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

Math of environment-particle entanglement

  1. Mar 24, 2015 #1
    I'd like to understand the math of how something specific in the environment (like photons) can be said to be entangled with a system (like dust particles).

    Im reading a newbie friendly intro to superposition and entanglement "Home of the Wave Function"... I'd like to visualize entanglement using purely Schroedinger wave equations. I read that in entanglement (is what follows right)?

    "We have already discussed the linear superposition of different quantum waves for the same system. But supposed we have two or more systems. Two or more quantum systems can share the same quantum wave. When this happens, we say that the systems are entangled with each other. The separate systems no longer have quantum waves of their own. The process "quantum connects" the two systems no matter how far apart they get from each other, provided that they are protected against collitions with other particles.

    Getting to Hilbert space. I know the Schroedinger Wave is a vector in Hilbert space.. but in the following passage where vanesh mentioned:

    "In the case that a quantum state of a multi-system is NOT one of those very special states which are "factorisable" (psi(p,q) = f(p) g(q)), then we say that the quantum state is an ENTANGLED state of the subsystems."

    What is meant by factorisable and what does this mean psi(p,q) = f(p) g(q)? what is f and g? I know vectors and components axis in Hilbert space... What is the equivalent of factorisable in pure Schroedinger wave equation?

    Can you give an actual example of how an actual specific thing in the environment can be entangled with a system in decoherence? like photon with dust? How would this form entanglement where it is not factorisable?
  2. jcsd
  3. Mar 24, 2015 #2


    User Avatar
    Science Advisor
    Gold Member

    The quantum state of a pair of particles is said to be entangled if you can't express it as a mixture of product states (as opposed to a superposition)

    If you have a single wavefunction for two particles that is non-factorizeable, that means you cannot express it as the product of two wavefunctions, one for each particle.

    if a photon scatters off an atom, the photon and atom interact in the scattering process. The quantum state describing the two of them together becomes entangled. The way that this can create entanglement can be seen in terms of conservation laws.

    The collision between atom and photon must conserve total momentum, but the uncertainties in the momenta of atom and photon need not change much in the collision. In order for the total momentum to be conserved, while still obeying the laws of quantum mechanics, the state describing the atom and photon together must be entangled.
  4. Mar 24, 2015 #3
    In decoherence.. what else comprise the environment that can entangle with the system besides photons? maybe Microwave Background Radiation? what else? It appears that it is hard to create entangled pair like in EPR or para down conversion in photons, yet in decoherence the environment can easily get entangled with the system, why is that?
  5. Mar 24, 2015 #4


    User Avatar
    Science Advisor
    Gold Member

    Any quantum object in "the environment" that is capable of interacting with a system at all is capable of being entangled with it.

    So, cosmic rays, space dust, photons of all energies (from radio, to microwave, to IR, to visible, to UV, X-rays, gamma rays, and beyond), particles in the solar wind, all could interact with a quantum system, and entangle with it.

    The difficulty is not so much in creating entanglement.
    Rather, it is in isolating a quantum system from the environment, so that a pair of entangled particles stay entangled and don't decohere.
    Photons in parametric downconversion are nice because they don't interact strongly with the environment. Other quantum systems are harder to isolate, and so are harder to work with.
  6. Mar 24, 2015 #5
    I'd like to understand entanglement using purely the image of waves, in superposition the book says

    "Now, suppose two waves are heading towards each other, with amplitudes in the range such that the superposition principle applies. When the waves meet, a complex wave is created according to the rule given above. If the waves then separate, the original amplitudes are retained (barring friction). That is, no distortion was introduced, and the wave is linear. Thus, a complex linear sstem can be decomposed without destroying the system. You can see that linear systems lend themselves to mathematical treatment because they can be reduced to a simple superposition of elements."

    What is the wave equivalent in entanglement such that the original amplitudes are no longer retained?
  7. Mar 24, 2015 #6


    User Avatar
    Staff Emeritus
    Science Advisor

    For simplicity, let's just consider one spatial dimension, [itex]x[/itex]. Then a general wave function for a two-particle state would be:

    [itex]\psi(x_1, x_2)[/itex]

    The meaning of this is that [itex]|\psi(x_1, x_2)|^2[/itex] is the probability density of finding the first particle at position [itex]x_1[/itex] and the second particle at position [itex]x_2[/itex].

