Selective nature of Schrodinger Equation

[sidharath]

When Schrodinger equation is applied to harmonic oscillator solution is possible

when particle is moving away from mean position because of positive potential energy therefore equation can be changed to Hermite polynomial but when particle is moving towards mean position because of negative potential energy solution is not acceptable because it cannot be changed to Hermite poltnomial which shows selective nature of Schrodinger equation.Do you agree that there is some flaw in equation and it needs to changed

[swansont]

No. You're doing it wrong.

 

A system with positive potential energy is not bound, and is not a harmonic oscillator. OF COURSE you get a different solution than with a negative potential.

[sidharath]

dear swansont while applying Schrodinger equation sign of potential energy has to be taken into account . In case of SHO it is only at the extreme position that potential energy is positive when particle has reached there while moving away from mean position otherwise total energy is partly potential and kinetic energy. Potential energy at the mean position is zero. If usual norm of calculating potential enrgy is adopted it is seen that at any point it is with positive sign when particle is moving away from mean position but you say that positive potential energy is not possible but equation with acceptable solution is possible only with positive sign of potential .When the particle is moving towards mean position under attractive force hence particle is doing work which leads to negative potential energy but equation cannot be solved when sign is negative., this is my contention.According to me your reply is not quite to the point because you say that positive potential energy is not possible but solution is possible only when potential energy is positive. i request you to elaborate it.

[studiot]

Forgive the obvious, but for a particle is the 'vibration' not transverse, so it is neither moving away from nor towards the centre?

 

Hermite polynomials have radial symmetry and apply to shells which do indeed vibrate in and out, but then shells are the other face of quantum mechanics and not particles.

Edited August 9, 2013 by studiot

[swansont]

dear swansont while applying Schrodinger equation sign of potential energy has to be taken into account . In case of SHO it is only at the extreme position that potential energy is positive when particle has reached there while moving away from mean position otherwise total energy is partly potential and kinetic energy. Potential energy at the mean position is zero. If usual norm of calculating potential enrgy is adopted it is seen that at any point it is with positive sign when particle is moving away from mean position but you say that positive potential energy is not possible but equation with acceptable solution is possible only with positive sign of potential .When the particle is moving towards mean position under attractive force hence particle is doing work which leads to negative potential energy but equation cannot be solved when sign is negative., this is my contention.According to me your reply is not quite to the point because you say that positive potential energy is not possible but solution is possible only when potential energy is positive. i request you to elaborate it.

 

A harmonic oscillator requires a bound system. The PE will always be negative if you have defined it to be zero far away from the source of the interaction. If you define it differently you should still get the same answer, though the math may be more difficult.

 

Whether the particle is moving toward or away from the mean position is moot, because the potential energy is defined by the position, and not the velocity. You will get different solutions if the form of the potential energy is different.

 

Do you have an actual example problem where this is an issue?

[sidharath]

dear swansont i need sign of potential energy when potential is used in Schrodinger equation at basic level while equation is used to find out the energy of particle of mass m executing harmonic oscillation.Th basic equation is (laplacian operator)psi+constant(E-V)psi=0 V is potential energy. While using this equation to find out value of energy E attention has to be paid to sign of V, as in the case of hydrogen atom sign of V is negative due to attractive force. In case of particle of mass m executing oscillations there is motion towards and away from mean position. When the particle is moving away from mean position work has to be done on it because the direction of force is opposite to displacement therefore according established rules potential V has got positive sign .The solution of the above equation with positive V is acceptable because it can be reduced to Hermite polynomial.When oscillating particle is moving towards mean position the displacement and force being in the same direction particle is doing worK or potential is decreasing therefore sign of potential energy is negative . When V is put as negative in the equation solution is not acceptable because equation cannot be changed to HermitE Polynomial, which shows that application of eqtion is selective in nature . This is my contention. i solicit your views . This is my problem

[studiot]

What you are proposing is tantamount to proposing that each quantum level is 'fuzzy' ie occupies a spread of energy values equal to the peak to peak variation.

 

What evidence do you have for this?

