Calculating the Fraction of Time Particles Spend Outside a Potential Well

[JD_PM]

Homework Statement

Please see image

Relevant Equations

Please see image

I want to compute the fraction of time both particles spend outside the finite potential well. All I can get is the probability to find them outside. The wavefunction outside the potential is:

$$\frac{d^2\psi}{dr^2} = -L^2 \psi$$

Where:

$$L = \sqrt{\frac{2mE}{\hbar^2}}$$

Solving the differential equation one gets:

$$\psi = Esin(Lr + \delta)$$

More details (starting 7:02):



After some Math you get the coefficients and the probability is just $\psi^2$.

But I am working all the time with the Time-independent Schrodinger equation so this approach, I'd say, is faulty so as to get the fraction of time.

May you give me a hint on this?

Thanks

 

[kuruman]

Suppose you had a way to measure whether the particles are outside the potential well. You repeat the measurement $N$ times and you find that $m$ times the particles were outside well and $N-m$ times they were inside the well. The ratio of the times the particles were outside the range to the total time is $m/N$. That's the fraction you are seeking. What does the area under the "exponential fall" in the figure represent?

 

[JD_PM]

Suppose you had a way to measure whether the particles are outside the potential well. You repeat the measurement $N$ times and you find that $m$ times the particles were outside well and $N-m$ times they were inside the well. The ratio of the times the particles were outside the range to the total time is $m/N$. That's the fraction you are seeking.

I agree. But how can I get a quantitative result? Should I go for calculating the transmission coefficient?

What does the area under the "exponential fall" in the figure represent?

The deuterium wavefunction outside the potential well is:

$$\frac{-\hbar^2}{2m}\frac{d^2\psi}{dr^2} = E\psi$$

$$\frac{d^2\psi}{dr^2} = \frac{-2mE}{\hbar^2}\psi$$

$$\frac{d^2\psi}{dr^2} = L^2 \psi$$

Applying boundary conditions one gets:

$$\psi_{II} = De^{-Lr}$$

More details:



The area under the "exponential fall" is just the probability of finding both particles outside the potential well: $\int |\psi_{II}(r)^2|dr$

 

[kuruman]

The area under the "exponential fall" is just the probability of finding both particles outside the potential well ...

How is this different from "the fraction of time (that) the neutron and proton spend beyond the range of their nuclear force"? My thinking is this: Suppose you measure the position of the particles at the rate of one measurement per minute and you do this for an hour (60 measurements total). Also suppose that you found the particles beyond the range in 15 of the measurements, randomly distributed in the 60 minute interval.
What is the probability that the particles will be found outside the range? Answer: 15 out of 60 measurements = 1/4.
What is fraction of time that the particles spend beyond the range? Answer: 15 minutes out of 60 minutes = 1/4.

Maybe I'm missing something which wouldn't be the first time ...

 

[JD_PM]

How is this different from "the fraction of time (that) the neutron and proton spend beyond the range of their nuclear force"? My thinking is this: Suppose you measure the position of the particles at the rate of one measurement per minute and you do this for an hour (60 measurements total). Also suppose that you found the particles beyond the range in 15 of the measurements, randomly distributed in the 60 minute interval.
What is the probability that the particles will be found outside the range? Answer: 15 out of 60 measurements = 1/4.
What is fraction of time that the particles spend beyond the range? Answer: 15 minutes out of 60 minutes = 1/4.

Maybe I'm missing something which wouldn't be the first time ...

I have this question in mind and will reply soon

 

[JD_PM]

How is this different from "the fraction of time (that) the neutron and proton spend beyond the range of their nuclear force"? My thinking is this: Suppose you measure the position of the particles at the rate of one measurement per minute and you do this for an hour (60 measurements total). Also suppose that you found the particles beyond the range in 15 of the measurements, randomly distributed in the 60 minute interval.
What is the probability that the particles will be found outside the range? Answer: 15 out of 60 measurements = 1/4.
What is fraction of time that the particles spend beyond the range? Answer: 15 minutes out of 60 minutes = 1/4.

I agree with your reasoning, and I think that it is equivalent to computing the transmission coefficient.

What I have tried to calculate it (I will go step by step):

The Schrodinger Equation inside the potential well is:

$$\frac{d^2\psi}{dr^2} = -\frac{2m(E-V_0)}{\hbar^2}\psi$$

Defining constant K to be a real term:

$$\frac{d^2\psi}{dr^2} = -k^2\psi$$

This is the classical simple harmonic oscillator. The solution of the differential equation:

$$\psi_I = Asinkr + Bcoskr$$

Applying boundary conditions ($\psi_I(0) = \psi_I(R) = 0$) one gets:

$$\psi_I(r) = Asinkr$$

The Schrodinger Equation outside the potential well is:

$$\frac{d^2\psi}{dr^2} = -\frac{2mE}{\hbar^2}\psi$$

Defining constant L to be a real term:

$$\frac{d^2\psi}{dr^2} = L^2\psi$$

The solution of the differential equation:

$$\psi_{II} = Ce^{Lr} + De^{-Lr}$$

First term grows exponentially (it is non-normalizable). Thus:

$$\psi_{II}(r) = De^{-Lr}$$

OK once I have both wavefunctions I should evaluate the coefficients A and D and then computing the transmission coefficient.

