Fourier Transform ,QM and "Particles"

Discussion in 'Pseudoscience Archive' started by Anamitra Palit, Aug 24, 2012.

  1. Anamitra Palit Registered Senior Member

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    102
    Consider the following transformation of an arbitrary well behaved function f(x,y), the transformation being given by:

    \(F(E,p,x,t)=\int_{-\infty}^{x}\int_{-\infty}^{t}f(x,t)e^{ia(Et-px)}dxdt\)

    \(\frac{\partial F}{\partial x}=\int_{-\infty}^{x}\int_{-\infty}^{t}[\frac{\partial f}{\partial x}e^{ia(Et-px)}-f(x,t)iap e^{ia(Et-px)}]dxdt\)

    \(\frac{\partial F}{\partial x}=\int_{-\infty}^{x}\int_{-\infty}^{t}[\frac{\partial f}{\partial x}-f(x,t)iap )e^{ia(Et-px)}]dxdt\)----(1)
    Similarly,
    \(\frac{\partial F}{\partial t}=\int_{-\infty}^{x}\int_{-\infty}^{t}[\frac{\partial f}{\partial t}e^{ia(Et-px)}+f(x,t)iaE e^{ia(Et-px)}]dxdt\)

    Or,
    \(\frac{\partial F}{\partial t}=\int_{-\infty}^{x}\int_{-\infty}^{t}[\frac{\partial f}{\partial t}+f(x,t)iaE )e^{ia(Et-px)}]dxdt\)----(2)
    If \(\frac{\partial F}{\partial x}=0\) we have from (1) as an option:

    \(\frac{\partial f}{\partial x}-f(x,t)iap =0\)
    or,\(f(x,t)=Ae^{iapx}\)---- (3)

    Again if \(\frac{\partial F}{\partial t}=0\)


    then,
    \(\frac{\partial f}{\partial t}+f(x,t)iaE=0\)
    Or,\(f(x,t)=Be^{-iaEt}\) ------------- (4)
    We may write,
    \(f(x,t)=Ce^{-ia(Et-px)}\) -------------- (5)
    And proceed as in the previous manner.

    Inserting (5) into(1) we have:

    \(F(E,p)=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}Ce^{-ia(Et-px)}e^{ia(E_0t-p_0x)}dxdt\)
    Or,
    \(F(E,p)=Const \times \del (E-E_0)\del (p-p_0)\) ------------ (6)

    We now consider,
    \(\psi (x,t)=\Sigma C_ie^{-ia(E_it-p_ix)}\) ----------- (7)

    By inserting (7) into (1) we have,
    \(F(E,p)= \Sigma C_i\del (E_i-E_0)\del (p_i-p_0)\) --------(8)

    F(E,p) is the "Transformation " of \(\psi (x,t)\), the transformation being defined on the first line of the post.It becomes athe Fourier Transform if x and t are allowed to tend to infinity.

    F(E,p) if integrated over the (E,p) domain , it counts the number of modes having \(E=E_0\) and \(p=p_0\)
    If the number of such nodes corresponding to some \((E_0,p_0)\) is divided by the total number of possible modes we get a probability picture in an expected manner.By Fourier inversion we should get the same picture on the (x,t) domain.
    The periodic nature of psi is clear from the expression (7) or (5).Psi is indeed a "wave function" By using suitable boundary conditions we restrict ourselves to discrete values of E and p.
    Observations:
    1. Equations (5) and (7) satisfy the Klein Gordon Equations.
    2 Invariance of the exponent in (5) or (7) may be achieved through the Lorentz transformations provide "a" is an invariant. This corresponds to the fact that \(\hbar\) is an universal constant.
    3.If x and t represent represent x-coordinate(spatial) and time respectively then momentum and energy would be suitable candidates for p and E if Et and px are to be dimensionally identical .

    Now let us consider an association between the variables\((t,x,y,z,x_1,x_2,x_3.....)\) and \((E,p_x,p_y,p_z,p_1,p_2,p_3.....)\). Here \(x_1,x_2,x_3 etc\) are some fundamental quantities which are "hidden" or "curled up " in nature---variables that cannot be accessed by presently available experimental methods.


