Daily Archives: February 7, 2022

Formula that describes the transition rate from one energy eigenstate of a quantum system into other energy eigenstates

In quantum physics, Fermi's golden rule is a formula that describes the transition rate (the probability of a transition per unit time) from one energy eigenstate of a quantum system to a group of energy eigenstates in a continuum, as a result of a weak perturbation. This transition rate is effectively independent of time (so long as the strength of the perturbation is independent of time) and is proportional to the strength of the coupling between the initial and final states of the system (described by the square of the matrix element of the perturbation) as well as the density of states. It is also applicable when the final state is discrete, i.e. it is not part of a continuum, if there is some decoherence in the process, like relaxation or collision of the atoms, or like noise in the perturbation, in which case the density of states is replaced by the reciprocal of the decoherence bandwidth.

Although named after Enrico Fermi, most of the work leading to the "golden rule" is due to Paul Dirac, who formulated 20 years earlier a virtually identical equation, including the three components of a constant, the matrix element of the perturbation and an energy difference.[1][2] It was given this name because, on account of its importance, Fermi called it "golden rule No.2".[3]

Most uses of the term Fermi's golden rule are referring to "golden rule No.2", however, Fermi's "golden rule No.1" is of a similar form and considers the probability of indirect transitions per unit time.[4]

Fermi's golden rule describes a system that begins in an eigenstate | i {displaystyle |irangle } of an unperturbed Hamiltonian H0 and considers the effect of a perturbing Hamiltonian H' applied to the system. If H' is time-independent, the system goes only into those states in the continuum that have the same energy as the initial state. If H' is oscillating sinusoidally as a function of time (i.e. it is a harmonic perturbation) with an angular frequency , the transition is into states with energies that differ by from the energy of the initial state.

In both cases, the transition probability per unit of time from the initial state | i {displaystyle |irangle } to a set of final states | f {displaystyle |frangle } is essentially constant. It is given, to first-order approximation, by

where f | H | i {displaystyle langle f|H'|irangle } is the matrix element (in braket notation) of the perturbation H' between the final and initial states, and ( E f ) {displaystyle rho (E_{f})} is the density of states (number of continuum states divided by d E {displaystyle dE} in the infinitesimally small energy interval E {displaystyle E} to E + d E {displaystyle E+dE} ) at the energy E f {displaystyle E_{f}} of the final states. This transition probability is also called "decay probability" and is related to the inverse of the mean lifetime. Thus, the probability of finding the system in state | i {displaystyle |irangle } is proportional to e i f t {displaystyle e^{-Gamma _{ito f}t}} .

The standard way to derive the equation is to start with time-dependent perturbation theory and to take the limit for absorption under the assumption that the time of the measurement is much larger than the time needed for the transition.[5][6]

The golden rule is a straightforward consequence of the Schrdinger equation, solved to lowest order in the perturbation H' of the Hamiltonian. The total Hamiltonian is the sum of an original Hamiltonian H0 and a perturbation: H = H 0 + H ( t ) {displaystyle H=H_{0}+H'(t)} . In the interaction picture, we can expand an arbitrary quantum states time evolution in terms of energy eigenstates of the unperturbed system | n {displaystyle |nrangle } , with H 0 | n = E n | n {displaystyle H_{0}|nrangle =E_{n}|nrangle } .

We first consider the case where the final states are discrete. The expansion of a state in the perturbed system at a time t is | ( t ) = n a n ( t ) e i E n t / | n {displaystyle |psi (t)rangle =sum _{n}a_{n}(t)e^{-iE_{n}t/hbar }|nrangle } . The coefficients an(t) are yet unknown functions of time yielding the probability amplitudes in the Dirac picture. This state obeys the time-dependent Schrdinger equation:

Expanding the Hamiltonian and the state, we see that, to first order,

( H 0 + H i t ) n a n ( t ) | n e i t E n / = 0 , {displaystyle left(H_{0}+H'-mathrm {i} hbar {frac {partial }{partial t}}right)sum _{n}a_{n}(t)|nrangle e^{-mathrm {i} tE_{n}/hbar }=0,} where En and |n are the stationary eigenvalues and eigenfunctions of H0.