Let's start evaluating the coefficients by applying boundary conditions:

The wavefunction must be continuous, which means that both $\psi_{I}$ and $\psi_{II}$ are equal to each other at R. Then:

$$Asin(kR) - De^{-LR} = 0$$

Its derivative as well. Then:

$$Akcos(kR) + LDe^{-LR} = 0$$

I know every term of these two equations but A and D. When I solve for A I get $A = 0$, which is a bad sign because the transmission coefficient is defined:

$$T = \frac{|D|^2}{|A|^2}$$

So I must be missing something here...

I know it is a long post, so let me know if you want more info related to a specific point of it.

Thanks.

 

[kuruman]

I am not sure all this makes sense. When $E<0$, you have bound states which you can get by solving the equation that you get by dividing the boundary condition equations,$$\tan(kR)=\frac k L$$. The transmission coefficient is meaningless when you have bound states.

 

[JD_PM]

I am not sure all this makes sense. When $E<0$, you have bound states

When $E<0$ we deal with bound states inside the well ($\psi_I(r) = Asinkr$) indeed.

But note that we also study the case outside the finite well.

I suspect we have quantum tunneling here. I mean, when we evaluate what is the wavefunction outside the well ($\psi_{II}(r) = De^{-Lr}$) and assume that E stays the same (i.e still negative), we are implying that both particles are leaking through the finite potential barrier. This, I have read, implies quantum scattering state (it is considered a bound state classically though).

Thus my point is that when $E<0$, we have both bound and scattering states.

If I am wrong, why?

 

[kuruman]

Thus my point is that when $E<0$ we have both bound and scattering states.

Can you provide a source that supports this point? You cannot have quantum mechanical tunneling here. If you did, the probability density would be non-zero at arbitrary distances. The wavefunction vanishes as $r \rightarrow \infty$.

 

[BvU]

When I solve for A I get A=0

No you don't. In the link in #3 the narrator works it all out !

@JD: the terms 'tunneling' and 'transmission coefficient' don't apply here. All that happens is that the wavefefunction is nonzero in the classically forbidden region.

 

[BvU]

leaking through the finite potential barrier

That would imply a third region with $V > 0$ between region I and region II

 

[JD_PM]

No you don't. In the link in #3 the narrator works it all out !

DrPhysics works out the probabilities of getting both proton and neutron at certain distances (for instance, at R it is 95%) but he doesn't work out the time both particles spend beyond the nuclear force range (i.e further than 2.1 fm).

I think I should evaluate the coefficients A and D using the equations:

$$Asin(kR) - De^{-LR} = 0$$

$$Akcos(kR) + LDe^{-LR} = 0$$I don't see why you don't get zero...

PS: He doesn't work out the coefficients either.

 

[BvU]

for instance, at R it is 95%

Watch out ! probability $\ne$ value of $\psi$

I don't see why you don't get zero...

$kR>\pi/2$

 

[bobob]

Consider the fact that the deuteron is and admixture of L=0 and L=2.

 

[JD_PM]

That would imply a third region with $V > 0$ between region I and region II

OK After studying alpha decay I completely see what you mean.

So at this point I am stuck ...

May you give me a hint on how to get the fraction of time both proton and neutron spend beyond the nuclear potential range?

 

[BvU]

I'm surprised you have to ask: already in #2 you had an expression for the probability ($\int |\psi_{II}(r)^2|dr$ ) and the wave function ($\psi_{II} = De^{-Lr}$). All you have to do is sweep things together like the guy in the video does around 41:45
(only you have $V_0$ given and need the value of $D$).

 

[JD_PM]

I'm surprised you have to ask: already in #2 you had an expression for the probability ($\int |\psi_{II}(r)^2|dr$ ) and the wave function ($\psi_{II} = De^{-Lr}$). All you have to do is sweep things together like the guy in the video does around 41:45
(only you have $V_0$ given and need the value of $D$).

I am afraid I am still missing something...

The probability of getting both particles outside the nuclear potential range is:

$$P = \int_{R}^{\infty} |\psi_{II}(r)^2|dr = \frac{D^2}{2L} e^{-2LR}$$

Here I have everything but $D$.

Using boundary conditions:

$$Asin(kR) = De^{-LR}$$

$$KAcos(kR) = -LDe^{-LR}$$

I should be able to get both constants because I have everything but $A$ and $D$.

Solving for constant $A$:

$$A(kcos(kR) + Lsin(kR)) = 0$$

Then:

$$A = 0$$

But this is wrong, because it would imply that $D = 0$.

I know I am missing something really basic here but don't see it... Thanks for your patience.