Energy Difference with a Two Delta Function Potential

[doggydan42]

Homework Statement

Consider a particle of mass m moving in a one-dimensional double well potential
$$V(x) = -g\delta(x-a)-g\delta(x+a), g > 0$$

This is an attractive potential with $\delta$-function dips at x=$\pm a$.

In the limit of large $\lambda$, find a approximate formula for the energy difference $\Delta E > 0$ between the ground state and the first excited state. (Hint: Start by finding approximate expressions for $\xi_{even}$ and $\xi_{odd}$ in the limit of large $\lambda$.)

Express your answer in terms of lambda for $\lambda$.

$$\frac{\Delta E}{E_0} = $$

Homework Equations

$$\kappa \equiv \sqrt{-\frac{2mE}{\hbar^2}}, xi \equiv \kappa a, \lambda \equiv \frac{mag}{\hbar^2}$$

$$\lambda_{even}= \frac{\xi}{e^{-2\xi}+1}, \lambda_{odd}= \frac{\xi}{1-e^{-2\xi}}$$

$$E_0 = \frac{\hbar^2}{2ma^2}$$

$$\frac{E}{E_0} = -\xi^2$$

The Attempt at a Solution

For large even lambda, $\lambda = \frac{xi}{2} \rightarrow \xi = 2\lambda$. So,
$$\frac{E_{even}}{E_0}=-\xi^2=-4\lambda^2$$
For odd lambda, $\xi = \lambda*0 = 0$, so
$$\frac{E_{odd}}{E_0}=-\xi^2=0$$

For the change in energe:
$$\frac{\Delta E}{E_0}=\frac{E_{even or odd}}{E_0}-\frac{E_0}{E_0} = \frac{E_{even or odd}}{E_0}-1$$

Frome here I was not sure how to identify whether to use even or odd E.

 

[stevendaryl]

I don't understand where your $\lambda_{even}$ and $\lambda_{odd}$ are coming from, or what they are supposed to mean.

The way that I would approach the problem is to start with an exact set of equations, and then see where you can simplify things by making approximations.

If the potential is zero every except at $x= \pm a$, then the solution will look like one of these possibilities:

1. Even solution: $\psi(x) = e^{\kappa x}$ for $x < -a$. $\psi(x) = A cosh(\kappa x)$ for $-a < x < +a$. $\psi(x) = e^{-\kappa x}$ for $x>a$.
2. Odd solution: $\psi(x) = - e^{\kappa x}$ for $x < -a$. $\psi(x) = A sinh(\kappa x)$ for $-a < x < +a$. $\psi(x) = +e^{-\kappa x}$ for $x>a$.

In each case, you can figure out what the constants $A$ and $\kappa$ must be from two different equations:

1. Continuity of $\psi$: $\psi(a_{-}) = \psi(a_{+})$. $\psi$ has the same value on both sides of $A$.
2. The limit as $\delta x \rightarrow 0$ of $\int_{a-\delta x}^{a+\delta x} \frac{-\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} + V(x) \psi(x) - E \psi(x) dx$ is 0

The last equation is the integration of Schrodinger's equation.

 

[doggydan42]

I don't understand where your $\lambda_{even}$ and $\lambda_{odd}$ are coming from, or what they are supposed to mean.

The way that I would approach the problem is to start with an exact set of equations, and then see where you can simplify things by making approximations.

If the potential is zero every except at $x= \pm a$, then the solution will look like one of these possibilities:

1. Even solution: $\psi(x) = e^{\kappa x}$ for $x < -a$. $\psi(x) = A cosh(\kappa x)$ for $-a < x < +a$. $\psi(x) = e^{-\kappa x}$ for $x>a$.
2. Odd solution: $\psi(x) = - e^{\kappa x}$ for $x < -a$. $\psi(x) = A sinh(\kappa x)$ for $-a < x < +a$. $\psi(x) = +e^{-\kappa x}$ for $x>a$.

