# Finding the Probability for a Particle in an Infinite Potential Well

• Irishdoug
In summary, the wavefunction for a particle in an infinite potential well of width a is ##\phi_n=\sqrt{\dfrac{2}{a}}\sin\left(\dfrac{n\pi x}{a}\right)## where n=0,2,4... as sine is an even function. The expansion coefficients for this wavefunction are ##c_{1}=\sqrt{\frac{15}{a}}##. The probability that the particle is found in the n=1 state when it's energy is measured is ##\frac{a\hbar^2}{m} \sqrt{\frac{15}{a}}.
Irishdoug
Homework Statement
Hi, I am self-studying QM and was wondering if you could check to see if my answers are right to an online exam I took. There are no solutions availabel
Relevant Equations
See below
Homework Statement: Hi, I am self-studying QM and was wondering if you could check to see if my answers are right to an online exam I took. There are no solutions availabel
Homework Equations: See below

A particle with mass m is in an infinite potential well of length x=0 to x=a
Q1a. Show explicitly that the wave-function ##\phi_{n}## = ##\frac{\sqrt{2}}{a}## sin(n##\pi##x/a) are eigenfunctions of the Hamiltonian of this system H= ##\frac{\hbar^2}{2m}## ##\frac{d^2}{dx^2}## and determine the corresponding eigen-values.

##\frac{\hbar^2}{2m}## ##\frac{d^2\phi}{dx^2}## = E##\phi##

##\frac{\hbar^2}{2m}## ##\frac{d^2 sin(n\pi x/a)}{dx^2}## = ##\frac{\sqrt{2}}{a}## ##\frac{-\hbar^2}{2m}## ##\frac{n \pi x}{a}## = ##\frac{\hbar^2 \pi^2 n^2 \sqrt{2}}{2ma^3}##

so, ##\frac{\hbar^2}{2m}## ##\frac{d^2\phi}{dx^2}## = E##\phi## where E=##\frac{\hbar^2 \pi^2 n^2 \sqrt{2}}{2ma^3}##

The eigen-values are thus E= ##\frac{\hbar^2 \pi^2 n^2 \sqrt{2}}{2ma^3}## were n = 0,2,4... as sine is an even function.

b. What is the physical interpretation of ##\phi^2##?

This is the probability of finding that particle in a certain location x, where x is between 0 and a.

c. What is the physical reason to demand that the wavefunction be normalised?

The physical reason is that QM is probabilistic in nature. A probability cannot be greater than 1. Therefore the wavefunction is normalized, i.e.

##\int_{-\infty}^{\infty} \phi^* \phi dx## = 1

to ensure this probability doesn't exceed one, which would be unphysical.

2. At time t=0 a particle is in an infinite potential well of width a prepared in a state ##\phi## = Nx(a-x)
a. Determine N so the wavefunction is properly normalized

##\int_{-\infty}^{\infty} \phi^* \phi dx## = 1 --> ##\int_{-\infty}^{\infty} N^2(ax-x^2)^2 dx##

Take N outside, the well is symmetrical so the limits of integration become 0 to a, and the integral is multiplied by 2.

End up with:
2##N^2## (##\frac{a^2a^3}{3} -\frac{aa^4}{2}+\frac{a^5}{5}##) = 2##N^2##*##\frac{a}{30}## so N = ##\sqrt{\frac{15}{a}}##

b. Although it's not an eigenfunction of the infinite potential well Hamiltonian, this wavefunction can be expanded in the eigenfunctions of this Hamiltonian

##\psi_{x} = \sum_{n=1}^{\infty} c_{n} \psi_{n}(x)##

where ##c_{n}## are the expansion coefficients and ##\phi_{n}## the infinite potential well wavefunctions. Calculate ##c_{1}##.

##\phi(x)## = ##\sqrt{\frac{15}{a}} x(a-x) ## ; ##c_{n} = \int_{-\infty}^{\infty} x\phi_{n}##

so, ##c_{n}## = ##\int_{-\infty}^{\infty} \sqrt{\frac{15}{a}} x^2(ax-x^2)##

Assume symmetrical so limits of integration go from 0 to a and integral is muliplied by 2

End up with 2##\sqrt{\frac{15}{a}}(\frac{a^5}{4}-\frac{a^5}{5})## = 2##\sqrt{\frac{15}{a}}(\frac{a^5}{20})## = ##\sqrt{\frac{15}{a}}(\frac{a^5}{10})## = ##c_{1}##

c. Find the probability that the particle is found in the n=1 state when it's energy is measured.

For this one I carried out the expectation value for E i.e.

<E> = ##\int_{-\infty}^{\infty} \frac{-\hbar^2}{2m} \frac{d^2}{dx^2}(\sqrt{\frac{15}{a}}x(a-x)dx##

I ended up with ##\frac{a\hbar^2}{m} \sqrt{\frac{15}{a}}##

For the last two questions I was more guessing at the methodology however. Thanks for any input.

