Does This Shape Function Satisfy the Continuity Requirements
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Continuity of the wave function
- Thread starter luisgml_2000
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H[tex]\Psi[/tex]=E[tex]\Psi[/tex]
Which idea is the right one?
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I've heard some people say that the wave function and its first derivative must be continuous because the probability to find the particle in the neighborhood of a point must be well defined; other people say that it's because it's the only way for the wave function to be physically significant. There is even another hypothesis, which states it's a consequence of the eigenvalue equationH[tex]\Psi[/tex]=E[tex]\Psi[/tex]
Which idea is the right one?
I don't think that the idea of continuity of wave functions plays any significant role in quantum mechanics. After all, Dirac delta functions (which are not continuous) have a prominent place there. Without delta functions one cannot define eigenvectors of many important operators of observables, such as position and momentum. So, I would prefer to keep delta functions and abandon the idea of continuity.
Eugene.
Every book I've seen until now just states that the wave function has to be square integrable (and I understand the reason for this), continuous and its first derivative a continuous function (except for discontinuous potentials), but the authors never give further explanations about it.
I get the jist of the post above. So when we consider the Schrodinger Equation, do we assume that a wavefunction must satisfy the Schrodinger Equation for ALL x (If so, why)? In this case, the above post is very convincing. If not, then surely we can make a wavefunction with lots of discontinuities so that the SE is satisfied by the wavefunction where the wavefunction is twice differentiable.
Edit: Do we usually assume that the wavefunction is twice differentiable for all x?
This may seem like a random post but I want to know a few things about the continuity of the wavefunction...I searched on google and up came this thread.I get the jist of the post above. So when we consider the Schrodinger Equation, do we assume that a wavefunction must satisfy the Schrodinger Equation for ALL x (If so, why)? In this case, the above post is very convincing. If not, then surely we can make a wavefunction with lots of discontinuities so that the SE is satisfied by the wavefunction where the wavefunction is twice differentiable.
Edit: Do we usually assume that the wavefunction is twice differentiable for all x?
If d/dx^2 doesn't exist then E is undefined and the particle doesn't exist.
It's nice because we choose wave functions that are continuously derivable. Sin/e^ix etc... A property of a wave is that it is ever changing. If your function stops changing at some point, it's not a wave function.
do we assume that a wavefunction must satisfy the Schrodinger Equation for ALL x
Yes.
It seems to me it would be rather difficult to use the SE to come up with useful predictions otherwise, unless we also had a good idea of exactly where it works and where it doesn't.
The analog of this question for classical mechanics is, "do we assume that Newton's Laws hold for all x (if so, why)?"
Ok thanks for your responses
No problem. Just remember, the probability of finding the particle in all of space is equal to 1. If you're not integrating over all of space with a continuous function then the the probability of finding the particle is no longer 1, however close to 1 it may be. Remember, quantum barriers only act to temporarily confine a particle. It can neither be stopped completely nor confined indefinitely.
Now, one might then ask, if we have a wavefunction with a discontinuity in either the function or its derivative, how would it evolve, according to the Schroedinger equation? But that's no problem either-- the infinitely fast rate of change of the wavefunction is not a problem if it only lasts for an infinitesmal amount of time, and this kind of thing can be solved by using integrated forms of the Schroedinger equation, instead of its differential form (analogous to how shocks are treated in hydrodynamics).
So why is it ever necessary to assume either the wavefunction or its derivative are continuous? It seems to me that the predictions of quantum mechanics involve doing integrals, so all that is really needed is that the necessary integrals exist, which is a less stringent requirement but perhaps more stringent ones are commonly adopted just out of pure convenience. Alternatively, perhaps what is really intended is that energy eigenstates have these continuity requirements, which is clearly true because they have to be stationary in time.
Or am I missing something?
Thanks
Bill
The question: where it came from. I think it was a numerical tool in early physics. Equations describing point objects (like point forces, point charges or point particles) came as taking the limit of some equation in zero length. The physical equations are smooth in general, so the resulting object was the limit of a series of smooth functions. Later, the mathematical distribution theory was applied to it.
So, ultimately the continuity/smoothness of wavefunctions, comes from experiment.
Distributions don't have problems with differentials and being continuous, but they do have problems with multiplication. In some cases, multiplication is not well defined on distributions, namely it's not associative. I've heard the very quantization procedure from the mathematical point of view may be seen as assigning meaning to distribution multiplication.
