Finding the probability of making a measurement E1

[Vitani1]

Summary:: How to solve this?

See attached. This is from Griffiths 3rd edition quantum mechanics textbook, problem 2.7:

This is the solution for an infinite square well. I am to find the probability of obtaining E1 where En = n^2pi^2hbar^2/2ma^2.

E1 = hbar^2pi^2/2ma^2

When finding the amplitude of this wavefunction from normalization you will see that it is independent of n. Therefore the probability of measuring E1 is the amplitude squared which is the same for all energy eigenvalues. Is this correct?

Thanks

 

[BvU]

When finding the amplitude of this wavefunction from normalization you will see that it is independent of n.

You mean $A$ is independent of $n$ ?
Is that the way to find $\Psi(x, t)$ ?

 

[Vitani1]

Typically the amplitude is dependent on n, but for this example this is not the case? To find psi(x,t) you will add a time-dependent complex exponential term which is dependent on n. To find the probability amplitude isn't it the same for if you squared the time-dependent and time-independent solutions?

 

[vanhees71]

Of course, it's independent of $n$. Why should the state depend on $n$. It's representing the state the particle is prepared in. The question now is, what's the probability to measure the energy eigenvalue $E_1$ when you measure the energy.

 

[vela]

Typically the amplitude is dependent on n, but for this example this is not the case?

The variable $A$ is a normalization constant, not a probability amplitude which is what you seem to be referring to.

 

[Vitani1]

I need to square this probability amplitude A (the coefficient in front of psi) and this is to give me the probability of obtaining the energy eigenvalue. When solving for the particle in an infinite square well there is typically a dependence on n as in Ancos(nx)+Bnsin(nx) = psi(x,0). To obtain the energy eigenvalue E1 you would use the equation Acos(x)+Bsin(x) (for example - this is not the exact solution). For this problem the solution is different since it is not a sum of trigonometric eigenfunctions and this coefficient An (or Bn) is independent of n. Therefore I'm concluding that squaring this amplitude A given in the problem is the correct solution giving the probability of a particle having energy E1 but I'm saying further that this same value is the probability for all energy values E (again due to no n-dependence). Is this correct?

 

[PeroK]

I need to square this probability amplitude A (the coefficient in front of psi) and this is to give me the probability of obtaining the energy eigenvalue. When solving for the particle in an infinite square well there is typically a dependence on n as in Ancos(nx)+Bnsin(nx) = psi(x,0). To obtain the energy eigenvalue E1 you would use the equation Acos(x)+Bsin(x) (for example - this is not the exact solution). For this problem the solution is different since it is not a sum of trigonometric eigenfunctions and this coefficient An (or Bn) is independent of n. Therefore I'm concluding that squaring this amplitude A given in the problem is the correct solution giving the probability of a particle having energy E1 but I'm saying further that this same value is the probability for all energy values E (again due to no n-dependence). Is this correct?

No, that's not correct. You need to express the wavefunction you are given in the energy eigenbasis. Or, at least, find the coefficient of $\psi_1$ when expressed in this basis.

Hint: Fourier's trick.

 

[Vitani1]

When you say eigenbasis do you mean express this psi1 in terms of the energy eigenstate? i.e. find the Fourier representation of x and use this? This is what I initially tried to do but this solution is a bit messy.

 

[PeroK]

When you say eigenbasis do you mean express this psi1 in terms of the energy eigenstate? i.e. find the Fourier representation of x and use this? This is what I initially tried to do but this solution is a bit messy.

Yes. QM gets messy sometimes.

 

[Vitani1]

This has been completed. I like this solution. I now have the time-independent SE in terms of the Fourier eigenstates. To get the time-dependent equation (hence an equation which is dependent on E) then I just multiply my argument by the exponential term including En. Since this is piecewise do I multiply both top and bottom solutions for both regions by the same exponential time-dependence term including the first term in the second region (which is a constant)? Also would it be okay to assume that the energy states for this wavefunction are the same as for a typical particle in a box?

 

[vela]

When you say eigenbasis do you mean express this psi1 in terms of the energy eigenstate? i.e. find the Fourier representation of x and use this? This is what I initially tried to do but this solution is a bit messy.

Show us what you did. The integral is pretty straightforward.

