Quantum Mechanics problem

[dkv]

Since I have my own opinions therefore I thought it would be better to post the thread here.I have a question. In Quantum Mechanics there is something called probability density which is a solution of Schrödinger equation.

For particle in a Box the solutions are very simple ,phi=sqrt(2/L)*sin(kx)

momentum p=nh/2L , energy =n^2*h^2/8mL^2

The initial conditions are given as at x=0, phi=0 and at x=L, phi=0

Particle in a box - Wikipedia, the free encyclopedia

 

However if I change the phi and initial conditions then will there be any change in the observables i.e the energy and momentum ?

 

For example if I let phi=Asin(kx)+B where B is a constant.

Initial conditions are given as x=0 phi=B and at x=L phi=B

i.e the probability densities are not exactly zero at x=0,L

 

Thanks

[dkv]

I do not find any change in the Energy and momentum at all. Moreover there are infinite wave function solutions with different initial conditions which satisfy the Schrodinger's equation and give the same energy and momentum..

The wave function remains normalizable.

Thus there is no definite shape of the probability distribution of particle in the box.

 

Energy and Momentum remain discreet.

[Erasmus00]

However if I change the phi and initial conditions then will there be any change in the observables i.e the energy and momentum ?

 

For example if I let phi=Asin(kx)+B where B is a constant.

Initial conditions are given as x=0 phi=B and at x=L phi=B

i.e the probability densities are not exactly zero at x=0,L

 

The probability needs to be 0 at x=0 and L or else your wave function isn't continuous (or fails to meet the boundary condition of 0 at infinity). You aren't free to add a constant (it solves the problem but not physical boundary conditions).

-Will

[dkv]

Hello Will,

The wave function remains continuous within the range.

My question is why do we need probability to be zero at x=0 when it is possible to find the particle at x=0?

For a large particle it is understandable but for small atomic ones this need not be the case because boudary is largely empty ..(more 90% empty)

 

Moreover uncertainity principle demands that we can not know the position of a particle and its momentum with any certainity more than h/2pi.

At x=0 we know that particle does not exist(i.e momentum is zero because mass is zero).

 

It is possible to have a finite probability at x=0 without changing the energy and momentum of the particle within the box.

[dkv]

Infact the Energy and Momentum remains unchanged in the Hydrogen Atom even if we change the probability density function by a constant...

Which means that there is no unique solution to the probability density.

[Jay-qu]

Im not sure where you are throwing this constant in, but you should note that the wavefunction remains unchanged when it undergoes rotation of phase. This invariance to a rotation done in U(1) provides the starting point from which QED can be derived.

[dkv]

In the integration of the wave function we can introduce the constant.I am not talking about the phase rotation.. I am talking about non-zero probabilities.

Meaning that the probabilities always remains greater than 0 at any point within the defintion of the problem.

The non-zero probabilities at the boundary conditions do not change the Energy or Momentum distribution inside the box or inside an atom.

For simplicity can you find any mistake in the box argument? No.

Solving for the particle in the box there are infinite number of wave function solutions ,with different non-zero initial conditions, giving the same discreet Energy and Momentum.

[Erasmus00]

Hello Will,

The wave function remains continuous within the range.

 

It needs to be continuous over ALL space. Outside of the box (if its an infinite well) the wavefunction must be 0 to solve Schroedinger. Therefore to be continuous at the boundary it must be 0.

 

Moreover uncertainity principle demands that we can not know the position of a particle and its momentum with any certainity more than h/2pi.

At x=0 we know that particle does not exist(i.e momentum is zero because mass is zero).

 

You cannot apply the uncertainty principle to a single point in space- its about distributions. Uncertainty is defined with standard deviation of a distribution and its mean.

 

It is possible to have a finite probability at x=0 without changing the energy and momentum of the particle within the box.

 

Which is fine, but it doesn't solve the Schroedinger equation with a continuous result.

-Will

[Erasmus00]

Infact the Energy and Momentum remains unchanged in the Hydrogen Atom even if we change the probability density function by a constant...

Which means that there is no unique solution to the probability density.

 

Your conclusion doesn't follow from your premise. A family of solutions can all have the same energy and momentum, which are differential operators. However, only one of the family satisfies the boundary conditions. You aren't free to set your boundary conditions haphazardly, they fall out from the requirement that schroedinger's equation must be satisfied in the wall of the box AND the wavefunction must be continuous.

-Will

[dkv]

Wave functions are only continous within the range.. It is not possible to create arbitarily continuos equations because potentials can change sharply. The known function of the particle in the Box is continuous within the range of conditions.

 

Changing the intial conditions doesnt make the function discontinuous. Can you give me an example where the equation becomes discontinuous ??

psi= A sin(kx) + b remains continuous at all points inside the box. (not outside the box because function is not defined outside the infinite potential)

You can still perform the differentiation and integration on the modified equation.