    In place of (1) we now have,
    \(K(E,p)=fg=ce^{-ia\Sigma(Et-p_1x_1-p_2x_2-p_3x_3-p_jx_j)}\)

    We may think of an extended Lorentz transformation for an arbitrary boost in the \(x_i-{x_i}'\) direction having exactly the sane form of our known Lorentz transformations.

    \(x'_i=\gamma(x_i-vt)\)
    \(t'=\gamma(t-v/c^2 x_i)\)
    Other components will not change.
    v is the n dimensional boost having a non-zero component only in the [latex]x_i-x'_i[[/latex] direction."c " is the speed of "n-dimensional " light. Incidentally we are working in an n+1 dimensional system (with the inclusion of time)
    If xi and x'i do not correspond to our known spatial coordinates x , y or z then two observers spatially at rest(relative motion being zero) will have different different clock rates if they have a relative motion in the xi-x'i direction.This corresponds to a situation in GR where two observers at rest at different potentials have different clock rates(example:GPS). Difference in clock rates at different points is a typical feature of curved space.

    Now our 3D "c" corresponds to light as we perceive in the known world. It matches with the concept of "photons"
    Can the n-Dimensional "c" match with something called the "Graviton".
    Can the n-Dimensional "c" lead to other type of "on s" that is to "Other-ons"?

    [In the multidimensional case there are certain modes(or oscillation modes) in the "hidden dimensions" that causes gravity according to our model. You may consider the difference in the clock rates as an indication]

    NB: The transformation given in the first line becomes a Fourier Transformation when both x and t tend to infinity
     
    Last edited: Aug 26, 2012
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  3. AlphaNumeric Fully ionized Registered Senior Member

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    I'm a little unsure what your point is for this thread. Are you going somewhere with all of the maths? You didn't need to do all of that to ask the questions you did.

    c is a scalar, which is quite different from the photon's electromagnetic field etc. The photon field, ie the electromagnetic field, has Lorentz properties in that it is not invariant under a Lorentz transform but rather its governing equations are.

    Now you've gone into vagary. You don't have a model of anything physical, you have explored some of the Fourier transform properties of simple oscillatory fields. Yes, you've picked some fields which we happen to know come up in quantum mechanics but there's a lot more to it than just \(\psi(x,t) \sim e^{i(Et-xp)}\). You certainly haven't shown you have a model of the photon, special relativity, gravitons or anything else. And where hidden dimensions come into it you really don't make clear. Please elaborate on what you're trying to do here and what you think you've shown.
     
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  5. Anamitra Palit Registered Senior Member

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  7. Anamitra Palit Registered Senior Member

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    We consider a system of "n" particles whose total energy E and net momentum P⃗ are fixed are fixed.There no net force on the system(assumed)

    \(\Sigma \epsilon_j=E(const)\)

    \(\Sigma \vec{p_j}=\vec{P}(Const)\)
    For an individual particle its momentum and energy remain constant for the time τ,the relaxation time(average time between successive collisions----a constant). That's an extra constraint for each particle [Radiational energy density at some point is assumed to be constant for some physically small time interval]

    The suffix "j" in the formulation above runs over the different particles.

    For such a situation, generally speaking, we get discrete solutions for energy and momentum.
    For an assemble of free particles we tend to be unmindful towards the fact that the total energy and total momenta may be constant or may be considered to vary between permissible or stipulated limits. Such “forgetfulness” may lead to the concept of continuous solutions.

    Let’s go a step further to solve the Eqn:
    \(f(E_j,p_j,t)=0\) ---------------- (1)
    Subject to:
    \(\Sigma E_j=C_1(t)\) ------------- (2)
    \(\Sigma p_{jx}=K_1(t)\) ------------ (3)
    \(\Sigma p_{jy}=K_2(t)\) ----------- (4)
    \(\Sigma p_{jz}=K_3(t)\) ------------ (5)
    Equation (1) is a description of the path of the system that evolves with time. Equations (2) through (5) are constraints. The suffix “j” considers the particles of the system

    Such equations in general will produce discrete solutions

    Thus Classical Physics favors discrete solutions to the problem in the general type of formulation.