    Such a state is "factorizable" if there are two functions, [itex]A(x)[/itex] and [itex]B(x)[/itex] such that:

    [itex]\psi(x_1, x_2) = A(x_1) B(x_2)[/itex]

    It's "entangled" if that's not possible. So for example, [itex]\psi(x_1, x_2) = \frac{1}{\sqrt{2}} (A_1(x_1) B_1(x_2) + A_2(x_1) B_2(x_2))[/itex] is entangled, if [itex]A_1[/itex] and [itex]A_2[/itex] are orthogonal states, and [itex]B_1[/itex] and [itex]B_2[/itex] are orthogonal states.
  8. Mar 24, 2015 #7


    User Avatar

    Staff: Mentor

    You have the two waves ##|A+B-\rangle## and ##|A-B+\rangle## propagating happily in the superimposed state ##\psi=\frac{\sqrt{2}}{2}(|A+B-\rangle+|A-B+\rangle)##, just as in your "waves heading towards each other" situation you have the superposition ##\psi=\frac{\sqrt{2}}{2}(\psi_L+\psi_R)## where ##\psi_L## and ##\psi_R## are the left-moving and right-moving waves.

    The superposition lasts until you perform a measurement.

    In the ##\psi=\frac{\sqrt{2}}{2}(\psi_L+\psi_R)## case, a measurement of the position yielding ##X_0## causes the wavefunction to collapse into ##\psi=\delta(x-X_0)##, which is not a superposition in the position basis (warning: there are a number of mathematical subtleties about the continuous spectrum of the position operator that I have glossed over). In the ##\psi=\frac{\sqrt{2}}{2}(|A+B-\rangle+|A-B+\rangle)## case, a measurement of either ##A## or ##B## (for definiteness, we can assume that these are measurements of the spin of two particles and ##|A+B-\rangle## is the state in which particle A is spin-up and its entangled partner B is spin-down) will cause the wavefunction to collapse into either ##|A+B-\rangle## or ##|A-B+\rangle##, neither of which is a superposition in this basis.

    (These post-measurement states are still superpositions in other bases. "Going North" is not a superposition if I've using North and East as my basis whereas "going Northeast" is (it's superposition of North and East). However, if I use Northeast and Northwest as my basis, it's the other away around).
  9. Mar 24, 2015 #8


    Staff: Mentor

    If anyone can explain it that way - I will tip my hat to them.

    The very concept of superposition and entanglement is mathematical. It is this - if |a> and |b> are states then c1*|a> + c2*|b> where c1 and c2 are complex numbers is also a state.

    Entanglement is an extension of superposition to different systems. Suppose two systems can be in state |a> and |b>. If system 1 is in state |a> and system 2 is in state |b> that is written as |a>|b>. If system 1 is in state |b> and system 2 is in state |a> that is written as |b>|a>. But we now apply the principle of superposition so that c1*|a>|b> + c2*|b>|a> is a possible state, The systems are entangled - neither system 1 or system 2 are in a definite state - its in a peculiar non-classical state the combined systems are in.

    It impossible, utterly impossible to explain it in English. You must do the math.

    I have previously given you a link that with a minimum of math explains all this stuff:

    Griffith's has been kind enough to make his book freely available. It is written not only for mathematicians and physicists, but philosophers as well. It avoids advanced math as much as possible - but since it cant be explained any other way some math is required.

    If you want to understand this you must put the effort in to come to grips with even a minimal amount of math. There is no other way. English and images is simply not up to the task.

  10. Mar 24, 2015 #9
    Yes. I wholeheartedly agree with you that one needs to understand the math. That's what I'm doing now. But it will take 3 month to read Griffith's book. That is why I just need to know something now which is why I made this thread. What is the meaning of "|b>"? , why is there a vertical line and an arrow to the right with b at middle? Is it differential algebra or differential geometry or what branch of math does that fall under? Also in the following I know that the "+" or plus in "c1*|a>|b> + c2*|b>|a>" is that there is interference. But what is c1*|a>|b>? I know c1 is some probability but what is |a>|b>? Does it mean you multiply |a> with |b>? Why. Can you put actual values into them so I know what exacty the equation is doing? That is all I need to know for know so in your reply pls. just answer the above without adding others so I won't be confuse. Thank you.

  11. Mar 24, 2015 #10


    Staff: Mentor

    Its called the Dirac Bra-Ket notation. Griffith explains it all:

    You are over thinking this - all its saying is the states form a vector space - that's it, that's all.

    After you anderstand that you can then come to grips with what it means physically.

    That such is the case ie the states form a vector space is one of the rock bottom essences of QM. It can be explained by deeper principles, but that is just by the by, to start with just accept its how QM is.

    Diracs explanation may help you:

    Last edited: Mar 24, 2015
  12. Mar 24, 2015 #11
    If c1*|a>|b> is dirac bra-ket notation. what is this [itex]\psi(x_1, x_2) = A(x_1) B(x_2)[/itex], what notation or math is that? linear algebra or differential algebra? just tell me why do you have to multiply [itex]A(x_1)[/itex] by [itex] B(x_2)[/itex], is it just to signify supeposition? I know the Schroedinger Equation and what [itex]\psi[/itex] means and squaring psi to get probability.
  13. Mar 24, 2015 #12


    Staff: Mentor

    Again Griffiths explains it - please make the time and effort to go through it.