[swansont]

dear swansont i need sign of potential energy when potential is used in Schrodinger equation at basic level while equation is used to find out the energy of particle of mass m executing harmonic oscillation.Th basic equation is (laplacian operator)psi+constant(E-V)psi=0 V is potential energy. While using this equation to find out value of energy E attention has to be paid to sign of V, as in the case of hydrogen atom sign of V is negative due to attractive force. In case of particle of mass m executing oscillations there is motion towards and away from mean position. When the particle is moving away from mean position work has to be done on it because the direction of force is opposite to displacement therefore according established rules potential V has got positive sign . The solution of the above equation with positive V is acceptable because it can be reduced to Hermite polynomial.When oscillating particle is moving towards mean position the displacement and force being in the same direction particle is doing worK or potential is decreasing therefore sign of potential energy is negative . When V is put as negative in the equation solution is not acceptable because equation cannot be changed to HermitE Polynomial, which shows that application of eqtion is selective in nature . This is my contention. i solicit your views . This is my problem

Positive and negative V describe different systems. The solutions are selective, the equation is not.

[studiot]

I am beginning to understand your difficulty.

 

A harmonic oscillator has a constant total energy and partitions this between kinetic and potential energy.

The potential energy is a minumum at the mean point and the kinetic energy a maximum.

 

At all points the total energy may be expressed as a function of velocity v; frequency [math]\nu [/math];

displacement from mean position,x ; amplitude a; force constant f

in an equation given by

 

[math]E = \frac{1}{2}m{\nu ^2} + \frac{1}{2}f{x^2}[/math]

Where the first term is kinetic and the second term potential energy.

 

Note that the potential energy is positive , whether x is away from the centre or towards it since the second potential energy term depends upon x squared. The constant f does not change with direction.

 

So when you come to substitute
[math]V = \frac{1}{2}f{x^2}[/math]
into the Schrodinger equation

There is no question of positive or negative signs, it is always positive.

 

 

You get to the Hermite solution by introducing a new variable s = x/a and allowing the simple constant of integration to become a function of s.

Have you seen this derivation?

Edited August 10, 2013 by studiot

[sidharath]

dear studiot thank you very much indeed for your reply. .My conclusion that sign of potential is postive when the particle is moving away and negative when the particle is moving towards mean position is based on some basic relations . I request you to go through that and tell me whether i am right or wrong.

U(final)-U(initial)=-W U and W are potential energy and work.When the particle is moving away from mean position and it is at x W=-1/2kx^2 potential energy at mean position is 0 therefore U=+1/2kx^2 . If the particle is moving towards mean position that is value of position x changes from x to o again W =-1/2kx^2 Work is calculated by using dot product

U(final)-U(initial)=0-U(initial)=-W=1/2kx^2 therefore U(initial)=-1/2kx^2 or the sign of potential energy is negative when the particle is moving towards mean position. i shall be thankful to you if you point out where i have gone wrong in my contention that sign of potential changes with change in direction My contention is based on fundamentals . It is another matter that to get solution negative sign is ignored but you see potential energy sign is not always positive ,it changes with direction of motion.Am i right?

Please allow me to give another example in support of my conclusion that sign of potential energy changes with change in direction.Suppose mass m is moving away from earth , when it is at height h potential energy is positive +mgh when it is moving down at height h potential energy is -mgh so there is change in sign with changed in directions. It is similar to oscillator where particle is moving alternatively in the opposite or same direction as that of restoring force and undergoes change in potential energy sign. please point out where i have gone wrong in my conclusion .You are at liberty to consider only positive sign while applying to Schrodinger equation to keep its general nature but actually it is not so .

So that Schrodinger equation is applicable when the particle is moving towards mean position it should be modified as under

(laplacian operator)psi+constant(E+V)psi=0 V is potential energy with negative sign .It shows that there are two schrodinger equations one when the particle is moving away from the mean position while other form given above is when the particle is moving towards mean position.

[swansont]

dear studiot thank you very much indeed for your reply. .My conclusion that sign of potential is postive when the particle is moving away and negative when the particle is moving towards mean position is based on some basic relations . I request you to go through that and tell me whether i am right or wrong.