This equation can be rewritten as a system of differential equations specifying the time evolution of the coefficients a n ( t ) {displaystyle a_{n}(t)} :

This equation is exact, but normally cannot be solved in practice.

For a weak constant perturbation H' that turns on at t = 0, we can use perturbation theory. Namely, if H = 0 {displaystyle H'=0} , it is evident that a n ( t ) = n , i {displaystyle a_{n}(t)=delta _{n,i}} , which simply says that the system stays in the initial state i {displaystyle i} .

For states k i {displaystyle kneq i} , a k ( t ) {displaystyle a_{k}(t)} becomes non-zero due to H 0 {displaystyle H'neq 0} , and these are assumed to be small due to the weak perturbation. The coefficient a i ( t ) {displaystyle a_{i}(t)} which is unity in the unperturbed state, will have a weak contribution from H {displaystyle H'} . Hence, one can plug in the zeroth-order form a n ( t ) = n , i {displaystyle a_{n}(t)=delta _{n,i}} into the above equation to get the first correction for the amplitudes a k ( t ) {displaystyle a_{k}(t)} :

whose integral can be expressed as

with k i ( E k E i ) / {displaystyle omega _{ki}equiv (E_{k}-E_{i})/hbar } , for a state with ai(0) = 1, ak(0) = 0, transitioning to a state with ak(t).

The probability of transition from the initial state (ith) to the final state (fth) is given by

It is important study a periodic perturbation with a given frequency {displaystyle omega } since arbitrary perturbations can be constructed from periodic perturbations of different frequencies. Since H ( t ) {displaystyle H'(t)} must be Hermitian, we must assume H ( t ) = F e i t + F e i t {displaystyle H'(t)=Fe^{-mathrm {i} omega t}+F^{dagger }e^{mathrm {i} omega t}} , where F {displaystyle F} is a time independent operator. The solution for this case is[7]

This expression is valid only when the denominators in the above expression is non-zero, i.e., for a given initial state with energy E i {displaystyle E_{i}} , the final state energy must be such that E f E i . {displaystyle E_{f}-E_{i}neq pm hbar omega .} Not only the denominators must be non-zero, but also must not be small since a f {displaystyle a_{f}} is supposed to be small.

Since the continuous spectrum lies above the discrete spectrum, E f E i > 0 {displaystyle E_{f}-E_{i}>0} and it is clear from the previous section, major role is played by the energies E f {displaystyle E_{f}} lying near the resonance energy E i + {displaystyle E_{i}+hbar omega } , i.e., when f i {displaystyle omega _{fi}approx omega } . In this case, it is sufficient to keep only the first term for a f ( t ) {displaystyle a_{f}(t)} . Assuming that perturbations are turned on from time t = 0 {displaystyle t=0} , we have then

The squared modulus of a f {displaystyle a_{f}} is

For large t {displaystyle t} , this will reduce to

a linear dependence on t {displaystyle t} .

The probability of transition from the ith state to final states lying in an interval d f {displaystyle dnu _{f}} (density of states in an infinitesimal element around E f {displaystyle E_{f}} ) is d w f i = | a f | 2 d f {displaystyle dw_{fi}=|a_{f}|^{2}dnu _{f}} . The transition probability per unit time is thus given by

The time dependence has vanished, and the constant decay rate of the golden rule follows.[8] As a constant, it underlies the exponential particle decay laws of radioactivity. (For excessively long times, however, the secular growth of the ak(t) terms invalidates lowest-order perturbation theory, which requires ak ai.)

Only the magnitude of the matrix element f | H | i {displaystyle langle f|H'|irangle } enters the Fermi's golden rule. The phase of this matrix element, however, contains separate information about the transition process.It appears in expressions that complement the golden rule in the semiclassical Boltzmann equation approach to electron transport.[9]

