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Quantum states and representation freedom

  1. Aug 10, 2014 #1
    Hello Forum,

    When a system is in a particular state, indicated by a |A>, we can use any basis of eigenvectors to represent it. Every operator that represents an observable has a set of eigenstates. I bet there are operators with only one eigenstate or no eigenstates. There are operators, like the evolution operator, that does not represent an observable.

    The position operator has an infinite spectrum of eigenvalues and infinite eigenstates. When we see the wavefunction written as Psi(x), the position representation is implicit.
    The energy operator (called Hamiltonian) also has an infinite spectrum and infinite eigenvectors.
    The momentum operator has an infinite spectrum of eigenvalues and infinite eigenstates.
    What about the angular momentum operator?

    The spin operator has only two eigenvalues and two eigenstates, correct? How is it possible to represent any general state using only two eigenvalues and two eigenvectors?

    So, in general, the generic state |A> can be written as a sum of any of those eigenvectors, correct?

    What is the difference between a pure state and a mixed state?

    Why do we call "1st quantization" the process of converting observables to operators? Does "2nd quantization" represent the conversion of the wavefunctions, i.e. states, to operators?

    Thanks,

    Fog37
     
  2. jcsd
  3. Aug 10, 2014 #2

    Matterwave

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    Let me see if I can answer your questions in a concise manner. They seem to be pretty broad in scope.

    1) An operator's spectrum depends on the Hilbert space on which it acts. For example, the momentum operator no longer has a continuous spectrum if you are in a bounded space (e.g. on a ring, or in the infinite square well where you are limited to operate on a closed interval). The angular momentum operator in regular 3-space has a countably infinite number of eigenvalues and eigenstates enumerated usually by ##\left|l,m\right>##.

    2) The state ##\left|\psi\right>## lives in some Hilbert space. It can be deconstructed into any valid set of vectors that form a basis for the Hilbert space it lives in. However, not all states live in the same Hilbert space. A state describing the spin of a spin 1/2 particle ##\left|s_z\right>##, for example, lives in a different Hilbert space than a momentum eigenstate ##\left|p\right>##. You can't deconstruct a state in one Hilbert space in terms of states in another Hilbert space.

    3) The difference between a pure and mixed state is a slightly subtle point. I'm not sure I can explain it very well in a concise manner. Basically a pure state has all the "interference" information between basis states, while a mixed state has lost these interference terms. One usually hears a pure state is a coherent mixture of states, while a mixed state is a statistical ensemble of states. In math language, a pure state is one which can be represented by a state vector ##\left|\psi\right>## while a mixed state must be expressed as a density matrix ##\rho=\sum_n p_n \left|\psi_n\right>\left<\psi_n\right|##. A pure state can also be represented by a density matrix ##\rho=\left|\psi\right>\left<\psi\right|##

    In this language, a pure state is one in which ##\rho^2=\rho##, while a mixed state does not have this equality hold.

    4) First quantization and second quantization is basically old language. Roughly it corresponded to 1st quantization being promoting observables to operators, and second quantization being promoting wave functions to operators (while demoting others). This language is not seen quite as much anymore.
     
  4. Aug 10, 2014 #3

    bhobba

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    You need to read the first two chapters of Ballentine - QM - A Modern Development:
    https://www.amazon.com/Quantum-Mechanics-A-Modern-Development/dp/9810241054

    But as an overview of mixed and pure states here is the skinny.

    First get out of your mind a state is an element of a vector space. It isn't - forget all you have read otherwise - its WRONG.

    In general its a positive operator of unit trace, P, such that the expected outcome of an observation with observable O is E(O) = Trace (PO). A state of the form |u><u| is called pure. A convex sum of pure states is called mixed. It can be shown all states are mixed or pure. The |u> in the pure state can be mapped to a vector space - but not uniquely ie |u><u| = |cu><cu| for any complex number c.

