A particle moving on the x-axis has a probability of $1/5$ for being in the interval $(-d-a,-d+a)$ and $4/5$ for being in the interval $(d-a,d+a)$, where $d \gg a$.

The normalised wave function for the "left" interval is $\phi_-$ and for the "right" interval is $\phi_+$. What is the normalised wave function $\phi_x$ for the particle.

From Atkins' Physical Chemistry; Chapter 7 Quantum Mechanics, International Edition; Oxford University Press, Madison Avenue New York; ISBN 978-0-19-881474-0; p. 234:

It's always possible to find a normalisation constant N such that the probability density become equal to $|\phi|^2$

This means that:

$$\begin{align} \int_{-d-a}^{-d+a}|\phi_-|^2 \,\mathrm{d}x &= \frac{1}{5} \tag{1} \\ \int_{d-a}^{d+a}|\phi_+|^2 \,\mathrm{d}x &= \frac{4}{5} \tag{2} \end{align}$$ $$$$

Since $d \gg a$, $$|\phi_-|^2 = \frac{1}{5 \cdot 2a}$$ and $$|\phi_+|^2 = \frac{4}{5 \cdot 2a}$$

Also we can say $\phi=c_1\phi_-+c_2\phi_+$, so $$\phi \cdot \phi^*=|\phi|^2$$


They have written the solution as $\phi = (1/\sqrt{5})\phi_-+ (2/\sqrt{5})\phi_+$. I'm not able to understand how they came to this result.


1 Answer 1


This problem can be thought of as a linear combination of atomic orbitals $\phi_-$ and $\phi_+$ to molecular orbital $\phi$ with broken symmetry (i.e. LCAO-MO and $c_1 \neq c_2$). The answer to it can be figured out as follows.

As stated in the conditions, the normalized atomic orbitals are $\phi_-$ and $\phi_+$ for the left and right intervals centered at $-d$ and $+d$, respectively. Since they are normalized, the integration of probability density of atomic orbitals in eqns. 1 and 2 should be equal to 1 for each.

When you integrate the probability density of the total wave function shown in the last equation, you don't need to consider the complex form. In addition, the first term can be integrated within $[-d-a,-d+a]$ to $c_1^2\int|\phi_-|^2 \,\mathrm{d}x = c_1^2 = 1/5$, the second term can be integrated within $[d-a,d+a]$ to $c_2^2\int|\phi_+|^2 \,\mathrm{d}x = c_2^2 = 4/5$, and the third term is integrated to zero due to the absence of overlap. Note that for simplicity, the open intervals $(-d-a,-d+a)$ and $(d-a,d+a)$ are changed to closed intervals $[-d-a,-d+a]$ and $[d-a,d+a]$, as the integration in open and closed intervals should lead to the same result (see Integrating on open vs. closed intervals on Mathematics.SE). This gives $c_1=1/\sqrt5$ and $c_2=2/\sqrt5$, which in turn means $\phi=(1/\sqrt5)\phi_- + (2/\sqrt5)\phi_+$.

The solution indicates that the total wave function has a constructive combination of the two $\phi_-$ and $\phi_+$ orbitals. This type of solution can be seen in the ground-state broken-symmetry solution of $\ce{H2}$ due to non-dynamic electron correlation, as the two H atoms are stretched to a bond length longer than the Coulson-Fischer point, where the two energy curves obtained from restricted and unrestricted (symmetric and broken-symmetry) wave functions start to bifurcate from each other. In addition, you can imagine that a node may appear in the lowest excited state, which means that the total wave function for that state has a destructive combination of the two atomic orbitals.

Of course, this problem is a simplified version of the practical problem because in reality there is an overlap between the two atomic orbitals unless the interatomic distance is stretched to very long where the overlap asymptotically approaches zero. However, I don't think the problem is aimed to teach about electron correlation or overlap but is used to familiarize students with LCAO-MO.


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