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

B Still confused about superposition and mixed states

  1. Jul 15, 2016 #1

    HAYAO

    User Avatar
    Gold Member

    I am confused about pure state or in mixed states. I've seen several threads on this forum, but I still can't get the grasp of it. I only have very little quantum chemistry to know what these means. So instead, I want to know the answer for specific examples so that I can get an idea.

    So I would like to know "Yes" or "No" to each numbered question in the following text where I attempt to derive wavefunctions and energies of new states produced by excitonic interaction between two identical molecules.

    Say for example, I have two identical molecules aligned close to each other. I can have three possible state:
    1) When both molecule is in ground state: [itex] \left | \phi _{0} \right \rangle = \left | \varphi _{1} \right \rangle \otimes \left | \varphi _{2} \right \rangle [/itex]
    2) When molecule 1 is in excited state: [itex] \left | \phi _{1} \right \rangle = \left | \varphi _{1}^{*} \right \rangle \otimes \left | \varphi _{2} \right \rangle [/itex]
    3) When molecule 2 is in excited state: [itex] \left | \phi _{2} \right \rangle = \left | \varphi _{1} \right \rangle \otimes \left | \varphi _{2}^{*} \right \rangle [/itex]
    Where [itex] \left | \varphi _{1} \right \rangle [/itex] and [itex] \left | \varphi _{2} \right \rangle [/itex] represents molecule 1 and 2 in ground state, respectively. The star means excited state. (Q1. Are all of these wavefunctions pure state?) So the total wavefunction of the entire system is:
    [tex]\left | \Phi \right \rangle = a_{0}\left | \phi _{0} \right \rangle + a_{1}\left | \phi _{1} \right \rangle + a_{2} \left | \phi _{2} \right \rangle [/tex]
    (Q2. Is this a pure state or mixed state?) The "true" hamiltonian H for deriving the energy of this system is given by:
    [tex] \textbf{H} = \textbf{H}_{0} + \textbf{H}_{J} [/tex]
    Where [itex] \textbf{H}_{0} [/itex] is the Hamiltonian in the absence of excitonic interaction, and [itex] \textbf{H}_{J} [/itex] is the excitonic interaction Hamiltonian. In the case of 2 identical molecules with excitonic interactions strength of J, ground state energy of Egr and excited state energy of Eex, the Hamiltonian is:
    [tex]\textbf{H} = \begin{pmatrix}
    E_{gr} & 0 & 0\\
    0 & E_{ex} & J \\
    0 & J & E_{ex}
    \end{pmatrix}[/tex]
    Solving the Schrodinger equation:
    [tex]\textbf{H}\left | \Phi \right \rangle = E\left | \Phi \right \rangle[/tex]
    through diagonalization gives eigenvalues of:
    [itex]E_{0} = E_{gr}[/itex]
    [itex]E_{1} = E_{ex}-J[/itex]
    [itex]E_{2} = E_{ex}+J[/itex]
    and eigenvectors of:
    [itex]\left | \Phi _{0} \right \rangle = \left | \phi _{0} \right \rangle[/itex]
    [itex]\left | \Phi _{1} \right \rangle = \frac{1}{\sqrt{2}} \left ( \left | \phi _{1} \right \rangle - \left | \phi _{2} \right \rangle \right )[/itex]
    [itex]\left | \Phi _{2} \right \rangle = \frac{1}{\sqrt{2}} \left ( \left | \phi _{1} \right \rangle + \left | \phi _{2} \right \rangle \right )[/itex]
    (Q3. Are these pure state or mixed state?) Finally, the wavefunction for the "new" entire system that takes account of excitonic interactions is given by:
    [tex]\left | \Psi \right \rangle = b_{0} \left | \Phi _{0} \right \rangle + b_{1} \left | \Phi _{1} \right \rangle + b_{2} \left | \Phi _{2} \right \rangle[/tex]
    (Q4. Is this pure state or mixed state?)

    Thank you.
     
  2. jcsd
  3. Jul 15, 2016 #2

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    All the states that you have mentioned are pure states. I think that it confuses the issue to consider composite states (although they are relevant for "improper" mixed states, described below). The difference between pure states and mixed states is best understood by considering just the wave function of a particular particle. The simplest case is a spin-1/2 particle where we only consider spin degrees of freedom. In that case, any pure state [itex]|\psi\rangle[/itex] is a superposition of the spin-up state [itex]|U\rangle[/itex] and the spin-down state [itex]|D\rangle[/itex]: ("spin up" and "spin down" is relative to a direction, which I'll take to be the z-axis in some coordinate system).