While the Golden rule is commonly stated and derived in the terms above, the final state (continuum) wave function is often rather vaguely described, and not normalized correctly (and the normalisation is used in the derivation). The problem is that in order to produce a continuum there can be no spatial confinement (which would necessarily discretise the spectrum), and therefore the continuum wave functions must have infinite extent, and in turn this means that the normalisation f | f = d 3 r | f ( r ) | 2 {displaystyle langle f|frangle =int d^{3}r|f(mathbf {r} )|^{2}} is infinite, not unity. If the interactions depend on the energy of the continuum state, but not any other quantum numbers, it is usual to normalise continuum wave-functions with energy {displaystyle epsilon } labelled | {displaystyle |epsilon rangle } , by writing | = ( ) {displaystyle langle epsilon |epsilon 'rangle =delta (epsilon -epsilon ')} where {displaystyle delta } is the Dirac delta function, and effectively a factor of the square-root of the density of states is included into | i {displaystyle |epsilon _{i}rangle } .[10] In this case, the continuum wave function has dimensions of 1 / {displaystyle 1/surd } [energy], and the Golden Rule is now

where i {displaystyle epsilon _{i}} refers to the continuum state with the same energy as the discrete state i {displaystyle i} . For example, correctly normalized continuum wave functions for the case of a free electron in the vicinity of a hydrogen atom are available in Bethe and Salpeter .[11]

The following paraphrases the treatment of Cohen-Tannoudji.[10] As before, the total Hamiltonian is the sum of an original Hamiltonian H0 and a perturbation: H = H 0 + H {displaystyle H=H_{0}+H'} . We can still expand an arbitrary quantum states time evolution in terms of energy eigenstates of the unperturbed system, but these now consist of discrete states and continuum states. We assume that the interactions depend on the energy of the continuum state, but not any other quantum numbers. The expansion in the relevant states in the Dirac picture is

where i = i / , = / {displaystyle omega _{i}=epsilon _{i}/hbar ,omega =epsilon /hbar } and i , {displaystyle epsilon _{i},epsilon } are the energies of states | i , | {displaystyle |irangle ,|epsilon rangle } . The integral is over the continuum C {displaystyle epsilon in C} , i.e. | {displaystyle |epsilon rangle } is in the continuum.

Substituting into the time-dependent Schrdinger equation

and premultiplying by i | {displaystyle langle i|} produces

where i = i | H | / {displaystyle Omega _{iepsilon }=langle i|H'|epsilon rangle /hbar } , and premultiplying by | {displaystyle langle epsilon '|} produces

We made use of the normalisation | = ( ) {displaystyle langle epsilon '|epsilon rangle =delta (epsilon '-epsilon )} .Integrating the latter and substituting into the former,

It can be seen here that d a i / d t {displaystyle da_{i}/dt} at time t {displaystyle t} depends on a i {displaystyle a_{i}} at all earlier times t {displaystyle t'} , i.e. it is non-Markovian. We make the Markov approximation, i.e. that it only depends on a i {displaystyle a_{i}} at time t {displaystyle t} (which is less restrictive than the approximation that a i {displaystyle a_{i}} 1 used above, and allows the perturbation to be strong)

where T = t t {displaystyle T=t-t'} and = i {displaystyle Delta =omega -omega _{i}} . Integrating over T {displaystyle T} ,

The fraction on the right is a nascent Dirac delta function, meaning it tends to ( i ) {displaystyle delta (epsilon -epsilon _{i})} as t {displaystyle tto infty } (ignoring its imaginary part which leads to an unimportant energy shift, while the real part produces decay [10]). Finally

which has solutions: a i ( t ) = exp ( i i t / 2 ) {displaystyle a_{i}(t)=exp(-Gamma _{ito epsilon _{i}}t/2)} , i.e., the decay of population in the initial discrete state is P i ( t ) = | a i ( t ) | 2 = exp ( i i t ) {displaystyle P_{i}(t)=|a_{i}(t)|^{2}=exp(-Gamma _{ito epsilon _{i}}t)} where

The Fermi golden rule can be used for calculating the transition probability rate for an electron that is excited by a photon from the valence band to the conduction band in a direct band-gap semiconductor, and also for when the electron recombines with the hole and emits a photon.[12] Consider a photon of frequency {displaystyle omega } and wavevector q {displaystyle {textbf {q}}} , where the light dispersion relation is = ( c / n ) | q | {displaystyle omega =(c/n)left|{textbf {q}}right|} and n {displaystyle n} is the index of refraction.

