Perturbation Theory: First-Order Energy Correction
The first-order correction to the energy of the ground state ($|0^{(0)}\rangle$) of a quantum system is given by the expectation value of the perturbation Hamiltonian ($H_1$) in the unperturbed ground state:
$ \Delta E_0^{(1)} = \langle 0^{(0)} | H_1 | 0^{(0)} \rangle $
In this case, the unperturbed Hamiltonian is that of a Simple Harmonic Oscillator (SHO), $H_0 = \frac{p^2}{2m} + \frac{1}{2}k x^2$. The perturbation is $H_1 = \alpha x + \beta x^3 + \gamma x^4$.
Analyzing Perturbation Terms
The ground state wave function of the SHO, $\psi_0(x)$, is known to be an even function of position ($x$). We analyze the contribution of each term in $H_1$ to the first-order correction:
- Term 1: $\alpha x$
The operator $x$ is odd under parity transformation ($x \rightarrow -x$). The ground state wave function $\psi_0(x)$ is even. The product $x \psi_0(x)$ is odd. The expectation value of an odd function over symmetric limits (from $-\infty$ to $+\infty$) is zero.
$ \langle 0^{(0)} | \alpha x | 0^{(0)} \rangle = \alpha \int_{-\infty}^{\infty} \psi_0^*(x) x \psi_0(x) \,dx = 0 $ - Term 2: $\beta x^3$
The operator $x^3$ is also odd under parity transformation. Similar to the $x$ term, the expectation value is zero.
$ \langle 0^{(0)} | \beta x^3 | 0^{(0)} \rangle = \beta \int_{-\infty}^{\infty} \psi_0^*(x) x^3 \psi_0(x) \,dx = 0 $ - Term 3: $\gamma x^4$
The operator $x^4$ is even under parity transformation ($x^4 \rightarrow (-x)^4 = x^4$). The product $x^4 \psi_0(x)$ is even. The expectation value of an even function is generally non-zero.
$ \langle 0^{(0)} | \gamma x^4 | 0^{(0)} \rangle = \gamma \int_{-\infty}^{\infty} \psi_0^*(x) x^4 \psi_0(x) \,dx \neq 0 $
Correction Dependence
Combining the results, the total first-order energy correction is:
$ \Delta E_0^{(1)} = \langle 0^{(0)} | \alpha x | 0^{(0)} \rangle + \langle 0^{(0)} | \beta x^3 | 0^{(0)} \rangle + \langle 0^{(0)} | \gamma x^4 | 0^{(0)} \rangle $
$ \Delta E_0^{(1)} = 0 + 0 + \gamma \langle 0^{(0)} | x^4 | 0^{(0)} \rangle $
Therefore, the first-order correction to the ground state energy depends only on the coefficient $\gamma$.
Important Questions from Perturbation Theory Time Independent Degenerate
- If the perturbation $V = \lambda x^3$ is added to the Hamiltonian of a one-dimensional harmonic oscillator, the matrix element $\langle m|V|0 \rangle$ is/are non-zero for which of the following states? Here, the eigenstates of the harmonic oscillator are denoted by $|n\rangle$.
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
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).The ground state energy of a particle of mass $m$ in an infinite potential well is $E_0$. It changes to $E_0(1 + \alpha \times 10^{-3})$, when there is a small potential bump of height $V_0 = \frac{\pi^2 \hbar^2}{50mL^2}$ and width $a = L/100$, as shown in the figure. The value of $\alpha$ is ________ (up to two decimal places).
- 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?