    [itex]|\psi\rangle = \alpha |U\rangle + \beta |D\rangle[/itex] where [itex]|\alpha|^2 + |\beta|^2 = 1[/itex]

    The hallmark of a superposition is that it leads to "interference". For example, suppose you start off with a particle in state [itex]|\psi\rangle[/itex] as above, and allow it to evolve under the influence of some Hamiltonian [itex]H[/itex]. Then at some time [itex]t[/itex] later, you test whether the particle has spin-up.

    We can compute the following two complex numbers:
    [itex]G(U,U,t) = \langle U | e^{-iHt} | U \rangle[/itex]
    [itex]G(U,D,t) = \langle U | e^{-iHt} | D \rangle[/itex]

    The significance of these numbers is this:

    [itex]P_{UU} = |G(U,U,t)|^2[/itex] is the probability of finding the particle in state [itex]|U\rangle[/itex], if it were originally in state [itex]|U\rangle[/itex].
    [itex]P_{UD} = |G(U,D,t)|^2[/itex] is the probability of finding the particle in state [itex]|U\rangle[/itex], if it were originally in state [itex]|D\rangle[/itex].

    But in our case, the original state is neither [itex]|U\rangle[/itex] nor [itex]|D\rangle[/itex] but a superposition: [itex]|\psi\rangle = \alpha |U\rangle + \beta |D\rangle[/itex]. Then what is the probability of finding the particle in state [itex]|U\rangle[/itex] at time [itex]t[/itex] if it were originally in state [itex]|\psi\rangle[/itex]?

    Well let [itex]G(U,\psi,t)[/itex] be the amplitude for this process. It is computed as follows:

    [itex]G(U, \psi, t) = \alpha G(U,U,t) + \beta G(U,D,t)[/itex]

    We square the amplitude to get the probability:

    [itex]P_{U,\psi} = |G(U,\psi, t)|^2 = |\alpha|^2 |G(U,U,t)|^2 + |\beta|^2 |G(U,D,t)| + (\alpha^* G(U,U,t)^* \beta G(U,D,t) + \alpha G(U,U,t) \beta^* G(U,D,t)^*)[/itex]

    If we let [itex]P_U = |\alpha|^2[/itex], and [itex]P_D = |\beta|^2[/itex], then

    [itex]P_{U,\psi} = P_U P(U,U,t) + P_D P(U,D,t) + (\alpha^* G(U,U,t)^* \beta G(U,D,t) + \alpha G(U,U,t) \beta^* G(U,D,t)^*)[/itex]

    Note the messy-looking "interference term" involving the [itex]G[/itex]s.

    Now, in contrast, if the original state were not in a superposition of [itex]|U\rangle[/itex] and [itex]|D\rangle[/itex], but a mixture, then the interference term disappears. Mathematically, a mixed state is represented not by a ket, but by a density matrix:

    [itex]\rho = P_U |U\rangle \langle U| + P_D |D\rangle \langle D |[/itex]

    This is the mixed state representing a probability [itex]P_U[/itex] of being in state [itex]|U\rangle[/itex] and a probability of [itex]P_D[/itex] of being in state [itex]|D\rangle[/itex]. You can think of (although this isn't completely correct) a mixed state as representing the situation of: The particle is either in state [itex]|U\rangle[/itex] or state [itex]|D\rangle[/itex], but I don't know which. In contrast, if the particle is in a superposition, you know precisely what the state is, and it's neither [itex]|U\rangle[/itex] nor [itex]|D\rangle[/itex].

    Okay, I promised to describe "improper mixed states", but this post is already pretty long, so I will make it a different post.
     
  4. Jul 15, 2016 #3

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    First of all one must stress that none of your kets represent a possible state of the two molecules, because they are not symmetrized (antisymmetrized) depending on whether you have bosons (fermions).

    If you write down a state vector, it's implied that it represents a pure state. You system is in such a pure state, if you have prepared this state, e.g., by an ideal filter-measurement of a complete set of compatible observables. This is a complete state determination, i.e., you have the full possible information about the state of the system. There's one more subtlety to keep in mind. The absolute phase of your vectors is irrelevant for the physical information it represents, i.e., you can multiply it by an arbitrary phase factor, and it still represents the same state. If ##|\psi \rangle## represents you state thus ##\exp(\mathrm{i} \varphi) |\psi \rangle## for all ##\varphi \in \mathbb{R}## represents the same state. A better representant of a pure state is thus the corresponding statistsical operator, which is the projection operator ##\hat{\rho}_{\psi} =|\psi \rangle \langle \psi|##, where I assume that the vector is normalized, ##\langle \psi|\psi \rangle=1##.

