Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Free Particle - Wave Packet

  1. Oct 4, 2014 #1
    If you solve TISE with V=0, you get a plane wave.

    This is not normalizable, so it's not a physically achievable state.
    But a linear combination does (?) I know why it does from the mathematics, a linear combination of plane waves is normalizable, but what does it really mean??? A free particle is associated to MORE than 1 plane wave??
    Therefore you create a wavepacket with a spread around a center.... From position to momentum, you switch with Fourier Transform etc etc

    I don't really get it.

    Why is it not normalizable from a "conceptual" point of view? A free particle has an infinite chance to be found anywhere? Is that the reason? If the free particle solution ( a single plane wave) isn't really meaning something, why does that form always come back (in solid state physics for example)

    And how do you construct it in real life, a free particle wavepacket. In textbooks they say: you get a lot of momenta so the uncertainty in momentum becomes greater, hence position gets more defined and vice versa.

    How can you give 1 particle more momenta to set its position right, and how can you make the uncertainty in position right to give it the right momentum. (In real life)
  2. jcsd
  3. Oct 4, 2014 #2
    x and p take continuous eigenvalue. Eigenstates are "normalized" as
    [tex]<x|x'>=\delta(x-x'),<p|p'>=\delta(p-p')[/tex] ,so
    [tex]<x|x>=\delta(0)=\infty,<p|p>=\delta(0)=\infty[/tex], not value 1.
    These eigenstates of continuous eigenvalues are designed so that superposed states of finite spectre width are normalized to unity. For example you can check Gaussian wavepackets are made of |x>s or |p>s in integral and normalized to 1. |x> and |p> are convenient mathematical tool but not realized in physics.
    Last edited: Oct 4, 2014
  4. Oct 4, 2014 #3


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    This is a common misunderstanding, because often quantum theory is introduced as if was "wave mechanics", which is only one of many representations of the general formalism and then they give you lots of energy-eigenvalue problems (i.e., the time-independent Schrödinger equation) to solve.

    You are very right, the plane wave is not square integrable and thus does not represent a state of a particle. The eigenvectors of eigenvalues of self-adjoint operators in Hilbert space, where the eigenvalue is not discrete but there's a continuous set of eigenvalues around it, are not true Hilbert-space vectors but distributions of a (dense) subspace of Hilbert space, where the self-adjoint operator is defined.

    To make the physicist's hand-waving way to deal with this mathematically rigorous the proper concept is to work with socalled "rigged Hilbert spaces". A good introduction can be found here:


    A wave packet is a square-integrable function that describes a non-stationary state. Its time evolution is given by the time dependent Schrödinger equation. The value of the generalized energy eigenfunctions is that you can express any such wave packet as a superposition of these generalized energy eigenfunctions, and thus the time evolution is easily found, given an initial condition.

    For a free particle, it's more convenient to use the generalized momentum eigenfunctions (let's take a particle on a line to keep things easy and set [itex]\hbar=1[/itex]). Then the momentum operator in the position representation reads
    [tex]\hat{p}=-\mathrm{i} \partial_x.[/tex]
    The generalized eigenfunctions of this operators are found by solving the eigen-value equation
    [tex]\hat{p}u_p(x)=p u_p (x) \; \partial_x u_p(x)=\mathrm{i} u_p(x) \; \Rightarrow \; u_p(x)=N \exp(\mathrm{i} p x).[/tex]
    The normalization constant is chosen such that the generalized normalization condition
    [tex]\int_{\mathbb{R}} \mathrm{d} x u_{p'}^*(x) u_{p}(x)=\delta(p-p'),[/tex]
    where [itex]\delta[/itex] is the Dirac distribution. From Fourier analysis we know that
    [tex]\int_{\mathbb{R}} \mathrm{d} x u_{p'}^*(x) u_p(x)=\int_{\mathbb{R}} \mathrm{d} x N_{p'}^* N_{p} \exp[\mathrm{i} (p-p')x]=2 \pi |N_{p}|^2 \delta(p-p').[/tex]
    This determines the normalization constant (up to an arbitrary phase which is unimportant for the following)
    [tex]u_p(x)=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x).[/tex]
    Now the Hamiltonian of the free particle is
    This implies that the momentum eigenfunctions are also the energy eigenfunctions with eigenvalue
    The time-dependent Schrödinger equation reads
    [tex]\mathrm{i} \partial_t \psi(t,x)=\hat{H} \psi(t,x).[/tex]
    The most general solution is thus given by
    [tex]\psi(t,x)=\int_{\mathbb{R}} \mathrm{d} p \exp \left(-\mathrm{i} \frac{p^2}{2m} t \right) u_p(x) A(p),[/tex]
    where [itex]A[/itex] is an arbitrary square integrable function. It is determined by the initial condition for the wave function,
    which must be a square integrable function too. Indeed we find
    [tex]\psi_0(x)=\int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi}} A(p) \exp(\mathrm{i} p x).[/tex]
    This implies
    [tex]A(p)=\langle p | \psi_0 \rangle =\int_{\mathbb{R}} \mathrm{d} x \langle p | x \rangle \langle x|p \rangle= \int_{\mathbb{R}} \mathrm{d} x \frac{1}{\sqrt{2 \pi}} \exp(-\mathrm{i} p x) \psi_0(x).[/tex]
  5. Oct 4, 2014 #4
    I'm not at the phase where I know what dirac normalization is yet. :(