    Technically the wave-function is the expansion of the state in terms of position eigenvectors. This leads to |a>|b> in the dirac bra-ket notation becoming [itex]\psi(x_1, x_2)[/itex].

    Me telling you that however will not make sense without the background to understand it. There is no short-cut - you must start from the start and progress.

    Please read Griffiths from the start and post with any questions - do not try to skip.

  14. Mar 24, 2015 #13
    Thanks. Last question before I finish reading it which would take a year in March 2016 (because I have to finish other books). It's great to know the relationship between |a>|b> and [itex]\psi(x_1, x_2)[/itex] being "Technically the wave-function is the expansion of the state in terms of position eigenvectors. This leads to |a>|b> in the dirac bra-ket notation becoming [itex]\psi(x_1, x_2)[/itex]." I know the meaning of eigenvectors being familiar with hilbert space. Now in Nugatory ##\psi=\frac{\sqrt{2}}{2}(|A+B-\rangle+|A-B+\rangle)##, what kind of math is that? why is there A + B -, how does it relate to both |a>|b> and [itex]\psi(x_1, x_2)[/itex]? It's ok if Nugatory or Stevendaryl (thanks to both too) will answer because Bhobba will just tell me to read the whole book and know the answer next year when just want a *rough* idea of it now.
  15. Mar 24, 2015 #14


    User Avatar
    Science Advisor
    Gold Member

    If you just had two particles, each described by their own wavefunction, and there was no interaction between the two, you would see the two waves interfere, and separate out again.

    In this non-interacting case, the two particles would be described independently by their own wave equations.

    In order for there to be entanglement, there must be an interaction between the two particles.

    To accommodate this interaction, you can no longer describe the behavior of the two particles by two different wave equations. You would have to have one bigger wave equation that can't separate into two because of extra terms describing the interaction between the two particles.
  16. Mar 24, 2015 #15
    I have written a popular book on this, that explains it with hardly any math, in conceptual terms. See my website rekastner.wordpress.com for info.
  17. Mar 25, 2015 #16


    Staff: Mentor

    Its vector space math.

    You might think you know what a Hilbert Space is, but if that is unclear you don't - its simple vector space math and Hilbert spaces are vector spaces:

  18. Mar 25, 2015 #17


    User Avatar
    Gold Member

    As the subject is "the math of entanglement", I am surprised to see that the word Hamiltonian does not appear!
    A two level system to be measured is prepared as a |s0> + b|s1>.
    A measurement device for this obsevable is in its rest state ##|S_0>## so we start with ##(a |s_0> + b|s_1) > \otimes |S_0>##
    We have to associate an hamiltonian H(t) so that this vector evolves to
    ##a |s_0>\otimes |S_0> + b|s_1 > \otimes |S_1>## Here S_1 is a state of the apparatus coupled to the second level of the system.
    This describes the premeasurement or entanglement.
    I have here a question. If the two levels are the eigenvectors of an hermitian operator, they are orthogonal. What about ##|S_0>## and ##|S_1>##?
    Decoherence will occur further. What has to be added so that it will choose this prefered basis (what about an orthonormal basis?)
  19. Mar 25, 2015 #18


    User Avatar
    Staff Emeritus
    Science Advisor

    It's a two particle wave function. A wave-function is a function on configuration space. Configuration space has one coordinate for each dimension of space and for each particle; so 2 particles in 3D space produces a 6-dimensional configuration space: [itex]x_1, y_1, z_1, x_2, y_2, z_2[/itex]

    No, it has nothing to do with superpositions. You've never worked with the Schrodinger equation for more than one particle? As I said, in general, the Schrodinger describes a wave function, which is a function on configuration space.

    Suppose you have two particles with a harmonic oscillator interaction. Then the Schrodinger equation for such a system would be given by:

    [itex]H \psi(x_1, x_2) = (-\frac{\hbar^2}{2m_1} \frac{d^2}{dx_1^2} -\frac{\hbar^2}{2m_2} \frac{d^2}{dx_2^2} + \frac{k}{2} (x_2 - x_1)^2) \psi(x_1, x_2)[/itex]
    Last edited: Mar 25, 2015
  20. Mar 25, 2015 #19


    User Avatar
    Gold Member

    Maybe the intrication between apparatus and environment does not only lead to projective measurements but to POVMs?
  21. Mar 25, 2015 #20
    some questions:

    Is entanglement transitive? If system A is entangled with system B, and which are coherent w.r.t. the environment, and then system C fully interacts with A, entangling with A, then does system B remain entangled to A, and if it does, then is B entangled to C?

    Does lack of entanglement imply superposition?

    Are the atoms in a molecule all entangled together?

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Math of environment-particle entanglement
  1. Entangled particles (Replies: 3)

  2. Entangled particles (Replies: 2)