 

[PeroK]

This has been completed. I like this solution. I now have the time-independent SE in terms of the Fourier eigenstates. To get the time-dependent equation (hence an equation which is dependent on E) then I just multiply my argument by the exponential term including En. Since this is piecewise do I multiply both top and bottom solutions for both regions by the same exponential time-dependence term including the first term in the second region (which is a constant)? Also would it be okay to assume that the energy states for this wavefunction are the same as for a typical particle in a box?

This is where mathematics is clearer than words. The energy eigenstates depend on the system, not on the initial wavefunction.

 

[Vitani1]

I would do this but I don't have a way to scan my papers. I have the correct result. I'll try and put it into "better" words. The time-dependent solution can be obtained by multiplying your independent solution by a complex exponential which includes hbar, i, and En. I've written the problem in terms of Fourier basis functions for psi(x,0). Now I would like to multiply both solutions in each region by this exponential to obtain the time-dependent SE. When I do this do I multiply the constant term (the first term in the second region between a/2 and a) by this exponential or just the second term since there exists on this term an explicit n-dependence?

 

[Vitani1]

Actually what you just said helps... this problem doesn't specify what E is so I guess I'll leave this in terms of E1.

 

[BvU]

this problem doesn't specify what E is so I guess I'll leave this in terms of E1.

This problem statement has no occurrence of the word or term $E$.

It seems to me you approximately know what needs to be done, but don't fully grasp the all-important concept of eigenfunctions

You have a $\Psi(x,0)$ given and you know the solutions for the time independent Schrödinger equation $H\Psi(x) = E\Psi(x)$. The solutions (eigenfunctions of $H$) have energy $E_n$, so $$H\Psi_n(x) = E_n\Psi_n(x)\ .$$
You know the time dependence of the eigenfunctions is $$ \Psi_n(x,t) = \Psi_n(x) \,e^{-i\omega_n t} $$ with $E_n = \hbar \omega_n$.

The Schrödinger equation is linear, so if you can write your given $\Psi(x,0)$ as $$\Psi(x,0)=\sum C_n\Psi_n(x)\ ,$$ then you have $$\Psi(x,t) = C_n\Psi_n(x) \,e^{-i\omega_n t} $$

You determine the $C_n$ from $\Psi(x,0)$ by evaluating the integrals $$\int_{-\infty}^{+\infty} \Psi_n^* \, \Psi(x,0) \; dx\ ,$$which in this exercise means evaluating Fourier coefficients.

Can I ask you to confirm that you understand all this ?
$\$
[edit] I notice I used all uppercase $\Psi$, where Griffith is more clear and uses lower case for the eigenfunctions.

 

[Vitani1]

Yes. I understand.

 

[BvU]

Excellent. So, where are we in this exercise ?

1. $A$ is determined
2. An expression for $\Psi(x,t)$ is found
3. The probablility to find $E_1$
4. <Energy> of $\Psi(x,t)$

I would do this but I don't have a way to scan my papers. $\qquad$ Don't bother unless you can do better and neater than a textbook. Exception: the sketch asked for in part (a) of the exercise

I have the correct result. I'll try and put it into "better" words. $\qquad$Don't bother -- no one will read the reams of text to comprehesively and unequivocally express even a few equations

The time-dependent solution can be obtained by multiplying your time independent solution by a complex exponential which includes hbar, i, and En. $\qquad$Correct only for eigenfunctions. For the given initial state (which is a sum of time-independent solutions), this multiplication has to be done term by term (each term has a different $E_n$).

I've written the problem in terms of Fourier basis functions for psi(x,0). $\qquad$Not the problem, but the given initial state, I hope . Do you think you can share the outcome with us (see below) ?

Now I would like to multiply both solutions in each region by this exponential to obtain the time-dependent SE. $\qquad$Both solutions ? each region ?

When I do this do I multiply the constant term (the first term in the second region between a/2 and a) by this exponential or just the second term since there exists on this term an explicit n-dependence?$\qquad$You have lost me (which is why I posted #15)

Free tip: instead of typing

obtaining E1 where En = n^2pi^2hbar^2/2ma^2.

learn a little bit of $\ \LaTeX \$ and type

$$E_n = \frac {n^2\pi^2 \hbar^2} { 2ma^2} $$​

​

to get $$E_n = \frac {n^2\pi^2 \hbar^2} { 2ma^2} $$Oh boy, what a long post

$\$

 

[Vitani1]

This homework is completed. Thank you.