    But I mean practically, what is a free particle anyway and what is a gaussian wave packet (in practice)?

    A plane wave is the solution of the TISE for V=0, it's not normalizable to 1. So it's not possible... U need a superposition of different solutions and there is a trade off between momentum and position.

    But what means it in real life? How can 1 particle be a superposition of different plane waves? So does a free particle exist yes or no? So to 1 particle there are different plane waves associated to it? And different momenta??

    Why do u need to integrate over a range of p's to get the uncertainty on x less? How would that be a manifestation in real life?

    How can one particle have more momentum?? What does it mean to integrate over a range of momentums in the fourier analysis of a wave packet

    Edit: @ Vanhees, I have to go now, I will read your answer later this day. Thx for the replies both of you
  6. Oct 4, 2014 #5
    One particle has one momentum value of observation (here I do not mind how sharp or blant the observation is) . QM says before observation it has no definite value but chance to take various values in general that is expressed by probabilty amplitudes[tex]\psi(p)[/tex] where [tex] |\psi(p)|^2dp[/tex] gives probabioity that we observe momentum in area [p,p+dp]. The probabipity to get the value a<p<b in observation is given by integral
    [tex]\int_a^b |\psi(p)|^2 dp[/tex]

    One particle has two or more momentum or plane wave? Yes, as probability amplitude before observation.
    One particale has two or more positions? Yes, as probability amplitude before observation.
    Last edited: Oct 4, 2014
  7. Oct 4, 2014 #6


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    I think, we have to go back and discuss the very foundations of quantum theory, and I use the minimal statistical interpretation since that's the physics and no more.

    The state of a particle in non-relativistic quantum theory can be described by a wave function, [itex]\psi(t,x)[/itex]. The wave function is a square-integrable function, i.e., you can choose a normalization such that
    [tex]\int_{\mathbb{R}} \mathrm{d} x |\psi(t,x)|^2=1.[/tex]
    Then the physical meaning of this wave function is that the probability to find a particle at a little position element [itex]\mathrm{d} x[/itex] around [itex]x[/itex] at time [itex]t[/itex] is given by
    [tex]P(t,x) \mathrm{d} x = |\psi(t,x)|^2 \mathrm{d} x,[/tex]
    i.e., [itex]P(t,x)[/itex] is the probability distribution function to find particle at [itex]x[/itex] when looking at time [itex]x[/itex].

    This probability distribution can be arbitrarily sharply peaked around a certain value [itex]x_0[/itex], but it will always have a finite width around this value, i.e., according to quantum theory you cannot determine the particle's position exactly. Further, according to quantum theory, a particle's state cannot be somehow "sharper" determined than knowing its wave function with this probabilistic interpretation.

    What does the knowledge of a probability distribution of the particle mean in practice? Here, I'm following the frequentist interpretation of probabilities, because that's the only one which makes sense to me, particularly when looking at what my experimental colleagues at RHIC, LHC, and GSI really do in heavy-ion experiments with particles: They let collide a lot of nuclei under as well prepared always equal circumstances (beam energy) they can and then measure the interesting quantities as accurate as they can. In this case they register the particles created in such a collision, figure out what kinds of particles (protons, pions, electrons, positrons,...) they are and measure their energies and momenta and make histograms, i.e., they count how many particles come in each momentum bin.

    Of course, you can think about experiments determining a particles position, e.g., by letting them hit a detector which registers their position. So the upshot is that you can check, whether quantum theory is correct in predicting a wave function, given a certain well-defined reproducible preparation process has been done on a large number of particles and then measuring their positions after a certain time under also well-determined circumstances (e.g., letting them run through a electric and/or magnetic field). Then you can make a historgram about the position of each particle in the so defined enemble and if you collect enough such results you find the statistical distribution of the particles and can compare it to the probability distribution [itex]P(x)[/itex] predicted by quantum theory.