    A mixed state is always represented by a statistical operator ##\hat{\rho}##, which is a positive semidefinite self-adjoint operator with trace 1 and which is not a projection operator as the statistical operator of a pure state, described above. You use such an operator, if you have only incomplete knowledge. An important example is a many-body system like a gas. There you know only a very few observables like the energy density of the entire system. In the important case that your gas is in thermal equilibrium you can describe it by the canonical ensemble. Then the statistical operator is
    $$\hat{\rho}=\frac{1}{Z} \exp(-\hat{H}/(k_{\text{B}} T)), \quad Z=\mathrm{Tr} \exp(-\hat{H}/(k_{\text{B}} T)), $$
    where ##T## is the temperature of the gas.
     
  5. Jul 15, 2016 #4

    kith

    User Avatar
    Science Advisor

    Mixed states basically arise in two cases:
    1) If already a compareable classical situation involves probabilities. In vanhees71's example of a gas at temperature [itex]T[/itex], you would need the Boltzmann distribution for the classical case. Another such situation would be to mix two ensembles with known pure states (this is where the name "mixed" state comes from).
    2) If two quantum systems are in an entangled state and you want to talk about the state of only one of the systems. These single system states are also mixed. They are sometimes called "improper" mixed states because there are correlations which wouldn't be present if we had reached the same state by mixing ensembles.
     
    Last edited: Jul 15, 2016
  6. Jul 15, 2016 #5

    HAYAO

    User Avatar
    Gold Member

    Thank you all. Unfortunately, your answers raised more questions than answers...

    1) vanhees71 mentioned that antisymmetry or symmetry is not taken into account. I am not sure how that can be done in the specific example that I have written: [itex]\left | \Phi \right \rangle = a_{0}\left | \phi _{0} \right \rangle + a_{1}\left | \phi _{1} \right \rangle + a_{2} \left | \phi _{2} \right \rangle [/itex]. Because they are fermions, the wavefunction needs to be antisymmetrized. But Even with "effective core potential" approximations and considering only the electrons involved in determining the relative energy of the molecules, along with bunch of other approximations, the hamiltonian is still a two-particle operator for each molecule. Now that the hamiltonian is made even more complicated due to tensor multiplication for effectively representing a system of two molecules, it is now a four-particle operator, I believe. Because the wavefunction consists of tens of orbital wavefunctions and spin functions of electrons, proton, and neutrons, symmetrization is a hellish process. I often see in textbooks and papers with no such antisymmetrization considered. So are the textbooks and papers wrong? Or is it sufficient without (anti)symmetrization?

    2) Thanks to you all, especially stevendaryl, now I know that all of the examples I have given is a pure state. My next question would be then, when people say "mixing of several states" like seen in Configuration Interaction and Spin-Orbit Coupling, are they really a mixed state? Specifically, people say intersystem crossing from singlet state to triplet state is an allowed process with enough Spin-Orbit coupling that causes singlet and triplet states "mixes". Or are they just another "new" pure state produced by such interactions?

    3) kith and vanhees71 mentioned that the classical situation that is in a mixed state. Would the zero-field splitting of excited triplet state in a molecule also be considered a "mixed state"? Although zero-field splitting is extremely small in molecules that they can be considered degenerate, at lower temperature, it arises to different emission of different lifetimes. That must mean that the distribution of the molecules that are in either of these three states can be written by Boltzmann distribution, right?

    4) Is the quantum state of electromagnetic field in a laser also a "mixed state"?
     
  7. Jul 15, 2016 #6

    kith

    User Avatar
    Science Advisor

    No, this kind of "mixing" is something entirely different. It is the result of adding terms to the Hamiltonian as a perturbation. You start with the unperturbed eigenstates and you get superpositions of them as eigenstates of the new Hamiltonian. (The situation is analogous to the one in Q3 in your OP)

    In this kind of mixing, you are improving your description of a single system by including interaction terms which you have neglected in your previous description. In the kind of mixing I was referring to, you have ensembles of systems in different states and mix these. This is analogous to putting balls of different colors in an urn. In both cases, you lose information.

    How would light coming from a laser and light coming from a light bulb be described classically?
     
    Last edited: Jul 15, 2016
  8. Jul 15, 2016 #7

    kith

    User Avatar
    Science Advisor

    Regarding your questions about the (anti-)symmetrization of states, it would be better if you opened a new thread.
     
  9. Jul 19, 2016 #8

    HAYAO

    User Avatar
    Gold Member

    Okay, thanks a ton. Now things are much more clear to me. I often found it strange that Configuration Interaction and Spin-Orbit coupling uses the term "mixed" and I thought it might had to do with something about "mixed state". But under the definition, I couldn't see how they are a "mixed state". Now I understand that they are NOT a "mixed state". I wonder why they used such a misleading term.
     
  10. Jul 19, 2016 #9

    kith

    User Avatar
    Science Advisor

    Although I didn't get confused over terminology in this case, I know this feeling well. So I am happy that I could clear this for you.

    Since you didn't comment on the laser example, let me give you the answer: the quantum state of a monochromatic, idealized laser is a so-called coherent state whis is pure and which corresponds to an electromagnetic wave of a certain wavelength. Superpositions of electromagnetic waves (like in the double slit) also correspond to pure states. A lightbulb, however, emits a thermal mixture of different wavelengths which cannot be written as a single electromagnetic wave. So there we have a mixed state in QM.
     
  11. Jul 19, 2016 #10

    HAYAO

    User Avatar
    Gold Member

    I see. Thank you very much kith!

    So black body radiation is also a mixed state?
     
  12. Jul 19, 2016 #11

    kith

    User Avatar
    Science Advisor

    Yes.
     
  13. Jul 22, 2016 #12

    Zafa Pi

    User Avatar
    Gold Member

    A pure state |a> is a unit vector in Hilbert space. The word pure is often omitted.If |a> and |b> are pure states then the superposition (|a> + |b>)/sqrt2 is a pure state.
    A mixed state is a state valued random variable. For example: We get |a> with probability p and get |b> with probability 1-p.
     
  14. Jul 22, 2016 #13

    bhobba

    Staff: Mentor

    Just to elaborate a bit.

    Although its sometimes omitted pure states are not strictly speaking unique elements of a vector space. That's because they have an interesting symmetry in that if c is any complex number c|a> is the same state as |a>. The reason for that is states are not really elements of a vector space but rather operators of unit trace. Pure states are those operators of the form |a><a| which is why they can be mapped to a vector space - but not uniquely. Mixed states are operators of the form ∑ pi|ai><ai|. It can be shown all states are mixed or pure and from the Born rule pi is the probability the mixed state is in pure state |ai><ai| if you observe it.

    Thanks
    Bill
     
    Last edited: Jul 23, 2016
  15. Jul 23, 2016 #14

    Zafa Pi

    User Avatar
    Gold Member

    Right you are. I could have said ray rather than vector. Yet I could have formulated it in the framework of a density operator D and stated that trace(D²) ≤ 1 with equality if and only if D is a pure state. But all that loses the forest for the trees. I think the OP's question was at a basic level and I tried to capture the answer in the simplest way possible. Note that I didn't even use ∑, but merely had the sum involving two states.
    However, please don't stop pointing out technical errors, they're great.
     
  16. Jul 24, 2016 #15

    kith

    User Avatar
    Science Advisor

    I don't think that the notion of a single random variable captures the concept of a mixed state completely. What about measuring in a different basis?
     
  17. Jul 24, 2016 #16

    Zafa Pi

    User Avatar
    Gold Member

    A pure state is a single quantum state |ψ⟩ or in or in the density operator formulation D = |ψ⟩⟨ψ|. A mixed state is a mixture or ensemble of pure states ={pi, |ψi⟩}, where pi is the probability of being in state |ψi⟩. In this case D = Σpii⟩⟨ψi|. Measurements are not involved with these definitions.
     
  18. Jul 24, 2016 #17

    kith

    User Avatar
    Science Advisor

    I take issues with your statement "A mixed state is a state valued random variable." in post #12. If I use a different basis than you, I get different states and different probabilities. This isn't captured by your random variable.

    Also your statement "[...] where pi is the probability of being in state |ψi⟩." (bolding mine) in post #16 seems questionable to me for the same reason. Your |ψi⟩ doesn't occur in my description and my probabilities sum to one.
     
  19. Jul 24, 2016 #18

    Zafa Pi

    User Avatar
    Gold Member

    You seem to understand what I've been saying, but disagree with it. I, unfortunately, don't understand what your saying. A state can have different representations in different bases, but it is still the same state. On matters such as these I refer to Nielsen & Chuang. I find it quite clear. What are your definitions of pure and mixed states? Do you consider your post #4 a definition?
    BTW my probabilities also sum to 1.
     
  20. Jul 24, 2016 #19

    kith

    User Avatar
    Science Advisor

    I mostly took issues with your statement about the mixed state being a random variable.

    The other point is the semantic subtlety of whether we should say that the system is in a certain pure state with a certain probability prior to a measurement. Ultimately, this is a question of interpretation.

    I just made a quick remark and don't intend to lead a long discussion about these minor issues. If we don't understand each other, it's not a big problem.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Still confused about superposition and mixed states
Loading...