    When people speak of states from a vector space they mean pure states. For pure states its simple to see E(O) = <u|O|u>. Now, just as an example, the observable of position that gives one for position x and zero otherwise is O = |x><x|. Then of course E(O) = probability of getting x. And from that its again easy to see E(O) = <u|x><x|u> = |<u|x>|^2 which is how the squaring rule comes into it.

    To understand eigenvalues and eigenvectors better you need to investigate the spectral theorem, but Ballentine gives you what you need to know about it for the purposes of QM.

    Thanks
    Bill
     
    Last edited by a moderator: May 6, 2017
  5. Aug 11, 2014 #4

    atyy

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    Is it wrong, or a matter of interpretation? Wrong is better, because then Many-Worlds will be immediately falsified :P
     
  6. Aug 11, 2014 #5

    tom.stoer

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    Let me correct: "When a system is in a particular state, indicated by a |A>, we can use any basis of eigenvectors of a self-adjoint operator to represent it."

    Could be, but if the the eigenstates do not span the full Hilbert space, then the operator is not self-adjoint and therefore does not represent an observable.

    It depends on the states you are interested in. If you want to describe spin s = 1/2 and nothing else, then the basis vectors are |s>, and everything is fine. If you want to describe spin and momentum, then the basis vectors read |p,s>, so for each p you have a two-dim subspace of the full Hilbert space.

    So in the latter case you have to use

    [tex]|A\rangle = \sum_p\,\sum_s\,a_{p,s}\,|p,s\rangle[/tex]
     
  7. Aug 11, 2014 #6

    bhobba

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    It is WRONG - not interpretationally wrong - but WRONG.

    The following state 1/2 (|a><a| + |b><b|) is not an element of a vector space, nor can it be mapped to one. It is a mixed state however.

    And it doesn't falcify MW - in fact MW very elegantly uses it.

    What decoherence does is transform a superposition like 1/root 2 (|a> + |b>) to a mixed state like 1/2 (|a><a| + |b><b|). |a><a| and |b><b| are each interpreted as their own world and each evolve as part of that state - no collapse - no nothing.

    Thanks
    Bill
     
    Last edited: Aug 11, 2014
  8. Aug 11, 2014 #7

    atyy

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    That's just terminology. I'm asking whether you object to Many-Worlds which is based on pure state evolution as fundamental. If pure state evolution is fundamental, then one should be able to base quantum mechanics on pure states which are vectors in a Hilbert space. Also if you object to pure states as fundamental, then one should also not accept Schroedinger's equation as fundamental. Mixed states do not generally evolve following Schroedinger's equation.
     
  9. Aug 11, 2014 #8

    bhobba

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    Atty - I think you are forgetting that the pure state of an entangled system, when you just observe one component, is a mixed state.

    This is one of the key areas detractors of MW harp on about - its dependant of that separation ie its the factoring problem.

    Thanks
    Bill
     
  10. Aug 11, 2014 #9

    atyy

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    I disagree. One is certainly entitled to define states in QM as pure states, and those are elements of a Hilbert space. It's like saying the projection postulate is wrong because one should use POVMs, or that POVMs are wrong because one should use the projection postulate. In fact whether one defines states as operators or states as elements in a vector space, the resulting physical theory and all its predictions are the same, so it is a matter of interpretation.
     
  11. Aug 11, 2014 #10

    bhobba

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    You can define anything as anything you like.

    But if you do that then what do you call P in the Born Rule which is Trace (PO) where P is any positive operator of unit trace?

    The usual approach taken is to start with elements of a vector space as states - see that it has phase invariance - which is rather interesting. Then you consider the situation where someone randomly presents you with a state to be observed and find that can't be a state. However if you define pi as that probability if P = ∑ pi |ui><ui| then E(O) = ∑ pi <ui|O|ui> = Trace (PO). Then you extend the definition of state to include these mixed states. And as Ballentine points out they occur in solving problems as well.

    You are only making things harder ignoring them.