    Of course, you can also measure other quantities, like momentum. Now as elaborated in my previous posting the wave function for this quantity is given by the Fourier transform of the position-wave function:
    [tex]\tilde{\psi}(t,p)=\frac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}} \mathrm{d} p \psi(t,x).[/tex]
    is the probability distribution for the momentum of the particle, given its preparation in the state described by the wave function [itex]\psi(t,x)[/itex] or equivalently by the momentum-wave function [itex]\tilde{\psi}(t,p)[/itex].

    Again, the interpretation is analogous as for the position measurement: You can measure the probability distribution by making experiments with a sufficiently large ensemble of equally and independently prepared particles and measuring the momentum (or any quantity you are interested in).

    Why this is the momentum distribution should be explained in your textbook. The full foundation of which observable is represented by which self-adjoint observable is a bit of math, which I cannot explain in a short forum posting. It's related to the symmetries of physical systems and the representation of symmetry groups as unitary ray representations on Hilbert space.

    It's an important point to switch to the more abstract Hilbert-space formulation of quantum theory to gain a proper understanding of what "state" and "observable" and their description by rays in Hilbert space and self-adjoint operators really means. Then it becomes clear that the position-wave functions or momentum-space functions are just different ways to describe a particles's state, pretty much like it doesn't make a real diffeence whether you describe a position in classical physics in terms of Cartesian or spherical coordinates.
  8. Oct 4, 2014 #7


    User Avatar

    Staff: Mentor

    A single plane wave corresponds to a particle with a definite, exact value of momentum: ##p = h/\lambda##. If such a state actually existed, you would be guaranteed to get that value of momentum if you measure it. But such a state is not possible, because, as you said, it's not normalizable.

    A superposition of plane waves with different wavelengths corresponds to a particle with different possible momenta. If you measure the momentum of that particle, you get one of those values of momentum, with a probability proportional to the square of the amplitude of the wave that has the corresponding wavelength.

    However, a superposition of a finite number of waves is not normalizable, either.

    A wave packet is a superposition of infinitely many waves, with a continuous (not discrete!) distribution of wavelengths and momenta. Such a packet is normalizable! :w The amplitude (or probability) of a particular momentum (or wavelength) peaks at some central value, and falls off towards zero as you go further and further away from the "center." You can define a range of momenta, ##\Delta p##, around the center by using the standard deviation.

    The sum (resultant) of all theses waves in terms of position x, is a wave function ##\Psi## that is peaked at some central position, and falls off as you go further away from the center. You can define the spatial "width" of the wave function, ##\Delta x##, by using, again, the standard deviation.

    The value of the product ##\Delta x \Delta p## depends on the "shape" of the wave packet (the particular distributions of possible momentum and position). However, it can be no smaller than a certain value: ##\Delta x \Delta p \ge \hbar / 2##. Does this look familiar? :D One way to derive this is to use the properties of Fourier transforms, because the position and momentum distributions are basically Fourier transforms of each other. There are also more abstract (and arguably more fundamental) ways of deriving it, using the properties of operators on Hilbert space.
    Last edited: Oct 5, 2014
  9. Oct 4, 2014 #8
    In post 6 there was used the frequentist intepretation. But is this one not invalidated by bell's theorem which assumes the statistics is obtained by averaging over a lot of results ?
  10. Oct 5, 2014 #9


    User Avatar

    Staff: Mentor

    Bell's theorem doesn't invalidate the idea of averaging over a lot of results, it does something completely different.

    It might be best to start a new thread for this question, or otherwise we'll totally hijack this one.
  11. Oct 5, 2014 #10


    Staff: Mentor

    Indeed it is.

    At the beginner level simply view stuff like the Dirac Delta and a plane wave function as an approximation introduced for mathematical convenience - they don't really exist so don't have to be square integrable etc. In general physically realisable functions are zero at infinity - which of course a plane wave isn't.

    But eventually you probably want to get to the bottom of it.

    For that I suggest starting out with distribution theory:

    Its important in its own right and should be part of any applied mathematicians armoury - the treatment of Fourier transforms is particularly elegant.

    To progress further you need to study functional analysis. A good start would be the following:

    After that you can delve into the full gory detail:

    But take your time, this is pretty advanced stuff it took me years to come to grips with.