Distributions don't have problems with differentials and being continuous, but they do have problems with multiplication. In some cases, multiplication is not well defined on distributions, namely it's not associative. I've heard the very quantization procedure from the mathematical point of view may be seen as assigning meaning to distribution multiplication.
That's true - but a lot of work has been done on the issue and evidently distributions can be further generalized so they can:
http://en.wikipedia.org/wiki/Colombeau_algebra
I like math but this is even getting over the top pure math wise for my tastes.
Thanks
Bill
Granted, one could simply exclude from the range of the function the domain where the potential is infinite, but one can imagine a limit where the potential height gets larger as its domain gets narrower, and in the limit becomes infinite over an infinitely narrow region. In an integral formulation, that would simply yield a kink in the first derivative of the wave function for an energy eigenstate, something we could easily model and address with experiment, but which we are hearing falls outside the realm of applicability of the current formulation of quantum mechanics. In practical applications, what we want is an integral constraint on a barrier like that, a constraint that simply sets the discontinuity in the slope of the eigenfunction.
I had heard, and I don't remember if it was from a textbook or from a professor that continuity of the wave function and it's derivative imply conservation of momentum across a boundary. Anyone care to elaborate or expel this idea?
Well the assumption of continuity and differentiation, ie the usual assumptions we make in physics, usually without even stating it, AND, the Born Rule gives probabilities independent of inertial frame (ie the laws of QM obey the POR) and the Galilean Transformations (ie classical mechanics - not relativity) yields the Schroedingers equation etc etc plus the conservation laws.
See Ballentine - Quantum Mechanics - A Modern Development.
But the real key is not the assumptions of differentiability etc it the symmetries imposed by the POR.
Thanks
Bill
Well the assumption of continuity and differentiation, ie the usual assumptions we make in physics, usually without even stating it, AND, the Born Rule gives probabilities independent of inertial frame (ie the laws of QM obey the POR) and the Galilean Transformations (ie classical mechanics - not relativity) yields the Schroedingers equation etc etc plus the conservation laws.See Ballentine - Quantum Mechanics - A Modern Development.
But the real key is not the assumptions of differentiability etc it the symmetries imposed by the POR.
Thanks
Bill
I read somewhere that as mod of the square of a wave function gives the probability of finding a particle hence it must be square integrable, must be finite everywhere i.e must be a bound state so that the probability is finite .As wave function satisfy Schrodinger equation , so it must be differentiated twice , for that its first order derivative must be continuous otherwise at the point of discontinuity we will not get its 2nd order derivative ,similarly the wave function must be ccontinuous everywhere else at the point of discontinuity , it s first order derivative would not exist .The above cases will be valid for potential wit finite discontinuity only.. But my question is what do we mean by finite and infinite discontinuity of potential? example of an infinite discontinuity is dirac delta function ? and why the above thing is not valid for potential with infinite discontinuity?
At a delta Funktion, only the first derivative is discontinuous. In #19, I die show how discontinuities can arise.
i.e its 2nd derivative is infinite which means its energy is infinite but i think that is not practically possible ... so does it mean that schorodinger equation cant be applied for this type of potentials ...
But there is no Problem with Schrödinger equation.
sir , 1 more thing .. will you please explain me that why do we need operators in quantum mechanics ? i mean i can understand that momentum and position cant be measured simultaneously... we cant get exact momentum value if position is defined and vice versa . But how does operator helps in this regard ?
sir , 1 more thing .. will you please explain me that why do we need operators in quantum mechanics ? i mean i can understand that momentum and position cant be measured simultaneously... we cant get exact momentum value if position is defined and vice versa . But how does operator helps in this regard ?
Here is the deep axiomatic reason:
https://www.physicsforums.com/threads/the-born-rule-in-many-worlds.763139/page-7
See post 137.
Thanks
Bill
actually i have my exam and i am really weak in quantum mechanics ... i would like to know another question sir ... i know the physical meaning of normalization , will you plz tell me the physical meaning of orthogonality ? i have one more small assignment too .. i tried but unable to do ... the question is that if H is an operator of an eigen function U1 with eigen value E1 and H is an operator of an eigen function U2 with eigen value E2 then show that H is also an operator of the eigen function U1+U2.
This really belongs in the homework help section.
But I will answer the orthogonality question by asking you to apply the Born rule to orthogonal states.
Since they are eigenfunctions H*U1 = E1*U1 and H*U1 = E1*U1 so H*(U1 +U2) = H*(E1 + E2) so E1 + E2 is an eigenvalue of H with eigenfunction U1+U2.
Thanks
Bill
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