    Ballentine, and me, simply deal with it from the start.

    Thanks
    Bill
     
  12. Aug 11, 2014 #11

    tom.stoer

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    I think we are far from the original question
     
  13. Aug 11, 2014 #12

    vanhees71

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    May be we are far from the original question, which we cannot answer in the forum. It's a good advice to point the poster to a good textbook, and Ballentine is among the best I know. So I strongly recommend to read the first parts of this book to get clear definitions about the fundamental objects of the theory without interpretational ballast you can add later, if you want, and in this book it's pretty well shown that most of it is in fact ballast you don't need to bother about as long as you stick to physics and don't want to enter philosophy or metaphysics. I'd consider this off topic in this forum.

    Now to the point concerning states: Pure states, i.e., states which provide maximal possible information about a quantum system, are represented by rays in Hilbert space, i.e., by a normalized Hilbert-space vector modulo a phase factor. It is of utmost importance to keep this in mind, because many important things are ununderstandable, if you simply define a state to be represented by a normalized Hilbert-space vector. E.g., you could not come to the idea that also the half-integer representations of the Lie algebra of the rotation group (angular-momentum algebra) has a physical meaning, and in fact it has, because all matter around us is composed of spin-1/2 particles (quarks and leptons). Alternatively you can define pure states as being represented by projection operators, [itex]\hat{R}=|\psi \rangle \langle \psi|[/itex] with a normalized Hilbert-space vector [itex]|\psi \rangle[/itex].

    In addition you need mixed states, which are represented by positive semidefinite self-adjoint operators with trace 1, the statistical operator, if you don't have complete information about a quantum system, i.e., if it is not known to be prepared in a state, where a complete compatible set of observables is determined. Of course, the pure state, represented by a projection operator is then only a special case of this general description by a statistical operator.
     
  14. Aug 11, 2014 #13

    tom.stoer

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    Why not???

    It has nothing to do with the Hilbert space states but with the fact that you have to use SU(2) instead of SO(3)
     
  15. Aug 11, 2014 #14

    vanhees71

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    Well, if you assume that the Hilbert space vector represents the states, then for half-integer spins a rotation around the 3-axis (quantization axis) immediately leads to [itex]\hat{D}_3(2 \pi) |\psi \rangle=-|\psi \rangle[/itex], and this would have to be considered as representing a different state than [itex]|\psi \rangle[/itex]. On the other hand, a rotation by [itex]360^{\circ}[/itex] should be the identity.

    Now, the states are to be considered equal if the representing state vectors differ only by a non-vanishing factor, but then [itex]|\psi \rangle[/itex] and [itex]-|\psi \rangle[/itex] represent indeed the same state, and thus you admit the half-integer irreps of the rotation Lie algebra. Reconstructing the group leads to representations of SU(2) which is the covering group of SO(3), which is obviously correct, because there are particles with half-integer spin which behave exactly as predicted by this interpretation.

    Another, even more drastic argument is, that you wouldn't even get a useful non-relativistic quantum theory that is compatible with the Galileo symmetry of Galileo-Newton spacetime! The fact that the rays represent states not the state vectors themselves leads to the conclusion that I do not need to realize the Galileo group by a unitary representation but a unitary ray representation is sufficient, and this admits both to use the irreps of the covering group as well as the central extensions of the group. Indeed, good old Heisenberg-Schrödinger non-relativistic quantum mechanics realizes a ray representation of a central extension of the Galileo group, with the mass as the central charge (and of course the SO(3) rotation subgroup substituted with its covering group SU(2)).

    Wigner and Inönü have shown in a famous paper that the unitary irreps of the original Galilei group do not lead to a sensible quantum dynamics of non-relativistic particles.

    http://link.springer.com/article/10.1007/BF02782239
    http://www.ncbi.nlm.nih.gov/pmc/articles/PMC527951/pdf/pnas00729-0073.pdf
     
  16. Aug 11, 2014 #15
    Thanks everyone. I am carefully going through your responses.

    I am reading the book "Essential Quantum Mechanics" by Gary Bowman. It is an introductory book.