    Right now simply realize your issues are answerable.

    Last edited by a moderator: May 7, 2017
  12. Oct 25, 2014 #11
    I've read all your posts a while ago and thanks for that. But I still haven't got an answer and wasn't sure how to really explain what I want to hear.

    I think I want to know what it means for the free particle to composed of more than 1 wave and how would you do it in a lab theoretically speaking. A free particle in a lab, how to create an electron that is free with a specified location. Theoretically speaking, because not a single one electron is free. So, how can u play around with momenta? How can u create a wavepacket in practice with enough different wavenumbers to localize it in reality?

    Conceptually and practically. Are we also still speaking about 1 particle if it's made up of more than 1 wave?

    Here's the worlds best scientific poster made in paint (lol):

  13. Oct 25, 2014 #12


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    The answer is simple: You cannot prepare an electron with a definite momentum, not even theoretically, because plane waves are generalized functions and not square integrable ones that represent states of real particles. What you can do, however, is to perpare particles with a quite well-defined momentum. The uncertainty relation tells you that this is at the cost of the precision in the electron's location.

    In practice, what's done is that you feed electrons into a particle accelerator and accelerate them with electromagnetic fields. What you get in reality are bunches of electrons with a quite well-defined momentum. You can use filters to filter away electrons whose momentum deviates too much from the value you like to prepare.

    Of course, you can describe any wave as a superposition of plane waves, the socalled Fourier-decomposition. The corresponding wave function is given by the Fourier integral (see my posting above)
    [tex]\psi(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \tilde{\psi}(\vec{p}) \exp[\mathrm{i} \vec{p} \cdot \vec{x}-\mathrm{i} E_{\vec{p}} t].[/tex]
    For convenience I use natural units with [itex]\hbar=1[/itex] in this posting. For non-relativistic electrons, you have
    To prepare particles with a well-defined momentum you use a wave-function in momentum space [itex]\tilde{\psi}(\vec{p})[/itex] that is sharply peaked around the value [itex]\vec{p}_0[/itex] you like to prepare. You can use, e.g., a Gaussian wave
    [tex]\tilde{\psi}(\vec{p})=N \exp \left (-\frac{(\vec{p}-\vec{p}_0)^2}{4 \Delta p^2} \right ).[/tex]
    The probability distribution for the momentum is
    [tex]\tilde{P}(\vec{p})=|N|^2 \exp \left (-\frac{(\vec{p}-\vec{p}_0)^2}{2 \Delta p^2} \right ),[/tex]
    with an average momentum [itex]\langle \vec{p} \rangle=\vec{p}_0[/itex] and a standard deviation [itex]\langle (\vec{p}-\vec{p}_0)^2 \rangle = \Delta p^2[/itex].
    You can calculate the Fourier integral exactly. What you'll get is a Gaussian distribution for the probability of the particles position with the expectation value
    [tex]\langle \vec{x}(t) \rangle=\frac{\vec{p}_0}{m} t,[/tex]
    and the standard deviation
    [tex]\langle \Delta x^2(t) \rangle=\frac{m^2+4 \Delta p^4 t^2}{4 \Delta p^2 m^2}.[/tex]
    As you see, the particle's position expectation value is moving like a classical paticle with constant velocity [itex]\vec{p}_0/m[/itex], but the position uncertainty grows with time, which is due to the uncertainty of the momentum.

    The Gaussian wave packet always fulfills
    [tex]\Delta x \Delta p \geq 1/2,[/tex]
    as it must be according to the Heisenberg uncertainty relation.
    Last edited: Oct 25, 2014
  14. Oct 25, 2014 #13


    User Avatar
    Science Advisor

    Are we still talking about a single electromagnetic wave when we form a pulse of electromagnetic radiation from a sum of Fourier modes obtained from Maxwell's equations? Of course we are. It's no different for single-particle solutions to Schrodinger's equation. In both cases we are just taking mathematically valid solutions to a linear DE that are not physically realizable (which is perfectly fine since not every solution to a DE has to be physically valid) and superposing these Fourier modes to get a solution that is physically reasonable i.e. a wave packet. In the case of Schrodinger's equation, since the wave packet is Gaussian in both spatial and momentum coordinates it is a very good approximation of a particle with sharply peaked momentum and position.
  15. Oct 25, 2014 #14


    User Avatar

    Staff: Mentor

    Any way you like. What we're saying is that no matter how you produce a particle, it must have a range of possible momenta (have a momentum uncertainty) and therefore be describable as a localized wave packet.