    With Reference to my initial posting we have,
    \(\psi (x,t)=\Sigma C_je^{-ia(E_jt-p_jx)}\) ------------ (6)
    Fourier Transform gives us:
    \(F(E,p)=\Sigma D_j \del(E_j-E_0)\del{p_j-p_0)\) ---------- (7)
    For a many-particle system we have:
    \(\psi (x,t)=\Sigma_n\Sigma_j C_je^{-ia(E_jt-p_jx_n)}\) ---------- (8)
    "j" runs over the number of states and "n" over the different particles.
    Fourier Transform:
    \(F(E,p)=\Sigma_n \Sigma_j M_{nj} \del_{nj}\del(E_{nj}-E_{n0})\del(p_{nj}-p_{n0})\) ---------- (9)

    The Physical Connection:
    Relation (8) is actually a solution for the set of equations from (1) through (5)

    Methodology: Select values of \(p_i\) and \(E_i\) consistent with relations from (2) to (5) and construct the function psi given by relation (8). In relation (8) exert your choice over the x_i to make \(f(E_n,p_n,t)=const\) and \(\psi(E_{nj},p_{nj})\) in relation (8) identical for the modes concerned.Incidentally "n" runs over particles while "j" over the djfferent "modes"
    [States or "modes" may be considered unchanging for times equal to the relaxation time \(\tau\).But may change for successive intervals of the said type].

    Incidentally \(\psi\) may be treated as a function of \(E_i\) and \(p_i\) though it contains the variables x and t.

    On some interval \((t_1,t_2)\) we choose \((E_i,p_i)\) in a manner consistent with constraints from (2) to (5). Then with suitable choice of the position coordinates we make f and \(\psi\) identical in the interval (t1,t2). You may have more than one choice for the xi. If you calculate the Fourier transform of \(\psi\) on the said interval allowing the position coordinates to vary continuously (and of course time to vary continuously)you will sieve out the values chosen values of (E_i, p_i). The same is done for the succeeding intervals.
     
    Last edited: Aug 26, 2012
  8. Tach Banned Banned

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    5,265
    No, it isn't. Amazing that no one else noticed that your RHS is riddled with mistakes. And they aren't typos.
    The rest is snipped since your whole post is nothing else but numerology based on bad math.
     
  9. Anamitra Palit Registered Senior Member

    Messages:
    102
    You may consider the following revisions

    \(F(E,p)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x,t)e^{ia(Et-px)}dxdt\)

    \(\frac{\partial F}{\partial x}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}[\frac{\partial f}{\partial x}e^{ia(Et-px)}-f(x,t)iap e^{ia(Et-px)}]dxdt\)
    \(\frac{\partial F}{\partial x}=\int_{-\infty}^{\infty}[\frac{\partial f}{\partial x}-f(x,t)iap )e^{ia(Et-px)}]dxdt\)----(1)
    Similarly,
    \(\frac{\partial F}{\partial t}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}[\frac{\partial f}{\partial t}e^{ia(Et-px)}+f(x,t)iaE e^{ia(Et-px)}]dxdt\)

    Or,
    \(\frac{\partial F}{\partial t}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}[\frac{\partial f}{\partial t}+f(x,t)iaE )e^{ia(Et-px)}]dxdt\)----(2)
    If \(\frac{\partial F}{\partial x}=0\) we have from (1) as an option:

    \(\frac{\partial f}{\partial x}-f(x,t)iap =0\) ------(3)
    or,\(f(x,t)=Ae^{iapx}\)---- (4)

    Again if \(\frac{\partial F}{\partial t}=0\)


    then,
    \(\frac{\partial f}{\partial t}+f(x,t)iaE=0\) ------------ (5)
    Or,\(f(x,t)=Be^{-iaEt}\) ------------- (6)
    We may write,
    \(f(x,t)=ce^{-ia(Et-px)}\) -------------- (7)
    And proceed as in the previous manner.
     
    Last edited: Aug 26, 2012
  10. Tach Banned Banned

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    5,265
    The above is pure nonsense. Even worse than what you posted before.
     
  11. Tach Banned Banned

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    5,265
    \(F\) is neither a function of \(x\), nor a function of \(t\) so, even that you edited your post after I pointed out your error, your post is still nonsense. You are simply posting numerology, there is no physics that can be derived from your posts.
     
  12. AlphaNumeric Fully ionized Registered Senior Member

    Messages:
    6,702
    Anamitra Palit, as Tach has pointed out, you're making a lot of mistakes here. Furthermore it isn't clear what you're trying to do. Your comments about photons, gravitons etc are unjustified because the properties of photons are quite different from scalar wavefunctions. There's more to modelling them than the Schrodinger equation.

    As far as I can tell you're going around in circles, converting delta functions to complex exponentials and back. Such solutions do come up in the Schrodinger equation but only when there is no potential, which isn't a very general case. In fact the Fourier transform needs infinite support, where the plane wave solution is non-normalisable and you have to be a lot more careful. Quantum mechanics doesn't work on the space of Fourier transformable functions (the Schwarz space), it works on a Riched Hilbert Space, which needs you to start considering Gaussian normalised functions.

    Please explain what the purpose of this thread (and your others) is because you're not going to be allowed to keep churning out equations without discussion, especially when they are wrong. Last chance before I kick this to pseudo.
     
  13. Anamitra Palit Registered Senior Member

    Messages:
    102
    You just consider the Fourier Transform of the function:
    \(f(x,t)=Ce^{-ia(Et-px)\)----------- (1)

    Fourier transform of f(x,t) is independent of either x ot t.
    Therefore \(\frac{\partial F}{\partial x}=0\) ---------- (2)

    and \(\frac{\partial F}{\partial t}=0\) ---------- (3)
    Partial differentiating (1) that is the expression for f(x,t),wrt "x" you get:
    \(\frac{\partial f}{\partial x}-iap=0 \) ---------- (4)
    And partial differentiating (1) wrt t youn get,
    \(\frac{\partial f}{\partial t}+iaE=0\) ----------- (5)

    Fourier Transform of f(x,t)=ce^{-ia{Et-px)} is given by:
    \(F(E,p)=const\times \del(E-E_0)\del(p-p_0)\)

    If f(E,p) is integrated on the entire domain (E,p) the value of F(E,p) we gat a value \(\times 1\)Cwill be proportional the number of \((E_0,p_0)\) modes.

    To count different modes for the same particle we use different values for \((E_0,p_0)\). We get\(C\times n[tex] where n is the number of possible modes for the same particle. The ratio of the above two quantities ive us the probability of occurrence of a particular mode for the same particle. For " n" particles: we may consider the function: [tex]\psi=\Sigma C_j e^{-ia(E_jt-p_jx_j)}dtdx_j\) ------------ (6)
    "j" runs over the particles:j=1,2,3...n
    The Fourier transform of psi:
    \(F(E,p)= \Sigma C_j\del(E_j-E_0)\del(p_j-p0)\)

    If F(E,p) is integrated over the entire (E,p) domain its value will be:\(\Sigma c_j\)

    With the above information we try to solve the following equations:

    \(f(E_j(t),p_j(t)=C(t)\) ----------- (7)
    \(\Sigma E_j=C_0(t)\) ---------- (8)
    \(\Sigma p_{jx}=K_1(t)\) ---------- (9)
    \(\Sigma p_{jy}=K_2(t)\) ------------- (10)
    \(\Sigma p_{jz}=K_3(t)\) --------- (11)

    "j" runs over the different particles.Equations (8) to (10) may be regarded as constraint equations.
    Generally speaking, we get discrete solutions for the above set of equations (8) to (11)

    We choose a discrete set \((E_j,p_j)\) consistent with the equations from (8) to (11) and form the psi function given by relation (6):
    \(\psi(E_j,p_j,x_j,t)=\Sigma C_je^{-ia(E_jt-p_jx_j)}dtdx_j\)
    We can always make f(xt) and \(\psi\) identical at the discrete points chosen by suitable adjustment of x_i and the constants C_j


    To Be Continued Shortly..........I will return to this thread will all required explanation at my earliest ...]
     
    Last edited: Aug 26, 2012
  14. AlphaNumeric Fully ionized Registered Senior Member

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    6,702
    You've gone around the houses to end up finding out how many particles there are, which you already knew because it's in the definition of the wave function! And your definition of the wave function includes the integration measures for some reason.

    You quoted me asking you to explain yourself and you ignored that and just carried on with this seemingly pointless Fourier transforming.

    You should have started with the explanation of the thread, not post lots and lots of error riddled maths which seems to go around in circles. You had the time to write out a long post with lots of maths in it but not to give a short overview of your work? I'm kicking this to pseudo, as I said I would. If you can demonstrate it is leading somewhere and you can correct the errors you've made then it can be moved back. You had your chance to keep the thread here, you passed on it.
     
  15. Anamitra Palit Registered Senior Member

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    102
    One of the important reasons have been highlighted(redlighted) for AlphaNeumeric. Others to follow. I mean other reasons to follow.
    Actually Quantum Mechanics is not necessary to understand why the assembly of particles should have discrete energy configurations
     
  16. Anamitra Palit Registered Senior Member

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    102
    With Reference to Post #8

    F(E,p) is neither a function of x or t. This simply means:\(\frac{\partial F}{\partial x}=0\) and

    \(\frac{\partial F}{\partial t}=0\)
     
  17. AlphaNumeric Fully ionized Registered Senior Member

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    6,702
    If you have a set of free particles then there's no need to consider more than one since they are just N separate systems. Since there's no exchange of energy or momentum between particles and the system is closed the energy and momentum are constant. What you've been doing has nothing to do wish the values of the E and p.

    Right, let's go through this. The Schrodinger equation with zero potential and support between x = -L/2 and x = L/2 is the 'particle in a box'. This has plane wave solutions similar to what you've written down. However, the spectrum of the operator is not continuous, it is discrete due to the finite size of the box. More specifically \(E_{n} \propto \frac{n^{2}}{L^{2}}\). Eigenfunctions look like \(\sin( \frac{n\pi x}{L} )\) in their x dependencies. Since the box is the only place where the wavefunction is non-zero all integrals are of the form \(\int_{-\frac{L}{2}}^{\frac{L}{2}}\). This compact support also means the wavefunction is normalisable, again with a dependency on L, \(\sqrt{\frac{L}{2}}\).

    Quantised spectrum is fundamental to quantum systems so continuous spectra are to be carefully considered. Clearly if \(L \to \infty\) then \(E_{n+1}-E_{n} \to 0\) and \(\int_{-\frac{L}{2}}^{\frac{L}{2}} \to \int_{-\infty}^{\infty}\). The normalisation factor is not longer valid though, since \(\sqrt{\frac{L}{2}} \to \infty\). It's at this point the use of Gaussian normalisation has to come in to deal with the issues of having a completely free particle. Continuous spectra also arise in systems with potentials, it's a matter of giving a particle enough energy to be completely above the potential well. This occurs when you input enough energy to an atom and it ionises.

    So what you're doing is still unclear. You are just messing with plane wave solutions when what you should be doing is messing with the operator which has eigenfunctions that are plane wave. That will then give you the values of E and p which are allowable because apriori you have no idea what they are. Much of quantum mechanics is devoted to extracting these values as they represent physical things like energy bands in transistors or emission spectra in gases. What you're doing still seems to be going around in circles.
     
  18. Anamitra Palit Registered Senior Member

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    102
    With the assembly of free particles considered in my postings,the total energy and total momentum(vector) are not changing with time.What ever "freedom " these particles own is with such constraints.

    I have already highlighted it--rather I put out a colorful display of this interesting fact in my previous posting.

    I would like to know what restriction would you like to impose on the "freedom" of these particles because of this constraint on energy and momentum.
     
  19. Anamitra Palit Registered Senior Member

    Messages:
    102
    You have a closed system in your considerations.And yet its energy goes on changing with time--just look at the different eigenvalues you are stating.This is not acceptable.

    But if you have a system of particles under an overall constraint/s ,in so far as the entire system is concerned,you have the possibility of exchange of energy between the different particles not only through collisions but otherwise also.This may be concluded even classically.
     
