Consider ground state of a hydrogen atom which influenced with external uniform weak electric field. What does mean the following statement? $$\left< 100|\quad Z\quad |100 \right> =\int { |{ \psi }_{ 100 }\quad(r)|^{ 2 } } Z\quad { d }^{ 3 }r$$

is zero, since $Z$ is odd under parity and ${ \psi }_{ 100 }\quad(r)$ has a definite parity. This means that there can be no correction term to the energy which is proportional to the electric field and hence there is no linear Stark effect.

Is it possible to explain the parity(definite parity) and also the reason for no correction explained in last latter sentence.


You haven't really explained what $Z$ is, so let me know in the comments if my definitions are OK with your problem.

For the ground state of the hydrogen atom, represented by $|nlm\rangle=|100\rangle$ (with a wavefunction $\psi_{100}$), we try to work out the first-order shift in the energy caused by the perturbation of an electric field along the $z$-axis. This field gives a perturbing potential $Z=e\Phi \hat{z}$ (where $e$ is the charge of the electron, $\Phi$ is the value of the field in the $z$ direction, and $\hat{z}$ is the position operator).

The first-order shift in the ground state energy eigenvalue is given by $$\Delta E_0=\langle 100|Z|100\rangle=\langle100|e\Phi\hat{z}|100\rangle=\int\psi_{100}^*(r)e\Phi\hat{z}\psi_{100}(r)dr^3=e\Phi\int z(r)|\psi_{100}(r)|^2dr^3$$ The last equality above is simply due to: $\hat{z}\psi=z\psi$.

Now, when we say that $\psi_{100}$ has a definite parity, what we mean is that it is either an even function or an odd function. A function with indefinite parity is one that is neither even nor odd. But whatever the parity of $\psi_{100}$ actually is, $|\psi_{100}|^2$ will be even (even$\times$even$=$even, odd$\times$odd$=$even). And of course, $z(r)$ is odd. Overall, the integrand is an odd function (even$\times$odd$=$odd). An an odd function integrated over the whole space is $0$.

Therefore $\Delta E_0=0$, there is no first-order shift in the ground state of hydrogen, i.e. there is no linear Stark effect.

(You could also try to show, by the same sort of reasoning, that there is no linear Stark effect for any atom in a non-degenerate energy eigenstate).

  • $\begingroup$ What you defined for $Z$ is exactly what I was looking for. Thank you for explanation. $\endgroup$
    – Sina
    Jan 18 '15 at 13:06

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