    How do you think one might prepare a particle that is not a localized wave packet, i.e. has a single definite momentum?
    Last edited: Oct 25, 2014
  16. Oct 25, 2014 #15


    Staff: Mentor

    Have you studied Fourier transforms?

    If you have then you know that any function, mathematically, can, in a sense, be considered to be composed of waves by taking it's Fourier transform. That's all it means - if you take the Fourier transform of a particles wavefunction, because physically it should not have a probability of being at infinity, its composed of a number of waves. If it was just one wave then the wave would extend to infinity, which is not actually realizable, but is introduced for mathematical convenience.

    In QM the Fourier transform basically transforms from the position representation to the momentum representation.

    If you have studied linear algebra its simply a change of basis.

  17. Oct 27, 2014 #16
    It basically takes the infinite time to measure the free electron momentum alone so those states don't really exist physically since even the whole Universe has the finite size. It also corresponds to the time the free localized Gaussian wave packet needs to spread near infinitely all over the whole space from the localized one while having the momentum phase factor. According to Copenhagen interpretation the wave function will collapse after the measurement to the eigenfunction of the momentum which is the pure plane wave and this is the only way to prepare the electron in this state. To see that one may take the Schiff handbook quantum mechanics method for the atom with the Doppler effect to measure the electron velocity from this effect and multiply by its mass. To measure the electron momentum from the Doppler effect one must use the electron internal degrees of freedom like spin since it is not the atom. The only way to measure between spin transition is to put electron in the magnetic field.
    This however will alter the electron trajectories to circles and it will not be free. One must use therefore the magnetic field infinitesimally close to 0.
    The energy between the spin energy levels will be close to zero too and so the frequency of the photon emission when the electron is
    originally prepared in the upper state. To measure the frequency shift of the near zero frequency it takes to measure its near infinite period
    so the momentum measurement time goes to infinity so physical electron can never collapse to the pure plane wave state. The standard trick for calculations however is to put stuff in the box so the state are exp^{i k r}/V^{1/2} so while itegrated over the whole volume V the norm is 1 and later go with V to infinity in results wherever the limits stay finite and have physical meaning.
    Last edited: Oct 27, 2014
  18. Oct 27, 2014 #17
    Hi Waxterzz. I enjoyed your free hand drawings.

    As for one particle- one wave, I would like to draw your attention that there is no unique rule on how to define unit and count waves. Fourier transform applies sinusoidal waves as units but it is mere one of possible rules. In vector mathematics, you can choose the direction of the given vector as your x-axis so that the vector lies on your chosen x-axis. Similarly even though your "wave packet" seems mixed waves and quite strange form it has, you can choose it as your "one" wave. I would like to add that unit length of this wave "vector" corresponds the number of particle one.
    Last edited: Oct 27, 2014
  19. Oct 27, 2014 #18
    Hi, Waxterzz As for position and momentum,
    if we realize a specified position with a specified error, automatically we know all about the momentum as QM allows as far. More precisely in other words coordinate wave function [itex]\psi(\mathbf r)[/itex] has all the information of the system including behavior of momentum.
  20. Oct 27, 2014 #19
    Hi Waxterzz. In momentum space wave packet of free particle is stable as you expect in superposion of linear equation solutuons, but in coordinate space wave packet collapses or diverges. Various probability of momentum is expressed as sum of sinusoidal waves. Uncertainty of speed causes broadening of position as time goes.
  21. Oct 27, 2014 #20
    The linear combination therefore means no more or less that that if the particle is in any free particle quantum state (may be Gaussian, exponential or any other as the function of position x) the measurement of momentum will result
    1) In obtaining the one real value of the measurement (most likely each time different like when tossing the dice) 2) That after each perfect momentum measurement the quantum state will collapse to or became exactly one plane wave. "The linear combination is normalizable" means here that when one repeats the momentum measurement many times on the same quantum state assuming it can be precisely cloned or replicated for the next and next measurement the probability to measure any value of momentum is equal to 1 in total or the modulus square of the expansion coefficient onto the plane waves called Fourier transform is the probability (density) to obtain the measurement value at that plane wave or with its momentum. It allows to calculate the average momentum as finite from all the measurements as the statistical average.
    Free particle wave packet can be constructed from its name by freeing the electron where there is no influence of the potential for example by kicking the ground state of the Hydrogen atom so strongly by the electromagnetic field (laser) that it will keep its original spike semi-Gaussian character but will leave far immediately.
    Last edited: Oct 27, 2014
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook