In this problem we obtain an analytic estimate of the energy...

In this problem we obtain an analytic estimate of the energy difference between the even- and odd-parity states of a double square well. Show that for large $\theta$, $\operatorname{coth} \theta-\tanh \theta \simeq 4 \mathrm{e}^{-2 \theta}$. Next letting $\delta k$ be the difference between the $k$ values that solve $$ \tan [r \pi-k(b-a)] \sqrt{\frac{W^{2}}{(k a)^{2}}-1}= \begin{cases}\operatorname{coth}\left(\sqrt{W^{2}-(k a)^{2}}\right) & \text { even parity } \\ \tanh \left(\sqrt{W^{2}-(k a)^{2}}\right) & \text { odd parity }\end{cases} $$ where $$ W \equiv \sqrt{\frac{2 m V_{0} a^{2}}{\hbar^{2}}} $$ for given $r$ in the odd- and even-parity cases, deduce that $$ \begin{array}{r} \left\{\left[\left(\frac{W^{2}}{(k a)^{2}}-1\right)^{1 / 2}+\left(\frac{W^{2}}{(k a)^{2}}-1\right)^{-1 / 2}\right](b-a)+\frac{1}{k}\left(1-\frac{(k a)^{2}}{W^{2}}\right)^{-1}\right\} \delta k \\ \simeq-4 \exp \left[-2 \sqrt{W^{2}-(k a)^{2}}\right] \end{array} $$ Hence show that when $W \gg 1$ the fractional difference between the energies of the ground and first excited states is $$ \frac{\delta E}{E} \simeq \frac{-8 a}{W(b-a)} \mathrm{e}^{-2 W \sqrt{1-E / V_{0}}} $$

In this problem we obtain an analytic estimate of the energy difference between the even- and odd-parity states of a double square well. Show that for large $\theta$, $\operatorname{coth} \theta-\tanh \theta \simeq 4 \mathrm{e}^{-2 \theta}$. Next letting $\delta k$ be the difference between the $k$ values that solve
$$
\tan [r \pi-k(b-a)] \sqrt{\frac{W^{2}}{(k a)^{2}}-1}= \begin{cases}\operatorname{coth}\left(\sqrt{W^{2}-(k a)^{2}}\right) & \text { even parity } \\ \tanh \left(\sqrt{W^{2}-(k a)^{2}}\right) & \text { odd parity }\end{cases}
$$
where
$$
W \equiv \sqrt{\frac{2 m V_{0} a^{2}}{\hbar^{2}}}
$$
for given $r$ in the odd- and even-parity cases, deduce that
$$
\begin{array}{r}
\left\{\left[\left(\frac{W^{2}}{(k a)^{2}}-1\right)^{1 / 2}+\left(\frac{W^{2}}{(k a)^{2}}-1\right)^{-1 / 2}\right](b-a)+\frac{1}{k}\left(1-\frac{(k a)^{2}}{W^{2}}\right)^{-1}\right\} \delta k \\
\simeq-4 \exp \left[-2 \sqrt{W^{2}-(k a)^{2}}\right]
\end{array}
$$
Hence show that when $W \gg 1$ the fractional difference between the energies of the ground and first excited states is
$$
\frac{\delta E}{E} \simeq \frac{-8 a}{W(b-a)} \mathrm{e}^{-2 W \sqrt{1-E / V_{0}}}
$$

Key Concepts

Dimensionless Parameters and Scaling

The introduction of dimensionless parameters, such as W, helps in reducing the complexity of the problem by encapsulating the physical parameters (mass, potential depth, well width) into a single quantity. Analyzing the limit where W is large allows for the extraction of the exponential dependence of the energy splitting on the barrier properties, providing clear insight into the scaling behavior of the tunneling effect.

Asymptotic Approximations

Asymptotic approximations are used to simplify complex expressions in limiting cases, such as when a parameter becomes very large. In this context, approximating hyperbolic functions for large arguments (e.g., coth ? - tanh ? ? 4e^(-2?)) is crucial for obtaining a tractable analytic expression for the energy splitting. This technique helps in identifying the dominant contribution to tunneling effects.

Hyperbolic Function Approximations

Hyperbolic functions like coth and tanh emerge naturally when solving the Schrödinger equation in regions where the potential is constant, leading to exponential solutions. Their approximations in the limit of large arguments are essential for simplifying the matching conditions at the boundaries of the potential wells and for deriving analytic formulas for the tunneling-induced energy splitting.

Parity of Wavefunctions

In the context of the double square well, eigenstates are classified by their symmetry properties as either even (symmetric) or odd (antisymmetric) parity. This classification simplifies the boundary condition matching and reveals that the energy difference between these states is directly linked to the tunneling amplitude. The even and odd parity solutions are a cornerstone in analyzing the spectral properties of symmetric potentials.

Double Square Well Potential

A double square well potential is a model used in quantum mechanics where a particle is confined in two adjacent potential wells separated by a barrier. This setup illustrates quantum tunneling, where the particle's wavefunctions in each well overlap, leading to distinct energy states that are nearly degenerate. The analysis of such a system helps in understanding phenomena like energy splitting, which arises due to the particle’s ability to tunnel between wells.

Quantum Tunneling and Energy Splitting

Quantum tunneling refers to the phenomenon where a particle penetrates through a potential barrier that it classically shouldn’t be able to cross. In the double square well model, this tunneling results in a slight splitting of energy levels between states of different parity. The energy difference, or splitting, is exponentially sensitive to the barrier’s properties (height and width), and is a direct measure of the tunneling probability.

Transcript

00:04 Okay, so for this problem, which is problem 58 in chapter 40 of university physics with modern physics 12th edition, we are looking at the wkv approximation.

00:29 And specifically with the wkv approximation, we are looking at the integral from one point to another point within a potential over the square root of twice the mass.

00:47 Times the difference between the current energy and the potential energy.

00:53 And this is an integral over the distance.

00:58 And in the wkp approximation, we can say that this is equal to n h divided by 2, where n is an integer.

01:10 And a couple relationships we're going to use at some point in this problem are that h is equal to 2 pi times h bar.

01:20 And also that k is equal to m omega squared and this comes from omega b from springs on spring systems so the first part of this question asks us to look at the classical turning points and remember the classical turning points are when e is equal to you when there's no more potential for the system to continue moving, it's going to turn around.

01:59 Think about a ball rolling up a hill, and once its kinetic energy at the bottom is equal to its potential energy, it will stop and turn around and fall back down.

02:09 In the case of a harmonic oscillator, we're looking at the potential one -half kx squared.

02:17 It's just the spring potential, and we would like to know that when e is equal to this quantity, what is x going to be? that's easy to multiply both sides by 2 over k and get that x squared is equal to 2e divided by k and therefore the turning points the classical turning points of the system are x is equal to the plus or minus square root of 2e over k part b is having us use these endpoints as are a and b in the system and so i'm going to rewrite our integral now instead of the integral from a to b we can do the integral from negative a to a because b and a are have the same magnitude overall of the square root of 2m times now the difference between e and our one -half k x squared number and that is an integral dx.

03:41 Simplifying things further, we know that the integral over an even function and remember x squared here is what we're looking at is the even part you can take the integral from negative a and turn that into twice the integral from zero to a of the square root of e minus one half k x squared d x d x and again that is the square rear of this okay so we're going to simplify the inside of the integrand now and say that this is equal to the integral of twice zero to of the square root of 2m .e minus m mk because the two is canceled x squared and that is an integral over x.

04:57 Okay.

04:59 Now if you remember from the standard integrals that one learns in an introductory calculus class, you can perform what's called a trigonometric substitution, but you can integrate something that has the form a squared minus x squared over an interval x the x and you'll wind up with one -half times quantity x times the square root of a squared minus x squared plus a squared times the inverse sign of x over the magnitude of a and quantity we're going to use this standard integral in this problem.

06:03 And to do so, that means we simply need to rewrite our integral on the left hand side as something that's resembling the integral on the right hand side.

06:13 You'll notice that it already looked pretty similar.

06:15 We simply need to get x squared by itself.

06:18 And we can do this by factoring out the square root of mk, and then we'll have the integral from 0 to a of the square root and now we're going to say 2m .e divided by mk the em are going to cancel and we will wind up with 2e over k a familiar quantity minus x squared.

06:51 Now we take the integral over this dx.

06:59 Okay so now we can simply plug this in to our standard interval we're using and we will get our 2 square root of mk from out front our pre -factor multiplied by 1 half times the quantity of x times the square root of a squared.

07:25 Now remember a squared was 2e over k as part of our substitution 2e over k is in the same spot as the a squared in our standard integral and so we are going to plug that in 2e over k minus x squared plus 2e over k times the inverse sign of x over the square root of 2e over k now remember this the bounds of our integrand were our integral were from 0 to a and that a quantity was our classical turning point so we are now going to plug those bounds in i'll actually write the classical turning point square of 2e over k.

08:25 So now when we plug this in, we will obtain two square root mk divided by two from the one half.

08:38 So this two is actually going to go away because here and here cancel.

08:45 And we will obtain the difference between x, which is now square root 2e over k, times the quantity of 2e over k minus 2e over k, which is 0, plus 2e over k times the inverse sign of the square root of 2e over k divided by the square root of 2a or k, which is 1.

09:17 Now we take that minus.

09:19 This is just a standard plugging in the bounds of an integral in calculus.

09:24 Plugging in 0 for x now will obtain 0 times a quantity, which we don't really care about because it's multiplied by zero, so it will go away, plus 2e over k times the inverse sign of 0 divided by something, which is just 0...

Please give Ace some feedback