    The book teaches that he system state |ψ⟩ embodies information about the system’s various observables, in various representations. As an abstract entity, the state itself is representation free. It exists independently of any language, any representation we choose to write it. Dirac notation is great because it symbolizes the representation-free object |ψ⟩.

    If a state is expanded in the eigenstates of some operator, and if that expansion consists of only one term, then the state is not a superposition in that representation. If there is more than one term in the expansion, the state is a superposition in that representation.

    Particle in the box problem: the energy eigenstates in introductory books are expresses as functions of the space variable x. The energy values are quantized. Each possible position, each x, corresponds to an eigenstate of the position operator x. The eigenstates and eigenvalues of H are discrete for this system while those for x are continuous.

    The eigenstates constituents are written as superpositions in the x representation. So which representation is the state |ψ⟩ written in? The H operator representation of the x operator representation? The answer seems to be the H representation since there is one coefficient for each H eigenstate and two coefficients for each eigenstate of x.

    The point seems to be that the particular state |ψ⟩ can be expanded either using the basis eigenstates of the H operator, x position operator, spin operator, etc...(all self-adjoint). What is strange is that the spin basis has only two eigenstates (and two eigenvalues, the + hbar/2 and -hbar/2). That does not seem to give flexibility to represent any particular vector....
     
  17. Aug 11, 2014 #16

    tom.stoer

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    You are still confusing two different systems, namely the particle in a box and a spin-1/2 system. They are two different systems!

    For a particle in an infinite square well potential confined to the [0,L] you may have an energy eigenstate |n>. The "energy-representation" is just this: |n>, or |En> if you like. For the x-representation you have to use the wave function ψ(x) which is a projection of |n> on the position-states <x|; so you get

    [tex]\psi_n(x) = \langle x|n\rangle = \sin k_nx;\;k_n = n\pi/L;\;n=1,2,\ldots[/tex]

    That means that the energy-eigenstate |n> is represented in x-rep. by the above mentioned wave function ψn(x).

    There is no spin! This particle does not carry spin! You cannot / must not use any spin variable for this particle b/c there is none!

    So for the particle in an arbitrary state |ψ> w/o spin you may use the following two representations.

    [tex]|\psi\rangle = \sum_{n=1}^\infty \psi_n\,|n\rangle[/tex]
    [tex]|\psi\rangle = \int_{-\infty}^{+\infty}dx\;|x\rangle\langle x|\psi\rangle = \int_{-\infty}^{+\infty}dx\;\psi(x)\;|x\rangle[/tex]

    (Note that for an energy eigenstate the first sum collapses to one single term)

    Now what you can do is to introduce an additional degree of freedom called spin. For a free particle where the Hamiltonian does not depend in the spin you find new states |n,s> where n is just the counting if the eigenvalues for the energy, and s is the new spin d.o.f.

    Then you find for an arbitrary state /w spin

    [tex]|\psi\rangle = \sum_s \sum_{n=1}^\infty \psi_{n,s}|n,s\rangle[/tex]
    [tex]|\psi\rangle = \sum_s \int_{-\infty}^{+\infty}dx\;|x,s\rangle\langle x,s|\psi\rangle = \sum_s \int_{-\infty}^{+\infty}dx\;\psi_s(x)\;|x,s\rangle[/tex]

    But now you are describing a system with spin! It's different from the first one. The spin does not replace the n or the x; it's an additional d.o.f. of your system.

    (Note that the new wave function ψs(x) is a two-component spinor for spin 1/2)
     
  18. Aug 11, 2014 #17
    I see now.

    My mistake was to put spin, an observable, on the same footing as energy H, position x, or momentum p and since those operators have a complete basis of eigenstates through which represent the particular state the system is in, I thought that the state expansion was also possible and valid using the spin eigenstates...
     
