100 boxes of Length L, an application of the famous Particle in A Box

[PhysicsTruth]

TL;DR

Particle in a box problem in infinite potential well with an ensemble of 100 boxes

Suppose I have 100 identical boxes of length L and the coordinates are x=0 at one end of the box and x=L at the other end, for each of them. Each has a particle of mass m. V=0 in [0,L], while it's equal to infinity in the rest of the regions. If I make a measurement on position of the particle on all the boxes at the *same time*, in how many boxes would the particle be expected to be found between x=0 and x=L/4 ?

My Approach: Probability of finding the mass m in a single box at a specific time is the integration of |psi(x)|^2 dx from x=0 to L/4. But how can we talk about the probability of 100 such identical boxes, measured at the same time? How does the probability of finding the mass m change in more than one box? How do I actually get a *number* of boxes in which the mass m can be expected to be found in [0,L/4]?

 

[Dale]

Once you have the probability, p, for one box then the probability distribution function for 100 boxes is just the binomial distribution with $B(n=100,p=p)$

 

[PeterDonis]

Each has a particle of mass m

In what state? The ground state for the given potential well?

If I make a measurement on position of the particle on all the boxes at the *same time*

Why does "at the same time" matter? If the particles are in stationary states, the probabilities for measurement results don't change with time, so it doesn't matter at what time you make the measurements on each particle.

 

[PhysicsTruth]

In what state? The ground state for the given potential well?
Why does "at the same time" matter? If the particles are in stationary states, the probabilities for measurement results don't change with time, so it doesn't matter at what time you make the measurements on each particle.

Yeah, in ground state.

 

[PhysicsTruth]

Once you have the probability, p, for one box then the probability distribution function for 100 boxes is just the binomial distribution with $B(n=100,p=p)$

How can you say that it's a binomial distribution, like I don't really get it intuitively why do I use a binomial distribution

 

[PhysicsTruth]

In what state? The ground state for the given potential well?
Why does "at the same time" matter? If the particles are in stationary states, the probabilities for measurement results don't change with time, so it doesn't matter at what time you make the measurements on each particle.

Yeah, same time doesn't matter, because they are the eigenstates and the wave-function collapses to a single eigen-state on measurement. Nothing more.

 

[Dale]

How can you say that it's a binomial distribution, like I don't really get it intuitively why do I use a binomial distribution

A Bernoulli trial is something that either happens (success) or does not happen (failure) with probability p of it happening and probability q = 1-p of it not happening. So in this case, each box is a Bernoulli trial. Either the particle is found in L/4 or it is not, and the probability for it being in L/4 is p. That is a Bernoulli trial.

The binomial distribution describes the number of successes in n Bernoulli trials. So you have 100 boxes, each is a Bernoulli trial, therefore the number of successes in 100 boxes is a binomial.

 

[PhysicsTruth]

A Bernoulli trial is something that either happens (success) or does not happen (failure) with probability p of it happening and probability q = 1-p of it not happening. So in this case, each box is a Bernoulli trial. Either the particle is found in L/4 or it is not, and the probability for it being in L/4 is p. That is a Bernoulli trial.

The binomial distribution describes the number of successes in n Bernoulli trials. So you have 100 boxes, each is a Bernoulli trial, therefore the number of successes in 100 boxes is a binomial.

According to my meagre knowledge, I know how to calculate the probability of *n* successes in a binomial distribution, or the *atleast* or *atmost* clauses, but in this case how do we find the actual number of boxes? Can you weigh in a bit more? Thanks.

 

[Vanadium 50]

but in this case how do we find the actual number of boxes

It's 100. You told us that.

If you mean something else, please explain what you mean.

 

[PhysicsTruth]

It's 100. You told us that.

If you mean something else, please explain what you mean.

No I meant the no. Of boxes where the particle would be expected to be found in [0,L/4]

 

[PhysicsTruth]

Ok,

According to my meagre knowledge, I know how to calculate the probability of *n* successes in a binomial distribution, or the *atleast* or *atmost* clauses, but in this case how do we find the actual number of boxes? Can you weigh in a bit more? Thanks.

Ok, I think I know it Is it nP, where n is the number of trials and P the probability of success in each trial?

 

[Dale]

According to my meagre knowledge, I know how to calculate the probability of *n* successes in a binomial distribution, or the *atleast* or *atmost* clauses, but in this case how do we find the actual number of boxes? Can you weigh in a bit more? Thanks.

It is a random number, so there is no way to know in advance the actual number. The expected value (mean) for a binomial variable $\text{mean}(B(n,p))=np$

You can find all of this information here:
https://en.wikipedia.org/wiki/Binomial_distribution