  20. AlphaNumeric Fully ionized Registered Senior Member

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    6,702
    Energy and momentum conservation in quantum mechanics follow from the form of the Hamiltonian and Noether's theorem. As I've repeatedly commented, you haven't even considered the Schrodinger equation, you're messing with plane wave solutions to it without considering the system itself.

    No, the energy doesn't change in time. The energy evolution is determined from the Hamiltonian. This follows via the Heisenberg equations, \(\frac{\partial A}{\partial t} = [A,H]\). If operator A is H then since [H,H]=0 we have \(\partial_{t}H = 0\) and thus it's expectation value is time independent (just convert to the Schrodinger picture), which is energy. What values the eigenvalues take is irrelevant, since both the eigenvalues and the time evolution 'flow' of the system are determined by the same thing, the Hamiltonian.

    You're now showing you don't understand the relevant quantum mechanics, showing I was right to move this to pseudoscience.
     
  21. Anamitra Palit Registered Senior Member

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    102
    The expectation value,which is the average value ,does not change with time. But the individual values of energy [value obtained from a SINGLE MEASUREMENT: you may consider several such trials and the separate values of measurement from the different trials]can and will change with time.For any operator that commutes with the Hamiltonian,the expectation ie the average value does not change with time. What about the results of the individual measurements of energy changing with time in a system you consider to be closed?
     
    Last edited: Aug 27, 2012
  22. AlphaNumeric Fully ionized Registered Senior Member

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    No, that isn't true. It isn't an average, it is the energy of the wave function.

    Let \(|\psi\rangle\) be a wave function with a decomposition written in terms of the Hamiltonian's eigenvalues and eigenstates, \(|\psi\rangle = \sum_{j}\alpha_{j}e^{-iE_{j}t}|\psi_{j}\rangle\). The energy of the wave function is defined as the Rayleigh quotient, \(E_{|\psi\rangle} = \langle H \rangle_{|\psi\rangle} = \frac{\langle \psi | H | \psi \rangle}{\langle \psi | \psi \rangle}\).

    Under time evolution a system evolves as \(|\psi(t)\rangle = e^{iHt}|\psi(0)\rangle\). You can easily confirm that \(|\psi\rangle\) indeed evolves like this given its expression in terms of the eigenvalues and eigenstates. So we want to see if the energy of the system varies in time, ie we want to compare \(E_{|\psi(t)\rangle} \) with \(E_{|\psi(0)\rangle} \). \(E_{|\psi(t)\rangle} = \frac{\langle \psi(t) | H | \psi(t) \rangle}{\langle \psi(t) | \psi(t) \rangle} \). The denominator is easy, \(\langle \psi(t) | \psi(t) \rangle = \langle \psi(0) | \psi(0) \rangle\) because the time evolution is unitary and the Hamiltonian is Hermitian. This is a necessary condition for probability conservation. So now we just consider the numerator. \(\langle \psi(t) | H | \psi(t) \rangle = \langle \psi(0) | e^{iH^{\dagger}t}He^{-iHt} | \psi(0) \rangle\). Since \(H = H^{\dagger}\) by Hermiticity and [H,H]=0 we can commute the first term through the central H, cancelling with the last term to give \(\langle \psi(0) | H | \psi(0) \rangle\). Therefore the time evolution doesn't change the Rayleigh quotient, it doesn't change the energy. You can trivially confirm this is the case for any linear combination of eigenstates, each eigenstate has its energy conserved individually and each eigenstate has a plane wave form (I'm using bra-ket Dirac notation) of the kind you've been throwing around. So your claim about individual energies changing is demonstrated false. Nothing I've done here is advanced, it is all stuff you learn in a foundation course in quantum mechanics. Your claims about quantum mechanics and the frankly sloppy nature of doing all the Fourier stuff suggests to me you didn't take such a course.
     
  23. Anamitra Palit Registered Senior Member

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    102
    Suppose you are making a measurement on the system. Do you get the value \(<\psi|H|\psi>\) as the result of a SINGLE measurement for energy.?
     

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