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B Mixed states V superposition V linear combinations?

  1. Apr 5, 2016 #1
    Can someone explain the difference using concrete examples. I will attempt to explain my current understanding by example;

    A H atom has different energy levels which can be exactly described by algebraic functions with quantum numbers n, l etc.

    An electron can be excited from say the ground state to an excited state.

    The electron is now in a linear combination of the two States or a mixed state or a superposition state???

    Thanks any explanation.
     
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  3. Apr 5, 2016 #2

    bhobba

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    That's not correct. In a hydrogen atom they are in pure states. Mixed states are something entirely different that goes right to the foundations of QM.

    First we will start with the axioms of QM:

    Axiom 1
    Associated with each measurement we can find a Hermitian operator O, called the observations observable, such that the possible outcomes of the observation are its eigenvalues.

    Axiom 2 - Called the Born Rule
    Associated with any system is a positive operator of unit trace, P, called the state of the system, such that expected value of of the outcomes of the observation is Trace (PO).

    Note - the state of a system is not an element of a Hilbert space - it is an operator. How they come into it I will explain.

    A state is called pure if its of the form |u><u|. A state is called mixed if its the convex sum of pure states ∑ pi |ui><ui|. It can be shown all states are either pure or mixed. Its the pure states that can be mapped to the Hilbert space - the u in |u><u| can be mapped to normalised vectors but there is an ambiguity in doing so because |cu><cu| = |u><u| if c is simply a phase factor. It can be extended further for all vectors in the Hilbert space by |u/||u||><u/||u||| - hence the states become rays in the Hilbert space - but the length is on no consequence. However always bear in mind the state is really an operator - this is simply a mapping.

    Like any vector the basis you choose to write the vector in is entirely arbitrary - it purely depends on utility - and in QM that often depends on the observable you are interested in.

    Basically mixed states are the sums of pure states when those states are expressed as operators. Physically its the same as if you presented pure states for observation with probability equal to its value in that sum ie if ∑pi |bi><bi| is the mixed state its observationally exactly the same as if you presented the pure states |bi><bi| randomly for observation with probability pi.

    Thanks
    Bill
     
  4. Apr 5, 2016 #3
    That will take some time to digest. How do we get from the wave functions for hydrogen to states in Hilbert space?
     
  5. Apr 5, 2016 #4

    bhobba

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    Its the other way around. You start with states and get wave-functions.

    Suppose you are given an orthonormal basis |bi>. Then from basic linear algebra any vector |u> can be expressed using that basis as |u> = ∑<bi|u>|bi> which follows trivially from ∑ |bi><bi| = 1. If you don't understand that you need to study up on linear algebra in Dirac notation:
    http://courses.cs.washington.edu/courses/cse599d/06wi/lecturenotes2.pdf

    <bi|u> is called the representation of |u> in that basis. Physically |<bi|u>|^2 is the probability of observing |u> to see if its in state |bi> and getting a yes outcome. This follows from the Born Rule. Now in QM you can also have a continuous basis |x> where x is something like position. For that the representation of |u> is <x|u>. If x is position <x|u> is called the wave function.

    Thanks
    Bill
     
  6. Apr 5, 2016 #5
    How do calculate anything if you don't have wave functions for operators to operate on, you just have symbols.
     
  7. Apr 5, 2016 #6

    bhobba

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    You calculate from the Born Rule.

    For the wave-function <x|u> |<x|u>|^2 is the probability density of the particle being observed at position |x>.

    Thanks
    Bill
     
  8. Apr 5, 2016 #7
    Sorry not following, without wave functions there is nothing g inside those brackets.
     
  9. Apr 5, 2016 #8

    bhobba

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    Not following your concern.

    As I said if <x|u> is the wave-function ie <x| is the continuous position basis, <x|u>|^2 is the probability density of the particle being observed at position |x>.

    QM is a theory about observables and states as per the two axioms I mentioned at the start. If X is the position observable its eigenvalues are a particles position. The associated eigenvectors |x> are the states of definite position with eigenvalue x which is the position. <x|u>|^2 is the probability density of doing a position measurement you get the position x.

    Thanks
    Bill
     
  10. Apr 5, 2016 #9

    atyy

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    The wave function ##\psi(x)## is a state in the Hilbert space for a single massive non-relativistic particle, represented in the position basis.

    The scalar product in the position basis is defined as ##\langle \phi | \psi \rangle = \int {\phi^{*}(x) \psi(x)} dx ##.

    The position basis is defined such that the position representation of a position eigenvector ##|x' \rangle## is ##\langle x| x' \rangle = \delta(x - x') ##.

    If we write the state as ##| \psi \rangle##, then its representation in the position basis is ##\langle x | \psi \rangle = \psi(x) ##.

    Similarly, the Fourier transform of the wave function will give the same state in Hilbert space, but represented in the Fourier transformed basis, also called the momentum basis.
     
    Last edited: Apr 5, 2016
  11. Apr 5, 2016 #10

    blue_leaf77

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    The electron is in a superposition state, which is a linear combination of those two states and it is not a mixed state. The concept of mixed state applied to a quantum system, in this case a hydrogen atom, becomes meaningful when you consider a virtual/conceptual ensemble of H atoms (i.e. it's not a real ensemble of H atoms existing in space at the same time) prepared through certain preparation procedure in which the results of the preparation contain some statistical randomness about the produced states, e.g. 10% it produces H atom with quantum numbers (1,0,0), 50% with quantum numbers (2,1,-1), and 40% with quantum numbers (3,2,0). Note that it is not a superposition of states and you cannot represent this virtual ensemble with a wavefunction either because the outcoming H atom can have various (pure) states. But the state of this ensemble can be expressed in terms of the density operator ##\rho = \frac{1}{10}|1,0,0\rangle \langle 1,0,0| + \frac{5}{10}|2,1,-1\rangle \langle 2,1,-1| + \frac{4}{10}|3,2,0\rangle \langle 3,2,0|##
    For a brief introduction to the difference between mixed and pure states, take a look at https://quantiki.org/wiki/states.
     
  12. Apr 5, 2016 #11
    Thanks all replies.

    So the state when the electron is excited from ground state |g> to excited state |e> and oscillating between the two States to emit radiation in the linear state a|g>+b|e> is a superposition state? so superposition is another word for linear combination of two or more states?

    Is that correct??
     
  13. Apr 5, 2016 #12
    That link to me to a website that requires membership, were you intending to link to a specific article??
     
  14. Apr 5, 2016 #13

    bhobba

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    Not really - but the answer relies on QFT - its entangled with the quantum vacuum. Best to leave that alone for a while - but its tied up with the phenomena of spontaneous emission:
    https://en.wikipedia.org/wiki/Spontaneous_emission

    Yes.

    Since the states form a vector space any state is a superposition of many other states in all sorts of ways.

    Thanks
    Bill
     
  15. Apr 5, 2016 #14

    blue_leaf77

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    If you imagine your H atom to exist in a world where QFT doesn't apply and hence there is no spontaneous emission, then yes that state you describe is a superposition state. In reality, however, if you are talking about energy levels of an atom, they are always coupled to the quantum vacuum of the electromagnetic field.
    However, just be careful that even the pure state ##|\psi_i\rangle## in a density operator of a mixed state ##\rho = \sum_i p_i |\psi_i\rangle \langle \psi_i|## can also be a superposition state if it's written in a basis in which it is... well in a linear combination of those basis.
    To summarize, a system is said to have a pure state if you know for certainty that this system must be in that state which you are told. Consequently, you know for certainty the probability distributions for every observables measured in that state. In contrast, a quantum system is in a mixed state if you don't know which state you may end up getting because this quantum system comes to you at random state, with its randomness specified by the statistical mixture of the individual pure states being mixed in. Consequently, in this state and prior to the measurement of observables, you don't know the probability distribution of the outcome of this measurement.
    If you are talking about my link, I don't think so, I am using a commercial internet provider (not university). So, I don't think I could have accessed that if it requires membership.
     
  16. Apr 5, 2016 #15
    I think my initial query is resolved.

    Related but different question how is superposition different to entangled states??
     
  17. Apr 6, 2016 #16

    atyy

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    A simple version of the idea is that entanglement is a particular superposition of states of two or more particles in which the resulting state cannot be written as a product state in any basis, ie. no matter how the basis is changed, the two particle state never has the form ##|v \rangle=|a_{1} \rangle|x_{2} \rangle##, instead it always has the form##|v \rangle=|a_{1} \rangle|x_{2} \rangle + |b_{1} \rangle|y_{2} \rangle##
     
  18. Apr 6, 2016 #17
    What gets a system jnto this state and what two state system fails to be entangled?

    I guess I am asking what is the test for entangled states, theoretically.
     
  19. Apr 6, 2016 #18

    vanhees71

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    It's worth to think about this a bit more. The general formalism without using a concrete realization of the Hilbert space can be very useful to understand general properties. It goes back to Dirac's formulation of quantum theory in 1926/27 and was made mathematically rigorous by von Neumann around the same time.

    There is a one-to-one mapping between the abstract (separable) Hilbert space ##\mathcal{H}## and the wave functions, which is the realization of this separable Hilbert space as functions ##\psi:\mathbb{R}^3 \rightarrow \mathbb{C}## that are square integrable, i.e., for which the Integral ##\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} |\psi(\vec{x})|^2## exists, so that you can normalize this integral to 1 such that ##|\psi(\vec{x})|^2## makes sense as a probability distribution for the position of the particle.

    In the abstract formalism the position and momentum of the particle are represented by self-adjoint operators in ##\mathbb{H}##, but they are not defined on the entire Hilbert space but only in a smaller space, because due to the canonical commutator relations ("Heisenberg algebra"),
    $$[\hat{x}_j,\hat{x}_k]=0, \quad [\hat{p}_j,\hat{p}_k]=0, \quad [\hat{x}_j,\hat{p}_k]=\mathrm{i} \hbar,$$
    they have no proper eigenvectors that are square integrable but only socalled generalized eigenvectors in the sense of distributions. Indeed these formal eigenstates are not normalizable but obey
    $$\langle \vec{x}|\vec{x}' \rangle=\delta^{(3)}(\vec{x}-\vec{x}').$$
    Further, these states are "complete" in the sense that
    $$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} |\vec{x} \rangle \langle \vec{x}|=\hat{1}.$$
    Now there is a one-to-one mapping between the abstract Hilbert-space vectors ##|\psi \rangle## and the wave function ##\psi(\vec{x})=\langle \vec{x}|\psi \rangle##. Given ##|\psi \rangle## that's how you get the wave function. If you have given the wave function, you can use the completeness relation to get back the abstract state
    $$|\psi \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} |\vec{x} \rangle \langle \vec{x}|\psi \rangle= \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} |\vec{x} \rangle \psi(\vec{x}).$$
    Also you get the scalar product of two vectors from the wave functions by inserting a completeness relation:
    $$\langle \psi|\phi \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \langle \psi|\vec{x} \rangle \langle \vec{x}|\phi \rangle = \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \psi^*(\vec{x}) \phi(\vec{x}).$$
    There are even a few cases, where you can solve the eigenvalue problem without ever referring to wave functions by only using the commutation relations. The usual examples are: Position and momentum, angular momentum, energy spectrum of the harmonic oscillator. Also the non-relativstic hydrogen atom can be solved without wave functions (in fact it was the first way it was solved by Pauli within matrix mechanics).
     
  20. Apr 6, 2016 #19
    What numbers go on the axes in Hilbert space. I get wave functions and how they are operated on to calculate real things like energy, position etc. That makes intuitive sense and practical sense, so you got this thingvthat looks like a vector in a multidimensional space analogous to a vector space, what can you calculate with it.

    It seems you have to already know a lot about the system in order to represent it in Hilbert space.

    I want to actually find about the system in the first place and it appears solving the SWE does that non relativistically. Can't see what you gain by using Hilbert space that you don't already know from calculating the wave functions and using the operators on them all via SWE.
     
  21. Apr 6, 2016 #20

    vanhees71

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    What do you mean by numbers? There are no numbers in ##|\psi \rangle## it's a vector in an abstract vector space ##\mathcal{H}##. Of course, you can use any representation you like and calculate everything, e.g., in the position representation ("wave mechanics").
     
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