U(final)-U(initial)=-W U and W are potential energy and work.When the particle is moving away from mean position and it is at x W=-1/2kx^2 potential energy at mean position is 0 therefore U=+1/2kx^2 . If the particle is moving towards mean position that is value of position x changes from x to o again W =-1/2kx^2 Work is calculated by using dot product

 

The potential energy depends on x. It's right there in the equation. There is no sign change for the opposite direction of motion.

[sidharath]

dear swansont i give here very simple example showing that though numerical value of potential energy remains same there is change in sign with direction of motion 'There is set rule for calculating magnitude and sign of potential energy , sign of potential energy is not arbitrary.Can you solve Schrodinger equation by taking electrostatic potential energy positive instead of negative .It is taken as negative because it is . Rules laid cannot be broken at will. if the mass m is moving down at height h potential energy is -mgh but if it is moving up at the same height potential energy is +mgh so there is change in sign with change in direction.

one way defining potential energy is U(final)-U(initial)=change in kinetic energy.suppose oscillating particle is at the extreme position with maximum potential energy and is about to move towards mean position. because it is moving towards mean position potential energy is-E using the above relation o-(-E)= positive kinetic energy if potential at the extreme position when the particle is about to move towards mean position is taken as positive+E the sign of kinetic energy will be negative which is wrong.i think that above discussion conclusively proves that in oscillator when particle moves towards mean position sign is negative .it is positive when the particle moves away from mean position

[studiot]

 

U(final)-U(initial

 

There is your error.

 

At a single point you cannot have final - initial, there is only one value.

 

The equations I gave provide the value for any point.

This diagram may help you understand.

 

I have sketched the kinetic and potential energies and the total energy, E of a harmonic oscillator v displacement graph.

 

You will note that at no time are any of the curves negative.

 

The KE and PE are parabolas, the total energy line is a constant straight line parallel to the horizontal axis.

 

Edited August 14, 2013 by studiot

[swansont]

dear swansont i give here very simple example showing that though numerical value of potential energy remains same there is change in sign with direction of motion 'There is set rule for calculating magnitude and sign of potential energy , sign of potential energy is not arbitrary.Can you solve Schrodinger equation by taking electrostatic potential energy positive instead of negative .It is taken as negative because it is . Rules laid cannot be broken at will. if the mass m is moving down at height h potential energy is -mgh but if it is moving up at the same height potential energy is +mgh so there is change in sign with change in direction.

No, this is flat-out wrong. The potential energy only depends on the position. If you define your zero to be at the bottom of the travel, then U = +mgh. If you define it at the top, them U = -mgh. You do not change definitions in the middle of a problem — as you say, you can't break the rules at will.

 

 

one way defining potential energy is U(final)-U(initial)=change in kinetic energy.suppose oscillating particle is at the extreme position with maximum potential energy and is about to move towards mean position. because it is moving towards mean position potential energy is-E using the above relation o-(-E)= positive kinetic energy if potential at the extreme position when the particle is about to move towards mean position is taken as positive+E the sign of kinetic energy will be negative which is wrong.i think that above discussion conclusively proves that in oscillator when particle moves towards mean position sign is negative .it is positive when the particle moves away from mean position

 

You have to decide, at the start of the problem, whether U(0) is zero, or U(x_max) is zero. U will depend only on x, and not on v.

 

KE is always positive. Let's pick the point where half of the energy is kinetic and half is potential. The kinetic energy is KE = E_tot/2

If the potential energy depends on the direction, in one case the energy is E_tot/2 and in the other it's -E_tot/2. Meaning the total energy changes from E_tot to zero, within the system that has no external work being done. Energy is not conserved. i.e. this is wrong.

[sidharath]

Potential energy is energy, so sign is not of much importance. But when sign is given it indicates the process of energy change. When a particle moves away from attractive force energy has to be spent for its motion or energy is gained accumulated later. That's why sign of potential energy is positive. When particle is moving in the same direction as the force energy is being liberated. Therefore negative sign is there. In case of SHO when particle reaches extreme position by overcoming attractive force it has accumulated +E energy or potential energy is positive. When particle starts moving from extreme position toward mean position it is to liberate same amount of energy,. That is why with change in direction in motion, +E will change to -E. When the particle is moving mean position at any point its potential energy is negative, which indicates that is by the time it reaches mean position that much amount of energy will be liberated.. When the particle is moving towards mean position the liberated potential energy is being converted to equal amount of kinetic energy. therefore at no point, sum of kinetic energy and potential energy cannot be zero. In case of a electron if its moves from infinity towards positive charge, when it reaches at any point its potential energy is negative. Which indicates that it has liberated that much amount of energy. same is the case of SHO. When the particle is moving towards mean position it is liberating energy so sign of potential energy is -, while inverse is the case when particle is moving towards mean postion. I dont think if there is any wrong in my presentation of view about the sign of potential energy. I can prove my point with equation or graphs also.

[swansont]

Potential energy is energy, so sign is not of much importance.

 

Conservation of energy is a fundamental concept of physics, with implications to the time-translation symmetry of the laws of physics, so the sign is of huge importance.

 

 

If you can't write down a basic energy balance equation, any solution is going to be wrong. That's the error. There is no reason to go blaming the Schrödinger equation when all you are presenting is an example of garbage in, garbage out.

[studiot]

 

is why with change in direction in motion, +E will change to -E

 

This is just plain nonsense.

 

I really wish it were not so, because I could then perform that trick on my bank balance, only from -£ to +£

 

go well.

 

[sidharath]

I am giving final reply in support of my contention which based upon fundamentals which are not of my own. I stand by the conclusions reached. I am amazed that even elementary things are not clear. In an S.H.O while the particle moves from mean position to extreme position kinetic energy is being changed to potential energy while reverse is the case when particle moves from extreme to mean position. Suppose u^ is the potential energy released. This potential energy is converted to t^ kinetic energy. Suppose particle is moving from extreme position towards mean position it has got potential energy = - 1/2(kx^2). When particle further moves from x to y position potential energy released u^ = U (final) - U (initial) = 1/2(k)(x^2 - y^2), where x > y. Therefore energy released is positive T^ = u^, where u^, therefore as expected kinetic energy increases at the expense of potential energy as particle moves towards mean position. If potential energy of the particle as it moves toward mean position the final result is t^ = 1/2(k)(y^2 - x^2), where y < x, therefore t^ is negative, which indicates that as particle moves towards mean position there is falling kinetic energy which is contrary to the fact. The above treatment conclusively proves without any doubt that correct result is got when the potential energy of the particle moving toward mean position is taken as negative. To get the correct results when the particle moves away from mean position, potential energy has to be taken as positive. Swansont has pointed out that at certain distance potential energy is taken as negative total energy E will be 0, but the following simple observation shows that even with negative potential energy it is not possible. Suppose particle moves from extreme position toward mean position with maximum potential energy - E. At a certain distance potential energy is U, sign negative is included in U. Change in potential energy to U - (- E) = U + E. The released potential energy is changed to kinetic energy T, where T = U + E, or E = T - U. Thus U = - U^, where U^ is positive. Therefore, E = E + U^, both the term on R.H.S are positive. Therefore, E can never be zero even with negative potential energy.

 

I request to contradict my contention in such a manner that replies are based upon true fundamental facts so that you may prove me wrong, but I emphatically I am not wrong at all. It is the ignorance about basic facts that your replies are based and your are ought negate my conclusion.

[studiot]

 

Suppose particle is moving from extreme position towards mean position it has got potential energy = - 1/2(kx^2).

 

You have been told repeatedly that this is not so.

 

Just claiming it or stating it does not make it so.

 

 

When particle further moves from x to y position potential energy released u^ = U (final) - U (initial) = 1/2(k)(x^2 - y^2), where x > y.

 

No energy is released by any particle executing SHM.

Energy is conserved.

 

I have tried to help but you keep repeating your misunderstanding, without reference to my comments, so I am just wasting my time.