Using the Coulomb gauge where A = 0 {displaystyle nabla cdot {textbf {A}}=0} and V = 0 {displaystyle V=0} , the vector potential of the EM wave is given by A = A 0 e i ( q r t ) + C {displaystyle {textbf {A}}=A_{0}{vec {epsilon }}e^{i({textbf {q}}cdot {textbf {r}}-omega t)}+C} where the resulting electric field is

For a charged particle in the valence band, the Hamiltonian is

where V ( r ) {displaystyle V({textbf {r}})} is the potential of the crystal. If our particle is an electron ( Q = e {displaystyle Q=-e} ) and we consider process involving one photon and first order in A {displaystyle {textbf {A}}} . The resulting Hamiltonian is

where H {displaystyle H'} is the perturbation of the EM wave.

From here on we have transition probability based on time-dependent perturbation theory that

where {displaystyle {vec {epsilon }}} is the light polarization vector. From perturbation it is evident that the heart of the calculation lies in the matrix elements shown in the braket.

For the initial and final states in valence and conduction bands respectively, we have | i = v , k i , s i ( r ) {displaystyle |irangle =Psi _{v,{textbf {k}}_{i},s_{i}}({textbf {r}})} and | f = c , k f , s f ( r ) {displaystyle |frangle =Psi _{c,{textbf {k}}_{f},s_{f}}({textbf {r}})} , and if the H {displaystyle H'} operator does not act on the spin, the electron stays in the same spin state and hence we can write the wavefunctions as Bloch waves so

where N {displaystyle N} is the number of unit cells with volume 0 {displaystyle Omega _{0}} . Using these wavefunctions and with some more mathematics, and focusing on emission (photoluminescence) rather than absorption, we are led to the transition rate

where c v {displaystyle {boldsymbol {mu }}_{cv}} is the transition dipole moment matrix element is qualitatively the expectation value c | ( charge ) ( distance ) | v {displaystyle langle c|({text{charge}})times ({text{distance}})|vrangle } and in this situation takes the form

Finally, we want to know the total transition rate ( ) {displaystyle Gamma (omega )} . Hence we need to sum over all initial and final states (i.e. an integral of the Brillouin zone in the k-space), and take into account spin degeneracy, which through some mathematics results in

( ) = 4 ( e A 0 m 0 ) 2 | c v | 2 c v ( ) {displaystyle Gamma (omega )={frac {4pi }{hbar }}left({frac {eA_{0}}{m_{0}}}right)^{2}|{vec {epsilon }}cdot {boldsymbol {mu }}_{cv}|^{2}rho _{cv}(omega )}

where c v ( ) {displaystyle rho _{cv}(omega )} is the joint valence-conduction density of states (i.e. the density of pair of states; one occupied valence state, one empty conduction state). In 3D, this is

but the joint DOS is different for 2D, 1D, and 0D.

Finally we note that in a general way we can express the Fermi golden rule for semiconductors as[13]

In a scanning tunneling microscope, the Fermi golden rule is used in deriving the tunneling current. It takes the form

where M {displaystyle M} is the tunneling matrix element.

When considering energy level transitions between two discrete states, Fermi's golden rule is written as

where g ( ) {displaystyle g(hbar omega )} is the density of photon states at a given energy, {displaystyle hbar omega } is the photon energy, and {displaystyle omega } is the angular frequency. This alternative expression relies on the fact that there is a continuum of final (photon) states, i.e. the range of allowed photon energies is continuous.[14]

Fermi's golden rule predicts that the probability that an excited state will decay depends on the density of states. This can be seen experimentally by measuring the decay rate of a dipole near a mirror: as the presence of the mirror creates regions of higher and lower density of states, the measured decay rate depends on the distance between the mirror and the dipole.[15][16]

See more here:

Fermi's golden rule - Wikipedia

This weeks Vicksburg Post Volunteer of the Week volunteers with the Purks-Golding YMCAs Rock Steady Program, Joel Horton. Rock Steady is an exercise program that helps those fighting Parkinsons disease improve their mobility, balance, strength and quality of life.

Horton was born in Grenada, Miss. and has lived in Vicksburg since 1983. He married Leslie Bell and raised two children. Horton also retired as a banker.

What was the inspiration to start the Rock Steady Program?

My wife, Leslie, was diagnosed with Parkinsons disease in 2012. I heard about Rock Steady and visited a class in Nashville with Leslie. I talked to some of the participants and heard how the program had helped them. I came back to Vicksburg and started to campaign to start a program at the YMCA.

What is your favorite memory while volunteering with Rock Steady?

Seeing some of the boxers improving and the smile on their faces while they are exercising.

What would you tell someone who is thinking about volunteering?

For me, its trying to put Jesuss Golden Rule into practice: Whatever you want others to do for you, do for them. I believe you receive a blessing from volunteering.

What kind of activities happen in the Rock Steady Program?

There are classes on Tuesday and Thursday. The program focuses on stretching, increasing strength and improving balance and coordination. There are some great instructors who keep the classes fresh with a lot of variety and fun activities.

What have you learned from volunteering with this organization?

The participants are such an inspiration because they have movement problems. Some come to class in wheelchairs but always with great attitudes. In addition to improving their physical condition, the camaraderie between the participants is obvious.

How has this changed you?

The participants have inspired me to be more grateful for my physical well-being.

Any additional comments?

Vicksburg is blessed to have the YMCA. The Rock Steady program couldnt function without the dedication of the Y instructors and the volunteers. I encourage anyone who has Parkinsons disease to come and check out the program.

If there is a volunteer who should be featured, please submit their name and contact information to volunteer@vicksburgpost.com.

More:

VOLUNTEER OF THE WEEK: Joel Horton helps fight Parkinson's disease - The Vicksburg Post - Vicksburg Post

February. The month of shattered dreams and ambitions. The trainers are gathering dust and chocolate bars have replaced protein bars. The gusto with which we attacked our new year resolutions is a vague memory.

If your motivation to stick to your resolution to exercise more this year is waning, youre not alone. Its suggested around 80% of people will have given up on their new year resolutions by February.

But the reason your motivation wanes might actually be because you chose the wrong motives and goals to begin with. And research shows us that choosing the right type of goal is the key to keeping us motivated over the long term.

Lower the effort

Many of us believe that we need to grimace, contort, sweat and pant our way to a healthier life. So at the beginning of January, we put in a load of effort to help us reach our goals.

Unfortunately, our brain encourages us to avoid physical effort. This is why the excessive effort we use when exercising will work against us in the long run leading us to feel less motivated to exercise by the end of January. Our brain is constantly monitoring our body for any changes from our resting state, which could mean danger to our health. The more physical effort we use, the more a signal is activated and our brain tells us that the activity just isnt worth the effort and potential risk.

This is why minimising the effort we need to put into exercise may actually better help us stick to our resolutions in the long-term. For example, if youre dreading even a fifteen minute jog, do five minutes instead. Or if you hate running but enjoy zumba, do that instead. The golden rule is that the activity youre trying to motivate yourself to do needs to be pleasurable. And research shows were much more likely to do something if it requires less effort especially when were starting new exercise regimes.

The same principle applies to reducing the psychological effort required to exercise, as our brains also encourage us to avoid it to such an extent that, when given the choice, we often prefer physical pain instead. It does this because it wants to save psychological effort for times of emergency.

When it comes to starting a new exercise regime in the new year, things like fitting workouts into our schedule or getting out of bed an hour earlier all require psychological effort. To reduce psychological effort, it may help to minimise needless decision-making. When its time to exercise, remove decisions like whether to walk or drive to exercise class, or put your trainers in the same place so you dont have to look for them.

Although these sound like small decisions to make, they can all add up to us feeling less motivated to exercise when were required to make them. Research even shows that when we think our goals require little effort to achieve, were more likely to achieve them.

Choose short-term goals

Another basic motivational mistake many of us made in January was to set our goals too far in the future. Many people start exercising with the aim to lose a few pounds perhaps in order to fit into their favourite jeans again. But when the outcome is far in the future, our brains dont associate the motivation (fitting into our jeans) with exercising so were less inclined to exercise.

By choosing a goal that has a more immediate outcome, our brains will associate the outcome positively with exercise because they occur simultaneously. For example, the mood-boosting benefits of exercise occur more quickly than physical health changes so this may be a better motivator for you to keep exercising well past January. In short, make the reason for exercise an immediate one you can achieve and the long-term benefits will follow.