 

What I said was nothing new.. Probability is never zero in the physical situations.

That is why a family of solutions can exist.

In the GR theory the solutions of einstein equations left the scope for a constant which Einstein incorporated initially but later called it the greatest mistake of his life.But today scientists are finding that such a constant can account for the virtual energy filled in the vacuum of the space.

Similarly by incorporating the missing constant in the wave function(no matter how small) we can predict the probabilistic interaction of particles at the boundary..

What happens when there is a minimum probability of at the boundaries? What happens if particles occasionaly interacts with the boundary?

For example if particle is photon then at the boundary what happens? Will it escape out of the box ? Will it get absorbed by the atoms on the boundary?

In my opinion if the boundary accepts the photon energy then sooner or later it will get absorbed and will leave the system.

These are some of interesting questions which arises due to this slight modification.

[Erasmus00]

Wave functions are only continous within the range.. It is not possible to create arbitarily continuos equations because potentials can change sharply.

 

Changing the potential sharply allows for a discontinuous first derivative of the wavefunction. The wavefunction must ALWAYS be continuous. Also, Schroedinger's equation must be valid everywhere the potential is defined (all space in the case of the particle in a box square well). This is a common problem discussed in any quantum textbook. You are simply wrong.

 

Changing the intial conditions doesnt make the function discontinuous. Can you give me an example where the equation becomes discontinuous ??

psi= A sin(kx) + b remains continuous at all points inside the box. (not outside the box because function is not defined outside the infinite potential)

 

This is incorrect. Schroedinger tells us

 

[math]-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi +V\psi = E\psi[/math]

 

Outside the well, where V heads to infinity, we must take psi to 0 to keep the equation valid. So the boundary conditions of an infinite well are that the wavefunction is continuous, at the boundary you have to match onto the wavefunction outside the well (which is 0 everywhere).

 

If you wish to discuss more realistic problems where the particle's probability will not head to 0 at the wall, look at particles in finite square wells. This is a slightly more difficult problem, but should also be in any intro quantum book.

-Will

[dkv]

I agree that the psi should be equal to 0 but it is not necessary.Solving the second order differential equation allows the scope for a constant...

As far as the original solution psi=Asin(kx) is concerned we integrate it over 0 to and L (length of box)...outside the box the porbability is simply assumed to be zero. It is assumed to be zero because the infinite potential doesnt allow the function to be defined continuously over the entire space.

In schrodingers equation V(x)=0 inside the box and outside it is infinite making it discontinous.

Why the probability outside is zero is a matter of defintion of the problem.. but at the boundary x=0 and x=L ,where V(x) is still equal to 0,we can assume some finite probability. The modified function remains well defined and integrable without changing the observables i.e energy and momentum.

[Erasmus00]

I agree that the psi should be equal to 0 but it is not necessary.Solving the second order differential equation allows the scope for a constant...

 

No it doesn't, psi appears directly in the equation, if we want to be allowed to make V infinitely large (more technically, if we start with large V and take a limit), the only way to have a well behaved equation is for psi to be 0.

 

As far as the original solution psi=Asin(kx) is concerned we integrate it over 0 to and L (length of box)...outside the box the porbability is simply assumed to be zero. It is assumed to be zero because the infinite potential doesnt allow the function to be defined continuously over the entire space.

In schrodingers equation V(x)=0 inside the box and outside it is infinite making it discontinous.

 

The probability isn't assumed 0, it is the solution to the equation! And yes, the potential is discontinuous, however, the wavefunction is not, and cannot be.

 

The modified function remains well defined and integrable without changing the observables i.e energy and momentum.

 

It isn't well defined, its discontinuous at the boundary (0 on the outside, non-zero on the inside of the well at the same point).

-Will

[dkv]

What stops me from putting x= 2L in the equation psi=sqrt(2/L)sin(kx)? Nothing but the definition of the problem..

At the boundary the V(x) is still equal to 0 because point x=0 is definable. The range in which V(x) is zero includes both x=0 and x=L. There is nothing which stops the probability density from taking any arbitary value at the boundaries..

In real number system infinities are left outside the real set because they are undefined. Therefore if we include x=0,L within the defined space the V(x) remains 0.I see no reason why the psi should must be equal to zero.

[Erasmus00]

Therefore if we include x=0,L within the defined space the V(x) remains 0.I see no reason why the psi should must be equal to zero.

 

Try this- solve a finite well with V not equal to infinity, and then take the limit when V approaches infinity. Maybe this will help you understand. Report back here with results and I will help you where I can.

-Will

[dkv]

At infinite potential the probability density becomes zero. However we can still introduce a constant in the exponential part of the solution. e^(-ax+:shrug: thereby introducing a finite probability density at any arbitary boundary...

This will not change the energy and momentum because energy and momentum depend only upon the alpha and k.

[dkv]

These equations will modify

Particle in finite-walled box