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Quantum states as normal vectors

  1. Apr 8, 2015 #1
    Are all quantum states represented by normal vectors?
     
  2. jcsd
  3. Apr 8, 2015 #2

    Nugatory

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    Every quantum state is represented by a vector in an abstract vector space.

    What exactly do you mean by "normal"?
     
  4. Apr 8, 2015 #3

    bhobba

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    Nugatory is correct about the usual states talked about in QM - they are called pure.

    But in more advanced work states are in fact positive operators of unit trace. Pure states are those of the form |u><u|. Mixed states are convex sums of pure states ie of the form ∑pi |bi><bi| where pi are positive numbers that sum to 1. It can be shown all states are mixed or pure.

    Thanks
    Bill
     
    Last edited: Apr 8, 2015
  5. Apr 8, 2015 #4
    Major slip of the brain there, I should have read my post properly. What I meant to ask is: Are all quantum states represented by unit vectors in some complex vector space?
     
  6. Apr 8, 2015 #5

    Nugatory

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    They are represented by vectors, not necessarily unit vectors. If I were to be picky about it, I would say that they are represented by rays, where a ray is a set of all vectors that are a complex constant multiple of one another (same "direction", different magnitudes).

    Of course if I'm going to calculate the probability of an outcome of an observation I'll want to normalize the vector to be a unit vector... so the answer to the question that I think you meant to ask is "yes". :smile:
     
  7. Apr 8, 2015 #6

    vanhees71

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    No, pure states are never represented by unit vectors but by unit rays (i.e. by unit vectors modulo an arbitrary phase factor) or, equivalently, by projection operators
    $$R_{\psi}=|\psi \rangle \langle \psi|$$
    where ##|\psi \rangle## is a normalized Hilbert-space vector.

    A general state is represented a positive semidefinite trace-class operator, the statistical operator (see also posting #3).
     
  8. Apr 9, 2015 #7
    Is this what is referred to as a global phase factor?

    So is the direction of a vector component the defining feature for a pure state? This makes sense because two rays with the same direction but different magnitude would give the same expectation value with respect to the same observable since they are identical unit vectors once normalized. Right?
     
    Last edited: Apr 9, 2015
  9. Apr 9, 2015 #8

    vanhees71

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    Roughly speaking, it's the global phase factor. It is an important point to make clear that pure states are represented as rays not as vectors in Hilbert space. Equivalently you can say that a statistical operator represents a general state, which is more simple to deal with in practice. From this point of view the pure states are defined as the special case of the statistical operator being a projection operator with trace 1.

    This issue is important, because it explains, e.g., why there are particles with half-integer spin. E.g., a spin-1/2 spinor changes sign under 360-degree rotations. Now, since not the spinor represents a spin state but the corresponding ray, the rotated state is the same as the original one as it must be for 360-degree rotations. If you'd falsely assume that the spinor itself would represent the state, you'd come to the conclusion that all half-integer representations of the rotation group are not allowed, which for sure is wrong, because all the matter around us consists of spin-1/2 particles (quarks and leptons).
     
  10. Apr 9, 2015 #9
    I must admit I've had a difficult time getting to grips with these global phase factors. Im currently writing a project on the basics of quantum information and quantum computing and they crop up everywhere. I understand that global phase factors are essentially irrelevant since they do not affect the expectation values for any observables. My proffesor has brushed over the toic when I hav mentioned it. So what exactly are they there for or why do they appear?
    I thought that pure states corresponded to points on the bloch sphere which are defined by unit vectors aren't they?
     
  11. Apr 9, 2015 #10

    bhobba

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    They appear because in reality states are positive operators of unit trace - not elements of a vector space. By definition pure operators are of the form |u><u|. The |u> are elements of a vector space and can, and usually are, thought of that way. But note |cu><cu| = |u><u| where c is any complex number of unit length ie a phase factor. Thus phase factors are unimportant.

    Why are states positive operators of unit trace? Usually its simply stated its an axiom and axioms don't have reasons beyond that. But in this case there is a deeper reason associated with a theorem called Gleason's theorem:
    http://en.wikipedia.org/wiki/Gleason's_theorem

    Thanks
    Bill
     
  12. Apr 9, 2015 #11

    vanhees71

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    Yes, that's the special case of a two-dimensional Hilbert space, realized, e.g., by the spin of a spin-1/2 particle or the polarization of a photon.

    Here, it is easy to characterize all possible Statistical operators. These are represented by postive semidefinite Hermitean (in finite-dimensional spaces equivalent to self-adjoint) ##\mathbb{C}^{2 \times 2}## matrices. All these matrices can be written as real linear combinations of the identity matrix and the three Pauli matrices:
    $$\hat{\rho}=\frac{1}{2} (\hat{1}+\vec{r} \cdot \hat{\vec{\sigma}}.$$
    Since the Pauli matrices are trace-less you have ##\mathrm{tr} \hat{\rho}=1## for any ##\vec{r} \in \mathbb{R}^3##.

    Further the eigenvalues must be positive semidefinite. From the explicit form of the Statistical Operator,
    $$\hat{\rho}=\frac{1}{2} \begin{pmatrix}
    1+r_3 & r_1-\mathrm{i} r_2 \\
    r_1+\mathrm{i} r_3 & 1-r_3
    \end{pmatrix},$$
    you find the eigenvalues
    $$\lambda_1=\frac{1}{2}(1+|\vec{r}|), \quad \lambda_2=\frac{1}{2}(1-|\vec{r}|).$$
    This implies that ##|\vec{r}| \leq 1##.

    This provides a one-to-one mapping of all possible states of a two-level system to the compact unit ball in ##\mathbb{R}^3##.

    For a pure state, the Statistical Operator must be a projection operator to a one-dimensional sub-space of ##\mathbb{C}^2##, i.e., you must have one eigenvalue to be 0, which implies that ##\hat{\rho}## represents a pure state for all ##\vec{r}## with ##|\vec{r}|=1##, and thus all possible pure states are parametrized by a unit vector in ##\mathbb{R}^3##.

    It's also easy to check that the projector property
    ##\hat{\rho}^2=\hat{\rho}##
    is fulfilled exactly for all ##\vec{r} \in S_2##.

    Note that the Bloch sphere characterizes pure states with their Statistical Operators not as vectors in ##\mathbb{C}^2##. Of course, again this characterization of a pure state as a trace-1 projection operator is equivalent to its characterization as the ray, determined by the eigenvector with eigenvalue 1 of this projection operator.
     
  13. Apr 9, 2015 #12

    atyy

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    Because a pure state is a ray, when you write a unit vector representing the state, there is more than one way to write it, and all states that differ by a global phase factor are different ways of writing the same state. It is analogous to the electric potential in electrostatics - only the potential difference is physical - physical situations described by potentials that differ by a global constant represent the same physical situation. Since the global phase factor of the wave function and the global constant to the potential can be freely chosen according to taste, convention and convenience.
     
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