In each case, you can figure out what the constants $A$ and $\kappa$ must be from two different equations:

1. Continuity of $\psi$: $\psi(a_{-}) = \psi(a_{+})$. $\psi$ has the same value on both sides of $A$.
2. The limit as $\delta x \rightarrow 0$ of $\int_{a-\delta x}^{a+\delta x} \frac{-\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} + V(x) \psi(x) - E \psi(x) dx$ is 0

The last equation is the integration of Schrodinger's equation.

The $\lambda$ for even an odd come from continuity at a for the even and odd $\psi$, and from the equation for the discontinuity of the derivative at a:
$$\Delta_a \psi' = -\frac{2mg}{\hbar^2}\psi(0)$$. Using this equation and teh continuity at a, $\lambda_{even}$ can be found using the even wavefunction, and $\lambda_{odd}$ with the odd wavefunction from the definition of $\lambda$.

 

[stevendaryl]

The $\lambda$ for even an odd come from continuity at a for the even and odd $\psi$, and from the equation for the discontinuity of the derivative at a:
$$\Delta_a \psi' = -\frac{2mg}{\hbar^2}\psi(0)$$. Using this equation and teh continuity at a, $\lambda_{even}$ can be found using the even wavefunction, and $\lambda_{odd}$ with the odd wavefunction from the definition of $\lambda$.

Okay, but since the delta function is not centered on $x=0$, it's centered on $x=\pm a$, so you don't use $\psi(0)$.

 

[doggydan42]

Okay, but since the delta function is not centered on $x=0$, it's centered on $x=\pm a$, so you don't use $\psi(0)$.

Sorry, that was a typo. I meant to type $\psi(a)$.

 

[stevendaryl]

Sorry, that was a typo. I meant to type $\psi(a)$.

Okay, but I still don't understand your even and odd cases. Where are you getting the even and odd wave functions from? What are they? And what is $E_0$? And where are you getting those expressions for $\lambda_{even}$ and $\lambda_{odd}$?

 

[doggydan42]

Okay, but I still don't understand your even and odd cases. Where are you getting the even and odd wave functions from? What are they? And what is $E_0$? And where are you getting those expressions for $\lambda_{even}$ and $\lambda_{odd}$?

The wavefunctions I got from solving the time independent Schrodinger equation on different intervals (x <-a, -a<x<a, and x> a). The solutions you gave we're essentially the ones I used.

$E_0$ was defined to be $E_0 = \frac{\hbar^2}{2ma^2}$

 

[stevendaryl]

The wavefunctions I got from solving the time independent Schrodinger equation on different intervals (x <-a, -a<x<a, and x> a). The solutions you gave we're essentially the ones I used.

$E_0$ was defined to be $E_0 = \frac{\hbar^2}{2ma^2}$

I don't understand what you are doing. Here are the equations that I get, for the even and odd cases: (In terms of your $\xi$ and $\lambda$:

Even case:

1. $A cosh(\xi) = e^{-\xi}$
2. $\xi (e^{-\xi} + A sinh(\xi)) = 2 \lambda A cosh(\xi) = 2 \lambda e^{-\xi}$

Odd case:

1. $A sinh(\xi) = e^{-\xi}$
2. $\xi (e^{-\xi} + A cosh(\xi)) = 2 \lambda A sinh(\xi) = 2 \lambda e^{-\xi}$

Using $cosh(\xi) = \frac{1}{2} (e^\xi + e^{-\xi})$ and $sinh(\xi) = \frac{1}{2}(e^\xi - e^{-\xi})$, we can rewrite these as:

1. $A/2 (e^{\xi} \pm e^{-\xi}) = e^{-\xi} \Rightarrow A = \frac{2e^{-\xi}}{e^{\xi} \pm e^{-\xi}}$
2. $\xi (e^{-\xi} + A/2 (e^\xi \mp e^{-\xi})) = 2 \lambda e^{-\xi} \Rightarrow \xi (e^{-\xi} + \frac{e^{-\xi}}{e^{\xi} \pm e^{-\xi}} (e^\xi \mp e^{-\xi})) = 2 \lambda e^{-\xi}$