Part (1a)
Your algebra needs improving and you need to be honest with it. For example, if you have/need an extra negative sign don't fudge it; go back and see where it was/was not introduced. Here you did not use ##-\frac{\hbar^2}{2m}\frac{d^2\phi}{dx^2}## in the Hamiltonian and you ended up with a negative sign in front of the energy which you swept under the rug.
The wavefunction that you posted is not normalized. It should be ##\phi_n=\sqrt{\dfrac{2}{a}}\sin\left(\dfrac{n\pi x}{a}\right)##. You should re-derive the energy; it looks like you made mistakes taking derivatives.
Irishdoug said:
... as sine is an even function.
The sine is an odd function (##\sin(nx)=-sin(-nx)##). The parity of the wavefunction does not enter the picture because you do not have a symmetric potential.

Part(1b)
Irishdoug said:
This is the probability of finding that particle in a certain location x, where x is between 0 and a.
Not quite. You need to clarify for yourself the difference between probability and probability density.

Part (1.c)
OK.

Irishdoug said:
... the well is symmetrical so the limits of integration become 0 to a
The limits of integration are 0 to a in the first place. Read the problem !
Irishdoug said:
A particle with mass m is in an infinite potential well of length x=0 to x=a ...
Please fix all of the above and re-post. Maybe someone will pick it up from there if you have more questions.

PeroK, Michael Price, berkeman and 1 other person
Brilliant thanks for your repsonse. Apologies, I missed out the negative sign when posting on this forum I did it in my actual work. Also the normalisation was a mistake I made in latex, again I have ##\sqrt{\frac{2}{a}}## in my actual work. I cannot edit the orginal post anymore. I'll be more careful next time however.
I'll look at my algebra and derivatives and repost.

Thanks again.

Irishdoug said:
I'll look at my algebra and derivatives and repost.

1. if you want to find coefficients ##c_n## such that ##\phi(x) = \sum_{n=1}^{\infty} c_{n} \psi_{n}(x)##, then ##c_n=\int \phi(x)\psi_n(x)dx##, not what you have in (2b).
2. if ##\phi(x) = \sum_{n=1}^{\infty} c_{n} \psi_{n}(x)## and ##\phi(x)## is normalized, it is true (please show the following if you have not already done so) that (a) ##\sum_{n=1}^{\infty} |c_{n}|^2=1## and (b) the expectation value for the energy is given by ##\langle E \rangle=\sum_{n=1}^{\infty} |c_{n}|^2E_n##.

berkeman
A particle with mass m is in an infinite potential well of length x=0 to x=a
Q1a. Show explicitly that the wave-function ##\phi_{n}## = ##\sqrt{\frac{2}{a}}## sin(n##\pi##x/a) are eigenfunctions of the Hamiltonian of this system H= -##\frac{\hbar^2}{2m}## ##\frac{d^2}{dx^2}## and determine the corresponding eigen-values.

-##\frac{\hbar^2}{2m}## ##\frac{d^2\phi}{dx^2}## = E##\phi##

-##\frac{\hbar^2}{2m}## ##\frac{d^2 sin(n\pi x/a)}{dx^2}## = ##\sqrt{\frac{2}{a}}## ##\frac{-\hbar^2}{2m}## ##\frac{n^2 \pi^2}{a^2}## -sin(##\frac{n \pi x}{a}##) = ##\frac{\hbar^2 \pi^2 n^2}{2ma^2}## ##\sqrt{\frac{2}{a}}##sin(##\frac{n\pi x}{a}##)

so, ##\frac{\hbar^2}{2m}## ##\frac{d^2\phi}{dx^2}## = E##\phi## where E=##\frac{\hbar^2 \pi^2 n^2}{2ma^2}##

The eigen-values are thus E= ##\frac{\hbar^2 \pi^2 n^2}{2ma^2}## were n = 0,1,3... as sine is an odd function.

Am I correct in my belief that the energy levels are the eigen-values as they are the scalars of the eigenfunctions and thus meet the correct definition of eigenvalues?

b. What is the physical interpretation of ##\phi^2##?

This is the probability of finding the system in a particular eigenstate, when an ensemble of measurements is made.

2. At time t=0 a particle is in an infinite potential well of width a prepared in a state ##\phi## = Nx(a-x)
a. Determine N so the wavefunction is properly normalized

##\int_{-\infty}^{\infty} \phi^* \phi dx## = 1 --> ##\int_{-\infty}^{\infty} N^2(ax-x^2)^2 dx##

Take N outside, the limits of integration are 0 to a.

End up with:
##N^2## (##\frac{a^2a^3}{3} -\frac{aa^4}{2}+\frac{a^5}{5}##) = ##N^2##*##\frac{a^5}{30}## so N = ##\sqrt{\frac{30}{a^5}}##

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kuruman said:

1. if you want to find coefficients ##c_n## such that ##\phi(x) = \sum_{n=1}^{\infty} c_{n} \psi_{n}(x)##, then ##c_n=\int \phi(x)\psi_n(x)dx##, not what you have in (2b).
2. if ##\phi(x) = \sum_{n=1}^{\infty} c_{n} \psi_{n}(x)## and ##\phi(x)## is normalized, it is true (please show the following if you have not already done so) that (a) ##\sum_{n=1}^{\infty} |c_{n}|^2=1## and (b) the expectation value for the energy is given by ##\langle E \rangle=\sum_{n=1}^{\infty} |c_{n}|^2E_n##.
Ok so in trying to find how to do this question, I found that it is taken straight from Griffiths Intro to QM, so when I do the math I can just look at that for this part, to ensure it's correct.

But I just want to be clear I understand what's going on. To find the expansion coefficient, I take my prepared state ##\psi## and multiply it by the wavefunction ##\phi## that corresponds to the expansion co-efficient I am looking for i.e. ##\phi_{1}## for ##c_{1}##, ##\phi_{2}## for ##c_{2}## etc. I than take the integral of that with the appropriate limits of integration.

This gives me ##c_{n}##, which when squared tells me the probability of measuring the observable ##E_{n}##, which is some number? Is this reasoning correct?

The last part of the question I will then simply be using <H>=##c_{n}^2####E_{n}## which should give me a numerical value less than 1?

Yes, this is a problem out of Griffiths.

Irishdoug said:
But I just want to be clear I understand what's going on. To find the expansion coefficient, I take my prepared state ψ\psi and multiply it by the wavefunction ϕ\phi that corresponds to the expansion co-efficient I am looking for i.e. ϕ1\phi_{1} for c1c_{1}, ϕ2\phi_{2} for c2c_{2} etc. I than take the integral of that with the appropriate limits of integration.
Your understanding is correct. Think of ##\phi(x)## as a vector and the ##\psi_n(x)## as unit vectors. Then the integral can be thought of as the component (or projection) of the vector on the unit vector. It's analogous to saying ##A_i=\vec A \cdot ~\hat x_i##.
Irishdoug said:
This gives me ##c_{n}##, which when squared tells me the probability of measuring the observable ##E_{n}##, which is some number? Is this reasoning correct?
Watch your language. Correctly said, ##|c_n|^2## is the probability of obtaining result ##E_n## when you measure the observable. The probability of measuring the observable depends on how determined you are to do the measurement, if you have more pressing business, to what extent you allow yourself to be distracted by phone calls, etc.
Irishdoug said:
The last part of the question I will then simply be using ##<H>=c_{n}^2E_{n}## which should give me a numerical value less than 1?
No. As I wrote in post #4,$$\langle H \rangle=\langle E \rangle=\sum_{n=1}^{\infty} |c_{n}|^2E_n.$$This is the expectation or average value for the energy. Just like any average it implies making the measurement many times and then taking the average. If you make ##N## measurements the well known form for writing the average is$$\langle E \rangle=\frac{\sum_{i=1}^{\infty} N_iE_i}{\sum_{i=1}^{\infty} N_i}=\frac{\sum_{i=1}^{\infty} N_iE_i}{N}$$where ##N_i## is the the number of measurements that yielded value ##E_i##. Knowing that out of a total of ##N## measurements, ##N_i## of them returned value ##E_i##, you would say that the probability that anyone measurement will result in value ##E_i## is $$p_i=\frac{N_i}{N}.$$This means that you can express the average as$$\langle E \rangle=\sum_i p_iE_i.$$Do you see how it works?

kuruman said:
Yes, this is a problem out of Griffiths.Your understanding is correct. Think of ϕ(x)ϕ(x) as a vector and the ψn(x)ψn(x) as unit vectors. Then the integral can be thought of as the component (or projection) of the vector on the unit vector. It's analogous to saying Ai=→A⋅ ^xiAi=A→⋅ x^i.

Watch your language. Correctly said, |cn|2|cn|2 is the probability of obtaining result EnEn when you measure the observable. The probability of measuring the observable depends on how determined you are to do the measurement, if you have more pressing business, to what extent you allow yourself to be distracted by phone calls, etc.

