For a system of two indistinguishable particles (say, electrons) the obvious requirement is that there should be no observable difference between the system in state $\psi(1, 2)$ and the system in state $\psi(2, 1)$, \begin{equation} |\psi(1, 2)|^{2} = |\psi(2, 1)|^{2} \, , \end{equation} where $1$ and $2$ stand for coordinates of particles. This implies that \begin{equation*} \psi(1, 2) = \pm \psi(2, 1) \, . \end{equation*}


And in order to satisfy this equality in what is often called the orbital approximation, where the state of a many electron system $\psi$ is represented as a product of a single-particle states $\psi_i$, the wave function $\psi(1, 2)$ should be actually written as not just a product of $\psi_1$ and $\psi_2$ (none of which, neither $\psi_1(1) \psi_2(2)$ nor $\psi_2(1) \psi_1(2)$, is symmetric or antisymmetric), but rather as the following linear combination of these products, \begin{equation*} \psi(1, 2) = \frac{1}{\sqrt{2}} \Big( \psi_1(1) \psi_2(2) \pm \psi_1(2) \psi_2(1) \Big) \, , \end{equation*} where $1 / \sqrt{2}$ is the normalization factor.


The choice of the sign in the expressions above is not arbitrary, but rather determined by the particles spin $s$, namely, \begin{equation*} \psi(1, 2) = (-1)^{2s} \psi(2, 1). \end{equation*} Particles with integer spins, called bosons (photons, alpha particles), are described by a symmetric wave function \begin{equation*} \psi_{\mathrm{s}}(1, 2) = \frac{1}{\sqrt{2}} \Big( \psi_1(1) \psi_2(2) + \psi_1(2) \psi_2(1) \Big) \, , \end{equation*} while the particles with half-integer spins, called fermions (electrons, protons, neutrons), are described by the anti-symmetric wave function \begin{equation*} \psi_{\mathrm{a}}(1, 2) = \frac{1}{\sqrt{2}} \Big( \psi_1(1) \psi_2(2) - \psi_1(2) \psi_2(1) \Big) \, . \end{equation*}

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  • $\begingroup$ thanks for the great answer; but why aren't those products the same, the terms of the functions depend on n,l,m, but are those quantum numbers independent of the position vector of the electrons, or has a distinct vector a given set of n,l,m ? $\endgroup$ – hanna Sep 29 '16 at 9:35
  • $\begingroup$ @hanna, well, physically, the $\psi_a(\vec{r}_{1}) \, \psi_b(\vec{r}_{2})$ and $\psi_a(\vec{r}_{2}) \, \psi_b(\vec{r}_{1})$ are the same in a sense that they describe the exact same state of a 2-electron system, namely, a state in which one lectron occupies $\psi_a$ spin orbital, while another occupies $\psi_b$. And, of course, since electrons are indistinguishable it does not matter which electron occupies which orbital: the state described by $\psi_a(\vec{r}_{1}) \, \psi_b(\vec{r}_{2})$ is equal to one described $\psi_a(\vec{r}_{2}) \, \psi_b(\vec{r}_{1})$ in every observable way. $\endgroup$ – Wildcat Sep 29 '16 at 10:13
  • $\begingroup$ @hanna, mathematically though, it is quite apparent that the $\psi_a(\vec{r}_{1}) \, \psi_b(\vec{r}_{2})$ product differs from the $\psi_a(\vec{r}_{2}) \, \psi_b(\vec{r}_{1})$ one, which is odd, since we have then two apparently different mathematical description of the same state! This already suggests that there is something wrong with describing the state by these individual products, they both should be included in the description. $\endgroup$ – Wildcat Sep 29 '16 at 10:23

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