The right side of 2. can simplify enormously to: $\frac{2\xi}{e^\xi \pm e^{-\xi}}$. So we have:

$\frac{2\xi}{e^\xi \pm e^{-\xi}} = 2 \lambda e^{-\xi}$

Or

$\xi = \lambda (1 \pm e^{-\xi})$

So the even case is the solution to

$\xi = \lambda (1 + e^{-\xi})$

and the odd case is the solution to $\xi = \lambda(1 - e^{-\xi})$

Those are impossible to solve exactly, but we can see that if $\xi \gg 1$, then $\xi \approx \lambda$. One approach to solving this self-referential type of equation is to iterate:

Let $\xi_0 = \lambda$.
Let $\xi_1 = \lambda (1 \pm e^{-\xi_0})$
Let $\xi_2 = \lambda (1 \pm e^{-\xi_1})$
Etc.

Just keeping the first approximation gives:

$\xi \approx \lambda (1 \pm e^{-\lambda})$

 

[doggydan42]

I don't understand what you are doing. Here are the equations that I get, for the even and odd cases: (In terms of your $\xi$ and $\lambda$:

$\xi = \lambda (1 \pm e^{-\xi})$

That is almost the result I got. I think you made a mistake from $\frac{2\xi}{e^\xi \pm e^{-\xi}} = 2\lambda e^{-\xi}$

That would give:

$$\xi = \lambda e^{-\xi} (e^\xi \pm e^{-\xi}) = \lambda (1 \pm e^{-2\xi})$$

This is the result I got.

It seems that you gave a way to approximate the equatin in the limit of large $\lambda$. I believe should now change to a factor of 2 in the exponent.

$\xi_0 = \lambda$
$\xi_1 = \lambda (1 \pm e^{-2\xi_0})$
$\xi_2 = \lambda (1 \pm e^{-2\xi_1})$

I still need to figure out how to find the energy difference $\frac{\Delta E}{E_0}$ between the ground state and the first excited state. I probably will have to use $\frac{E}{E_0} = -\xi^2$. I am confused on how to identify which of these two states will be even or odd, and if they have different $\lambda$, how it will be possible to write this energy difference in terms of only $\lambda$ .

 

[stevendaryl]

Just keeping the first approximation gives:

$\xi \approx \lambda (1 \pm e^{-\lambda})$

That is almost the result I got. I think you made a mistake from $\frac{2\xi}{e^\xi \pm e^{-\xi}} = 2\lambda e^{-\xi}$

That would give:

$$\xi = \lambda e^{-\xi} (e^\xi \pm e^{-\xi}) = \lambda (1 \pm e^{-2\xi})$$

This is the result I got.

You're right.

It seems that you gave a way to approximate the equatin in the limit of large $\lambda$. I believe should now change to a factor of 2 in the exponent.

$\xi_0 = \lambda$
$\xi_1 = \lambda (1 \pm e^{-2\xi_0})$
$\xi_2 = \lambda (1 \pm e^{-2\xi_1})$

I still need to figure out how to find the energy difference $\frac{\Delta E}{E_0}$ between the ground state and the first excited state. I probably will have to use $\frac{E}{E_0} = -\xi^2$. I am confused on how to identify which of these two states will be even or odd, and if they have different $\lambda$, how it will be possible to write this energy difference in terms of only $\lambda$ .

I think you have your answer. The approximation $\xi \approx \lambda (1 \pm e^{-2 \lambda})$ gives the even case as

$\xi \approx \lambda (1 + e^{-2\lambda})$

and the odd case as

$\xi \approx \lambda (1 - e^{-2\lambda})$

The plus comes from $\psi(\xi) = A cosh(\xi)$ (an even function) and the minus comes from $\psi(\xi) = A sinh(\xi)$ (an odd function).

Then using $\frac{E}{E_0} = -\xi^2/2$ gives you the two possible energy values.