I'd to read it a few times but I understood what you meant eventually haha. I'll be more careful going forward. Cheers.

kuruman said:
No. As I wrote in post #4,

⟨H⟩=⟨E⟩=∞∑n=1|cn|2En.⟨H⟩=⟨E⟩=∑n=1∞|cn|2En.​

This is the expectation or average value for the energy. Just like any average it implies making the measurement many times and then taking the average. If you make NN measurements the well known form for writing the average is

⟨E⟩=∑∞i=1NiEi∑∞i=1Ni=∑∞i=1NiEiN⟨E⟩=∑i=1∞NiEi∑i=1∞Ni=∑i=1∞NiEiN​

where NiNi is the the number of measurements that yielded value EiEi. Knowing that out of a total of NN measurements, NiNi of them returned value EiEi, you would say that the probability that anyone measurement will result in value EiEi is

pi=NiN.pi=NiN.​

This means that you can express the average as

⟨E⟩=∑ipiEi.⟨E⟩=∑ipiEi.​

Do you see how it works?

Ok, apologies if I'm coming across as dumb. I understand the maths you gave but I'm confused about it finding the probability it's in state n=1 as per the question:

c. Find the probability that the particle is found in the n=1 state when it's energy is measured.

I presume this is the same as asking for the probability it is observed in state ##E_{1}##, thus I just need to calculate ##|c1_{1}|^2##? Is this correct?

Whereas with the expectation value, it is not a probability but an average value of E whose value is dependent on the number of measurements made. It differs from a "normal" average in the sense that it does not tell us the most likely value we will get if we make one single measurement, and can differ from the value of the allowed energy levels.

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Irishdoug said:
I presume this is the same as asking for the probability it is observed in state ##E_{1}##, thus I am just need to calculate ##|c_{1}|^2##? Is this correct?
Correct.

berkeman
kuruman said:
Correct.

berkeman
Irishdoug said:
A particle with mass m is in an infinite potential well of length x=0 to x=a
Q1a. Show explicitly that the wave-function ##\phi_{n}## = ##\sqrt{\frac{2}{a}}## sin(n##\pi##x/a) are eigenfunctions of the Hamiltonian of this system H= -##\frac{\hbar^2}{2m}## ##\frac{d^2}{dx^2}## and determine the corresponding eigen-values.

-##\frac{\hbar^2}{2m}## ##\frac{d^2\phi}{dx^2}## = E##\phi##

-##\frac{\hbar^2}{2m}## ##\frac{d^2 sin(n\pi x/a)}{dx^2}## = ##\sqrt{\frac{2}{a}}## ##\frac{-\hbar^2}{2m}## ##\frac{n^2 \pi^2}{a^2}## -sin(##\frac{n \pi x}{a}##) = ##\frac{\hbar^2 \pi^2 n^2}{2ma^2}## ##\sqrt{\frac{2}{a}}##sin(##\frac{n\pi x}{a}##)

so, ##\frac{\hbar^2}{2m}## ##\frac{d^2\phi}{dx^2}## = E##\phi## where E=##\frac{\hbar^2 \pi^2 n^2}{2ma^2}##

The eigen-values are thus E= ##\frac{\hbar^2 \pi^2 n^2}{2ma^2}## were n = 0,1,3... as sine is an odd function.
There is nothing in the mathematics that restricts ##n## to odd values. The relationship you found still works when ##n=2## for instance.

Am I correct in my belief that the energy levels are the eigen-values as they are the scalars of the eigenfunctions and thus meet the correct definition of eigenvalues?
Yes, the energies are the eigenvalues of the Hamiltonian.

Irishdoug
Hi, one final question then I'll leave the thread be.

I've done another question, again it's an infinite potential well. One part asks: "Are there energy Eigen-values that will never be observed?"

I
I answered: In theory there is a probability that all energy Eigenvalues can be observed, but there is a limit to the amount of time one can spend taking measurements meaning in a practical sense not all Energy eigen-values will be observed?

Right, wrong, not quite correct?

Still trying to wrap my head around it all.

Irishdoug said:
Hi, one final question then I'll leave the thread be.

I've done another question, again it's an infinite potential well. One part asks: "Are there energy Eigen-values that will never be observed?"
You say another question. Ideally we require that "other" questions be posted on separate threads but OK this time if your wavefunction is the same ##\psi(x)=Nx(x-a)##. In general, to determine whether an eigenvalue will or will not be observed, you need to look at the coefficients of the expansion, ##c_i##. For example, if the wavefunction of a particle in a box is $$\psi(x)=\sqrt{\frac{2}{a}}\left[\frac{1}{2}\sin\left(\frac{2\pi x}{a}\right) +\frac{\sqrt{3}}{2}\sin\left(\frac{4\pi x}{a}\right)\right],$$then the energies that will not be observed as a result of a measurement are all except ##E_2## and ##E_4##. In other words any ##E_i## for which ##c_i=0## in the expansion will not be observed. Why? Because its probability ##|c_i|^2## of being observed is zero. Makes sense, no? So to answer the question properly, you need to determine which coefficients (if any) in your expansion are zero.

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Irishdoug
It's a different wave-function but I understand the principle behind how to answer it based off your answer. Thankyou.

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