Hydrogen Atom Energy Shift in Electric Field
The problem asks for the leading order change in the energy of a hydrogen atom in its ground state when placed in a uniform electric field $\vec{E} = E_0\hat{z}$. This phenomenon is known as the Stark effect.
First-Order Perturbation Analysis
The interaction Hamiltonian due to the electric field is $H' = e \vec{E} \cdot \vec{r}$. For $\vec{E} = E_0\hat{z}$, this becomes $H' = e E_0 z$. According to perturbation theory, the first-order energy correction is:
$ \Delta E^{(1)} = \langle \psi_{1s} | H' | \psi_{1s} \rangle = \langle \psi_{1s} | e E_0 z | \psi_{1s} \rangle $
Here, $\psi_{1s}$ is the ground state wave function of the hydrogen atom. The ground state is spherically symmetric ($l=0$). Due to this spherical symmetry, the expectation value of the position operator $z$ is zero:
$ \langle z \rangle = \int \psi_{1s}^* z \psi_{1s} \, dV = 0 $
Therefore, the first-order energy correction is zero:
$ \Delta E^{(1)} = e E_0 \langle z \rangle = 0 $
Second-Order Perturbation Analysis
Since the first-order correction is zero, the leading order energy change arises from the second-order perturbation correction:
$ \Delta E^{(2)} = \sum_{k \neq 1s} \frac{|\langle \psi_k | H' | \psi_{1s} \rangle|^2}{E_{1s} - E_k} $
The matrix element $\langle \psi_k | e E_0 z | \psi_{1s} \rangle$ involves the operator $z$. For electric dipole transitions, parity must change. The ground state ($1s$) has even parity ($l=0$), so transitions occur to states with odd parity, such as $p$ states ($l=1$).
The numerator involves the square of the matrix element, which includes $(e E_0)^2$. The denominator ($E_{1s} - E_k$) is non-zero and finite for excited states $k$. Thus, the second-order energy shift is proportional to the square of the electric field strength:
$ \Delta E^{(2)} \propto (E_0)^2 $
Conclusion
The leading order change in the energy of the hydrogen atom's ground state in a uniform electric field is proportional to $(E_0)^2$. Therefore, the value of the exponent $n$ is 2.
Important Questions from Perturbation Theory Time Independent Degenerate
A particle of mass $m$ in an infinite potential well of width $a$ is subjected to a perturbation, $V' = \frac{h^2}{40ma^2}$ as shown in figure, where $h$ is Planck's constant.
The first order energy shift of the fourth energy eigenstate due to this perturbation is
$(\frac{h^2}{Nma^2})$
The value of $N$ is ____________ (in integer).- A particle of mass $m$ in the x-y plane is confined in an infinite two-dimensional well with vertices at $(0, 0)$, $(0, L)$, $(L, L)$, $(L, 0)$. The eigenfunctions of this particle are $\Psi_{n_x,n_y} = \sin(\frac{n_x\pi x}{L}) \sin(\frac{n_y\pi y}{L})$. If perturbation of the form $V = Cxy$, where $C$ is a real constant, is applied, then which of the following statements are correct for the first excited state?
A two-level quantum system has energy eigenvalues $E_1$ and $E_2$. A perturbing potential $H' = \lambda \Delta \sigma_x$ is introduced, where $\Delta$ is a constant having dimensions of energy, $\lambda$ is a small dimensionless parameter, and $\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. The magnitudes of the first and the second order corrections to $E_1$ due to $H'$, respectively, are
Consider a particle in a one-dimensional infinite potential well with its walls at $x = 0$ and $x = L$. The system is perturbed as shown in the figure
The first order correction to the energy eigenvalue is
Consider the Hamiltonian $\hat{H} = \hat{H}_0 + \hat{H}'$ where
\[\hat{H}_0 = \begin{pmatrix} E & 0 & 0 \\ 0 & E & 0 \\ 0 & 0 & E \end{pmatrix}\]and $\hat{H}$ is the time independent perturbation given by
\[\hat{H}' = \begin{pmatrix} 0 & k & 0 \\ k & 0 & k \\ 0 & k & 0 \end{pmatrix}\]where $k>0$. If, the maximum energy eigenvalue of $\hat{H}$ is 3 eV corresponding to $E=2$ eV, the value of $k$ (rounded off to three decimal places) in eV is ________.