[sidharath]

i am writing very simple and conclusive reply to support m contention that sign of potential energy particle moving away from mean position is positive while when it moves away potential sign is negative

The basic relation ia dU=-dF.dx.When the particle moves opposite to restoring force work done is negative

dU=W=-dT There is fall in kinetic energy because of positive work dU=-dT U=-T When the particle moves towards mean position in the restoring force direction work done is positive dU=-W=dT dU-dT orU=T IT is so because of liberation of energy there is increase in kinetic energy. suppose particle with maximum potential +E starts moving towards mean position .at certain point potential energy is reduced to U Change in potential energy IS U-E is converted to kinetic energy U-E=T or E=U-T WHICH IS WRONG. SWANSOT can shout here that E-0 MIDWAY it is concluded that particle with positive potential energy while moving towards mean position gives wrong result. if particle with maximum negative energy-E MOVES towards mean position and U IS POTENTIAL ENERGY at certain point change in potential energy U-(-E)=U+E-T

E=T-U THE sign of potential U is negative E=T+U^ U=-U^ SO CORRECT RESULT IS GOT . numerical value of potential is used There is now no doubt

[sidharath]

as proved in my last mail as particle moves towards mean position dU=dT. suppose is negative u--1/2x^2

dT=--Kxdx which shows as expected that when x decreases T increases. if u is positive as particle moves towards position U=1/2Kx^2 dT=+kx dx which shows that as x decreases t decreases which is contrary to reality .hence potenteial energy of particle moving towards mean position cannot be positive. apart from the direct derivation of potential potential energy as particle moves away or towards mean position by using basic relaton du =-Fdx=-w where the sign of potential energy as moves towards mean position is found out to be negative. indirect results using the same basic relation also prove my contention.If contention is to be refuted it should be based on fundamentals otherwise you are free to give your opinion

sorry there was a little error in above calculation/ As the particle moves towards mean position through distance dx work done w= -2kdx therefore du=-F dx=-w=2kdx=dT hence du=dT The rest of matter is the same . i am again feeling very sorry for minor unintentional lapse.

again minor mistake please add, as particle moves from position x to x-dx towards mean position work w has been calculated for this shift

[studiot]

 

as proved in my last mail

 

All you have proved is that you don’t know your subject properly.

 

 

The basic relation ia dU=-dF.dx.When the particle moves opposite to restoring force work done is negative

dU=W=-dT There is fall in kinetic energy because of positive work dU=-dT U=-T When the particle moves towards mean position in the restoring force direction work done is positive dU=-W=dT dU-dT orU=T IT is so because of liberation of energy there is increase in kinetic energy. suppose particle with maximum potential +E starts moving towards mean position .at certain point potential energy is reduced to U Change in potential energy IS U-E is

 

You seem to have mixed up U and E here. What is U? and What is E?, you have called them both potential energy in the above quote.

 

Why not just use conventional notation then no one will get confused?

 

When the particle moves opposite to restoring force

 

 

At no time whatsoever does the particle move opposite to the restoring force, during the execution of SHM. This is fundamental to the definition of SHM.

 

 

because of liberation of energy

 

At no time whatsoever is energy “liberated”, during the execution of SHM. This is fundamental to the definition of SHM. Energy is conserved.

 

The remainder of post#20 is just plain nonsense.

 

***************************************************************************

 

I will have one more try to offer an explanation.

 

If you do not respond to my comments, but simply repeat your eroneous claims ther is nothing more anyone can do.

 

It is convention that energy input to a system is positive and energy extracted from a system is negative. This goes in other branches (eg thermodynamics) of science as well for the sake of consistency.

 

So consider a system that is not yet executing SHM or indeed any motion.

To exclude gravity, a mass is attached to a fixed point by a spring and sits on a frictionless table.

 

At this point the system is at rest and has zero kinetic or potential energy (since gravity is excluded).

 

The mass is given an initial pull away from the fixed point and released.

As a result of the spring it then begins executing SHM.

 

Work has been done on the system – energy has been input.

 

This energy is the system energy that is subsequently exchanged back and fore between PE and KE. It is never “liberated” from this system.

 

At the time of release the motion is zero so the KE is zero.

The energy input to the system is all positive PE.

 

Do you understand this so far?

 

If you understand this part we can proceed to examine the definition of restoring force.

Edited August 19, 2013 by studiot