  19. Aug 11, 2014 #18

    tom.stoer

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    It simply depends on the size of your Hilbert space.

    Let's look at a 3-dim. eigenvalue problem (A - a)|a> = 0 with a 3*3 hermitean matrix A. Of course A has 3 lin. indep. eigenvectors |a>, and of course they span 3-space. Now Let's look at a 2-dim. eigenvalue problem (B - b)|b> = 0. Of course the 2 eigenvectors |b> do not span 3-space.

    Now you could think of 3-space as 2-space plus a 3rd direction, i.e. somehow enlarging 2-space to 3-space. For B and A, i.e. |b> and |a> this could work.

    But for position space and spin-space the situation is different. In the presence of spin you habe to use a tensor, something like |n,s> in the energy-rep. This is b/c position and spin are two entirely different physical quantities.

    In the example with A and B this would mean you solve for the 3 eigenvalues a with 3 eigenvectors |a>. Then you do the same for b and |b>. Now the important step is NOT to consider 2-space as a subspace of 3-space but to introduce the 2*3 = 6-dim. tensor product space

    [tex]|a,b\rangle = |a\rangle \otimes |b\rangle[/tex]

    Look at energy and spin; the states are

    [tex]|E_n,s\rangle[/tex]

    That means that you do not describe the system either in terms of energy or in terms of spin (which you do for position and momentum: you use either position of momentum), but you describe a system using both energy and momentum.
     
  20. Aug 11, 2014 #19
    Thanks for the patience Tom!

    Let me "try" to summarize some points:

    So the "state" that is a system is in is represented mathematically by a vector. The system is characterized by its observables which can be extracted from the state vector itself. The observables are energy, angular momentum, spin, position, linear momentum, etc. (which are linear operators in QM).

    The particular state the [tex] |psi\rangle [/tex] can be expanded into a linear superposition of eigenvectors that belong to an operator (observable). The state [tex] |psi\rangle [/tex] could expanded using only the eigenstates of the position operator, only the eigenstates of the momentum operator, only the eigenstates of the energy operator, etc but that is the expansion may not possible in the case of certain other operators.

    It also seems possible (see your example of energy and spin eigenvectors) to represent the particular state [tex] |psi\rangle [/tex] the system is in using eigenvectors of two operators at the same time: the same state can have a representation using only one set of eigenvectors or two set of eigenvectors, i.e. it can be represented using two observables....

    But if the system does not have a specific observable (like spin, for example, in the case of the particle in the box), then it is not possible to expand the state using spin eigenstates.

    At the beginning of the thread it was said that "1st quantization" means that the observables become operators in QM. Quantization means that the measurable values of the observables come up in discretely separated amounts. What is the connection between being an operator and the values of measurable observables being quantized?
    Is quantization implied when an observable becomes an operator?
     
  21. Aug 11, 2014 #20

    tom.stoer

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    yes, exactly

    It's a bit more complicated than that; but yes, quantization essentially means to translate obervables as functions on phase space into linear, self-adjoint operators acting on a Hilbert space.

    No, that's not true. After having constructed an operators A representing an observable one can solve the eigenvalue problem; then math tells you whether the spectrum is discrete or continuous, but this is not clear from the beginning. In many cases the spectrum of an observable contains both a discrete and a continuous part, i.e. the bound states of the Hydrogen atom have discrete energies whereas the scattering states have a continuous spectrum.

    As I said, it's the result of the eigenvalue problem on a specific Hilbert space. Look at the momentum operator -i∂x: it has a continuous spectrum when evaluated on the real line (which is quite messy b/c the plane waves are non-normalizable "generalized eigenfunctions"), but it has a discrete spectrum when evaluated for the infinite square well. The difference between these two cases is clearly not the mathematical expression for the operator (in both cases it's -i∂x) but also its domain (real line or finite inteval) and the boundary conditions (in the latter case following from the physical condition that the wave functions must have zeros at the boundaries).
     
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