Focus on being instead of having

The final motivational fix is switching the type of goal you have. So-called have goals serve little purpose for our motivational brain, which focuses on more important things such as being effective at what we do and making social bonds. An example of a have goal would be exercising so that you can have a better body. This type of goal is viewed as less important by our brain because it does not help us meet essential goals that help us thrive.

On the other hand, the types of goals that are more likely to keep us motivated are be goals. An example of a be goal would be exercising to be healthy, or to be more athletic. Be goals are superior because humans tend to want to bond with other like-minded people based on our identities. This motivation is thought to have developed in our ancestral past, as forming bonds helped us to survive. So someone may find exercise easier to stick with if theyre doing it as a way to demonstrate their athleticism, for example. As a result, people do a better job of sticking to be goals, compared to other types of goals.

Even if you have fallen off the wagon slightly by the end of January, that doesnt mean you have to give up on your goals entirely. But making some tweaks to them and your approach to exercise may help you better stick to your goals for the rest of the year.

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Go here to read the rest:

Three tips to help you stay motivated to keep exercising all year long - The Indian Express

Can you really save money at the gas pump just by filling up on a certain day? A recent study by GasBuddy, which tracks gas prices across the country, determined the best and worst days of the week to fill your tank and save the most money.That study found that Monday is the best day of the week to buy gasoline in 17 states, which is on par with the company's previous studies in 2017, 2018 and 2019.A new finding, however, is that Friday was one of the best days to buy gas in 2021."This trend refutes 2019s results, which placed Friday as one of the most expensive days of the week at the pump, and can perhaps be attributed to the ongoing COVID-19 pandemic and the emerging prominence of work from home lifestyles," a news release from GasBuddy says.The study determined that Thursday was the worst day of the week to buy gas in 28 states. "In 2021, the middle of the week became far more expensive to fill up than on Mondays or Fridays," the release says. "While the weekend previously held the title for the most expensive prices, Wednesday now follows Thursday as the most expensive day to fill up." When it comes to saving money at the pump, Monday becomes more than the dreaded end of the weekend. GasBuddy analyzed gas price data and found that the first day of the workweek offers the lowest average gas price in 17 states, making it yet again the best day to fill-up, said Patrick De Haan, head of petroleum analysis at GasBuddy. Though there is variation in daily gas prices across different states, the consensus is that filling up at the beginning or end of the workweek, on Monday or Friday, is the best way to save money."If you can't make it to a gas station on Monday, De Hann said Sunday is the next cheapest day to fill up."But even if you cant always time your fill-ups, the golden rule is to always shop around before filling up," De Haan said. Analysis of gas prices in individual states can be found here.

Can you really save money at the gas pump just by filling up on a certain day?

A recent study by GasBuddy, which tracks gas prices across the country, determined the best and worst days of the week to fill your tank and save the most money.

That study found that Monday is the best day of the week to buy gasoline in 17 states, which is on par with the company's previous studies in 2017, 2018 and 2019.

A new finding, however, is that Friday was one of the best days to buy gas in 2021.

"This trend refutes 2019s results, which placed Friday as one of the most expensive days of the week at the pump, and can perhaps be attributed to the ongoing COVID-19 pandemic and the emerging prominence of work from home lifestyles," a news release from GasBuddy says.

The study determined that Thursday was the worst day of the week to buy gas in 28 states.

"In 2021, the middle of the week became far more expensive to fill up than on Mondays or Fridays," the release says. "While the weekend previously held the title for the most expensive prices, Wednesday now follows Thursday as the most expensive day to fill up."

When it comes to saving money at the pump, Monday becomes more than the dreaded end of the weekend. GasBuddy analyzed gas price data and found that the first day of the workweek offers the lowest average gas price in 17 states, making it yet again the best day to fill-up, said Patrick De Haan, head of petroleum analysis at GasBuddy. Though there is variation in daily gas prices across different states, the consensus is that filling up at the beginning or end of the workweek, on Monday or Friday, is the best way to save money."

If you can't make it to a gas station on Monday, De Hann said Sunday is the next cheapest day to fill up.

"But even if you cant always time your fill-ups, the golden rule is to always shop around before filling up," De Haan said.

Analysis of gas prices in individual states can be found here.

Read more from the original source:

Is there a best or worst day to find gas? This study may have found the answer - WAPT Jackson