Given that a system's Hamiltonian is of the form H=(p^2)/(2 me)+V(𝐱) show that [x,[H, x]]=ħ^2 /

Given that a system's Hamiltonian is of the form H=(p^2)/(2...

Given that a system's Hamiltonian is of the form $$ H=\frac{p^{2}}{2 m_{\mathrm{e}}}+V(\mathbf{x}) $$ show that $[x,[H, x]]=\hbar^{2} / m_{\mathrm{e}}$. By taking the expectation value of this expression in the state $|k\rangle$, show that $$ \sum_{n \neq k}|\langle n|x| k\rangle|^{2}\left(E_{n}-E_{k}\right)=\frac{\hbar^{2}}{2 m_{\mathrm{e}}} $$ where the sum runs over all the other stationary states. The oscillator strength of a radiative transition $|k\rangle \rightarrow|n\rangle$ is defined to be $$ f_{k n} \equiv \frac{2 m_{\mathrm{e}}}{\hbar^{2}}\left(E_{n}-E_{k}\right)|\langle n|x| k\rangle|^{2} $$ Show that $\sum_{n} f_{k n}=1$. What is the significance of oscillator strengths for the allowed radiative transition rates of atoms?

Given that a system's Hamiltonian is of the form
$$
H=\frac{p^{2}}{2 m_{\mathrm{e}}}+V(\mathbf{x})
$$
show that $[x,[H, x]]=\hbar^{2} / m_{\mathrm{e}}$. By taking the expectation value of this expression in the state $|k\rangle$, show that
$$
\sum_{n \neq k}|\langle n|x| k\rangle|^{2}\left(E_{n}-E_{k}\right)=\frac{\hbar^{2}}{2 m_{\mathrm{e}}}
$$
where the sum runs over all the other stationary states.
The oscillator strength of a radiative transition $|k\rangle \rightarrow|n\rangle$ is defined to be
$$
f_{k n} \equiv \frac{2 m_{\mathrm{e}}}{\hbar^{2}}\left(E_{n}-E_{k}\right)|\langle n|x| k\rangle|^{2}
$$
Show that $\sum_{n} f_{k n}=1$. What is the significance of oscillator strengths for the allowed radiative transition rates of atoms?

Key Concepts

Commutators in Quantum Mechanics

Commutators quantify the non-commutativity of operators, which is a key feature in quantum mechanics reflecting the uncertainty principle. The commutator of two operators A and B, defined as [A, B] = AB - BA, provides insight into the underlying algebraic structure of quantum observables and helps in evaluating dynamical equations and transition matrix elements between states.

Oscillator Strength

The oscillator strength is a dimensionless quantity that measures the probability of a radiative transition between quantum states. It combines the energy difference between states and the square of the transition dipole moment. The fact that the sum of oscillator strengths equals one reflects a conservation of total transition probability, ensuring that the overall strength of an electron’s interaction with electromagnetic radiation is distributed among all possible transitions. This concept is fundamental in determining the allowed transition rates and absorption/emission spectra in atomic physics.

Sum Rules in Quantum Mechanics

Sum rules, like the Thomas-Reiche-Kuhn (TRK) sum rule, provide essential constraints on the properties of quantum systems by linking the total strength of allowed transitions to fundamental constants. They emerge from the basic commutation relations and conservation laws, ensuring that the integrated transition probability over all states conforms to a fixed value, thereby serving as a check on theoretical predictions and experimental measurements.

Hamiltonian

In quantum mechanics, the Hamiltonian operator represents the total energy of the system. It includes both kinetic and potential energy terms, and its structure not only dictates the energy spectrum of the system but also its time evolution through the Schrödinger equation. Understanding the Hamiltonian is crucial as it forms the basis for calculating expectation values, commutators, and transition amplitudes between states.

Double Commutator

The double commutator, such as [x,[H,x]], is a useful tool in quantum mechanics for deriving sum rules and other properties related to operator expectation values. It encapsulates higher order commutation relations and can be used to extract fundamental constants or constraints, such as establishing relations between energy differences and transition matrix elements.

Spectral Decomposition and Expectation Values

Spectral decomposition involves expressing operators in the basis of their eigenstates, allowing any state of the system to be represented in terms of these eigenstates. Taking expectation values in specific states and inserting resolutions of the identity leads to sum-over-states expressions which are instrumental in connecting microscopic operator dynamics with observable quantities.

Transcript

00:01 Suppose that we have a three state hamiltonian whose matrix elements relative to certain bases are given over here.

00:10 That is h is equal to 1 -0 -0 -0 -2 -0 -0 -0 -0 -0 -0 -2.

00:23 We have also two other observables, a and b, represented by the following matrices over a is equal to lambda 0100 sorry 1 00 002 and b is equal to mu 2000 2 001 010 fine so where lambda omega omega and mu are positive real numbers so number part a is that we have to solve for the again values of h so that is e1 is equal to a w e2 is equal to e3 is equal to 2h w and the again vectors would be h1 is equal to one zero zero eight two is equal to zero one zero eight three is equal to zero zero zero one now for a the again values can be determined as a negative a times i is equal to zero so this means that when we'll plug in we'll get negative a, lambda 0, lambda, negative a 0, 0, 0 to lambda, negative, a negative, a, that is equal to a square to lambda, negative, a, negative to lambda, negative, a, lambda square, that is equal to 0.

03:10 Written as 2 lambda negative a times a square negative lambda square is equal to 0.

03:20 So this can be written as 2 lambda negative a.

03:25 A negative lambda, a positive lambda is equal to 0.

03:31 And from here we'll get all of these terms in the parenthesis.

03:34 So a 1 is equal to 2 lambda, a2 is equal to lambda and a 3 is equal to negative limit.

03:43 So the next step is that the eigenvectors can be found as is equal to this.

04:07 So know that this is the agon vector.

04:23 So we have lambda 0101000002 times alpha beta gamma is equal to alpha alpha beta gamma fine so from here it yields at lambda beta beta lambda lambda is equal to a times alpha lambda alpha is equal to alpha beta and two lambda gamma gamma is equal to alpha gamma now solve these equations are using a1, a2 and a3.

05:14 First we substitute a1 that is equal to 2 lambda and then solve for alpha, beta and gamma in order to write the eigen vector as lambda beta is equal to 2 lambda alpha and beta is equal to 2 alpha from here.

05:40 Now, then we have lambda alpha is equal to 2 lambda beta.

05:47 This is that alpha is equal to 2 beta.

05:51 And over here we have 2 lambda gamma.

05:55 That is equal to 2 lambda gamma.

05:58 And from here we can clue that alpha is equal to beta is equal to 0.

06:05 So it means that the a -again vector is a 1 a1 is equal to 0 -0 -1 now for a 2 is equal to lambda we get that lambda beta is equal to lambda alpha is that beta is equal to alpha and lambda alpha is equal to lambda alpha is equal to lambda beta that implies that alpha is equal to beta and lastly we have to lambda gamma that is equal to lambda gamma that implies that lambda is equal to 0 so this gives us the agon vector for a 2 is equal to 1 over under root 2 1 110 so fine so the next the third step is that we have to solve for a 3 is equal to negative lambda so for that we get that lambda beta is equal to negative lambda alpha that implies that beta is equal to negative alpha then lambda alpha is equal to negative limda beta that is equal to negative beta then we have two lambda gamma is that is equal to negative lambda gamma and this implies that gamma is equal to zero and again vector a 3 is equal to one over under root to 1 negative 0...

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