Perturbation Analysis of Harmonic Oscillator
We are considering a one-dimensional harmonic oscillator with a perturbation $V = \lambda x^3$. We need to find the states $|m\rangle$ for which the matrix element $\langle m|V|0 \rangle$ is non-zero, starting from the ground state $|0\rangle$.
Operator Representation
The position operator $x$ for a harmonic oscillator can be expressed using the creation ($a^\dagger$) and annihilation ($a$) operators:
$x = \sqrt{\frac{\hbar}{2m\omega}} (a + a^\dagger)$
where $m$ is the mass, $\omega$ is the angular frequency, and $\hbar$ is the reduced Planck constant.
Perturbation Operator $V$
Substituting the expression for $x$ into the perturbation $V$:
$V = \lambda x^3 = \lambda \left(\sqrt{\frac{\hbar}{2m\omega}} (a + a^\dagger)\right)^3$
Expanding $(a + a^\dagger)^3$ gives:
$(a + a^\dagger)^3 = a^3 + 3a^2a^\dagger + 3aa^\dagger a^\dagger + (a^\dagger)^3$
So, the perturbation $V$ contains terms involving $a^3$, $a^2a^\dagger$, $aa^\dagger a^\dagger$, and $(a^\dagger)^3$.
Matrix Element Calculation $\langle m|V|0 \rangle$
The matrix element $\langle m|V|0 \rangle$ is non-zero only if the operator $V$ can connect the state $|0\rangle$ to the state $|m\rangle$. We use the properties of the creation and annihilation operators:
- $\langle n'|a|n\rangle \propto \delta_{n', n-1}$
- $\langle n'|a^\dagger|n\rangle \propto \delta_{n', n+1}$
We examine the effect of each term in $V$ on the ground state $|0\rangle$:
- $a^3|0\rangle$: Applying $a$ three times to $|0\rangle$ results in $|-3\rangle$, which is zero. Thus, $\langle m|a^3|0 \rangle = 0$.
- $a^2a^\dagger|0\rangle$: This leads to state $|1\rangle$ after applying $a^\dagger$ once, but the $a^2$ operator acting on $|1\rangle$ results in zero. The net effect is zero. Alternatively, the number of creation operators minus the number of annihilation operators must match the state difference: $1-2 = -1$. So, $\langle m|a^2a^\dagger|0 \rangle = 0$.
- $aa^\dagger a^\dagger|0\rangle$: Applying $a^\dagger$ twice results in state $|2\rangle$. Applying $a$ once yields state $|1\rangle$. The number of creation operators minus the number of annihilation operators is $2-1 = 1$. Thus, $\langle 1|aa^\dagger a^\dagger|0 \rangle \neq 0$.
- $(a^\dagger)^3|0\rangle$: Applying $a^\dagger$ three times to $|0\rangle$ results in state $|3\rangle$. The number of creation operators minus the number of annihilation operators is $3-0 = 3$. Thus, $\langle 3|(a^\dagger)^3|0 \rangle \neq 0$.
The terms $aa^\dagger a^\dagger$ and $(a^\dagger)^3$ can transition the state $|0\rangle$ to $|1\rangle$ and $|3\rangle$ respectively.
Conclusion on Non-Zero Matrix Elements
The matrix element $\langle m|V|0 \rangle$ is non-zero when $m=1$ (due to the $aa^\dagger a^\dagger$ term) and when $m=3$ (due to the $(a^\dagger)^3$ term).
Therefore, the matrix element $\langle m|V|0 \rangle$ is non-zero for $|m = 1\rangle$ and $|m = 3\rangle$.
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 ________.