 

[stevendaryl]

You're right.



I think you have your answer. The approximation $\xi \approx \lambda (1 \pm e^{-2 \lambda})$ gives the even case as

$\xi \approx \lambda (1 + e^{-2\lambda})$

and the odd case as

$\xi \approx \lambda (1 - e^{-2\lambda})$

The plus comes from $\psi(\xi) = A cosh(\xi)$ (an even function) and the minus comes from $\psi(\xi) = A sinh(\xi)$ (an odd function).

Then using $\frac{E}{E_0} = -\xi^2/2$ gives you the two possible energy values.

If I haven't made a mistake somewhere, then the odd function is lower energy than the even function. That's unusual, because in most familiar problems, such as the harmonic oscillator and the square well, the ground state is even.

 

[doggydan42]

You're right.

Then using $\frac{E}{E_0} = -\xi^2/2$ gives you the two possible energy values.

Why would I use $\frac{E}{E_0} = -\xi^2/2$? Where is the factor of 1/2 coming from?

 

[stevendaryl]

Why would I use $\frac{E}{E_0} = -\xi^2/2$? Where is the factor of 1/2 coming from?

That was an error. I just wrote down the wrong equation. Yes, it should be $E/E_0 = -\xi^2$

 

[doggydan42]

That was an error. I just wrote down the wrong equation. Yes, it should be $E/E_0 = -\xi^2$

So using that relation:

$$\frac{\Delta E}{E_0} = \frac{E_1}{E_0} - \frac{E_0}{E_0} = -\xi_1^2+\xi_0^2 = -\lambda^2 (1-e^{-2\lambda})^2+\lambda^2 = -\lambda^2 (1-2e^{-2\lambda}+e^{-4\lambda})+\lambda^2 = \lambda^2 (2e^{-2\lambda}-e^{-4\lambda}) = \lambda^2 e^{-2\lambda} (2 - e^{-2\lambda})$$

So the final answer would be $\lambda^2 e^{-2\lambda} (2 - e^{-2\lambda})$?

 

[stevendaryl]

So using that relation:

$$\frac{\Delta E}{E_0} = \frac{E_1}{E_0} - \frac{E_0}{E_0} = -\xi_1^2+\xi_0^2 = -\lambda^2 (1-e^{-2\lambda})^2+\lambda^2 = -\lambda^2 (1-2e^{-2\lambda}+e^{-4\lambda})+\lambda^2 = \lambda^2 (2e^{-2\lambda}-e^{-4\lambda}) = \lambda^2 e^{-2\lambda} (2 - e^{-2\lambda})$$

So the final answer would be $\lambda^2 e^{-2\lambda} (2 - e^{-2\lambda})$?

That's why I was confused by your use of $E_0$. The ground state is not $E_0$. It is $E_{even}$. So what you need to calculate is

$\Delta E = E_{odd} - E_{even}$

 

[doggydan42]

That's why I was confused by your use of $E_0$. The ground state is not $E_0$. It is $E_{even}$. So what you need to calculate is

$\Delta E = E_{odd} - E_{even}$

So if I ignore a factor of lambda squared,

$$-(1-e^{-2\lambda})^2 + (1+e^{-2\lambda})^2 = -1 + 2e^{-2\lambda} - e^{-4\lambda} + 1 + 2e^{-2\lambda} + e^{-4\lambda} = 4e^{-2\lambda}$$

So overall,
$$\frac{\Delta E}{E_0} = 4\lambda^2 e^{-2\lambda} $$

 

[stevendaryl]

So if I ignore a factor of lambda squared,

$$-(1-e^{-2\lambda})^2 + (1+e^{-2\lambda})^2 = -1 + 2e^{-2\lambda} - e^{-4\lambda} + 1 + 2e^{-2\lambda} + e^{-4\lambda} = 4e^{-2\lambda}$$

So overall,
$$\frac{\Delta E}{E_0} = 4\lambda^2 e^{-2\lambda} $$

I think that's correct.