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

I Spin eigenvectors

  1. Sep 12, 2016 #1

    dyn

    User Avatar

    When considering the 2 eigenvectors of the Sz operator the | + > eigenvector points in the positive z direction and the | - > points in the negative z direction ; so is it correct to write | + > = - | - > ?
    And similarly for the eigenvectors of the Sx and Sy operators ?
    Thanks
     
  2. jcsd
  3. Sep 12, 2016 #2

    DrClaude

    User Avatar

    Staff: Mentor

    No, that's not correct. Otherwise, you would have <+|-> = -1, so they would not be orthogonal. |+> can, in no way, be expressed in terms of |->. There is a limit to the classical representation of |+> as a vector pointing along +z.
     
  4. Sep 12, 2016 #3
    You should not think of spin states as actual vectors in our three dimensional space. I suggest you do some readig on the subject, if you can derive the eigenstates using the pauli matrices you can gain rather valuable intuition. Here is a wiki link in case you just want to see what the states are:
    https://en.m.wikipedia.org/wiki/Eigenspinor
     
  5. Sep 12, 2016 #4

    dyn

    User Avatar

    Thanks for your replies. Is the following equation correct ?
    | - > = e- iπSy | + >
    This seems to express the down state in terms of rotating the up state by π about the y-axis
     
  6. Sep 12, 2016 #5

    DrClaude

    User Avatar

    Staff: Mentor

    That's correct. But again, you can't treat the state |+> as simply a vector pointing along +z. Actually, the state -|+> is the same physical state as |+>, since the difference is only a global complex phase.
     
  7. Sep 13, 2016 #6

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    To make it very clear the bras and kets in the Hilbert-space are not vectors in configuration space.
     
  8. Sep 13, 2016 #7

    dyn

    User Avatar

    So if I measure spin in the z-direction and the system is in the | + > state I get the value ħ/2 ( the eigenvalue) but the associated eigenvector | + > does not have a direction in 3-d space ?
     
  9. Sep 14, 2016 #8

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    No, it's a vector in a two-dimensional complex unitary vector space, the Hilbert space for spin 1/2!
     
  10. Sep 17, 2016 #9
    The connection between complex vector space and 3d
    https://en.wikipedia.org/wiki/Pauli_matrices
     
  11. Sep 17, 2016 #10
    That is not necessarily a correct statement. Pauli matrices exist in their two dimensional Hilbert space, and they do not form a direct link with our own three dimensional real space. The connections they form are between one spin state and another, which all exist in the two dimensional complex vector space.
     
  12. Sep 17, 2016 #11
    lol
    Sigma X means x direction in 3d space...
     
  13. Sep 17, 2016 #12
    So...? That doesn't mean Pauli Matrices are the connection between the two vector spaces. They relate spin states, which are defined in the two dimensional Hilbert space, with our three dimensional space through acting on the spin states and producing the eigenvalues. The eigenvalues relate to our real space.

    They are an indirect link between the two spaces, not a direct transformation operator. You don't act on a space with the Pauli matrices to get another space.

    My point is, yes of course the Pauli Matrices relate - through some interpretations and definitions of the eigenvalues - spin states with the basis vectors in our real space, however that doesn't necessarily mean that they are the mathematical objects that transform one space to another.
     
  14. Sep 17, 2016 #13

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    A note about "vector spaces". A vector space is a set whose members have certain standard properties. These properties, in a sense, generalise the properties of 3D space. But, all sorts of mathematical objects have these properties: vectors of many descriptions (finite dimensional, infinite dimensional), functions (continuous functions, integrable functions) and QM spin states etc.

    Now, one approach would be to use the term "vector" only when it means a vector in normal 3D space. And, to some extent this is done. We sometimes talk about a "function" space and about "eigenfunctions" or about "spinors"; rather than just calling them all "vectors".

    But, in general, "vector" is used for both the normal 3D vector and any generalised vector in whatever vector space is under consideration. For example:

    eigenfunction in a function space ##\equiv## eigenvector in a vector space of functions

    One author might use the term "eigenfunction" and another use the term "eigenvector".

    So, you need to be careful not to interpret "vector" every time you see it as a 3D vector. Always look at and understand the context in which something is a vector.
     
  15. Sep 28, 2016 #14

    dyn

    User Avatar

    Hi. So I understand that | + > and | - > are vectors in a 2-D Hilbert space and not in 3-D space. I have the following example of the ket | j=1/2 , mj = 1/2 > represented by ( 1 0 )T so this is a 2-D Hilbert space again. But a rotation operator can be applied to the state to rotate the state around each of the 3 axes : x , y , z by angles α , β , γ . So we have a 2-D state yet the ket can be rotated around any of 3 orthogonal axes ? I'm confused about these 3 angles of rotation around the 3 orthogonal axes. Do these 3 angles and orthogonal axes exist in the 2-D space ? If the ket is not a vector in the standard 3-D world how can it be rotated around each of the 3 standard axes ? That would imply the ket has a direction in 3-D space !
    Thanks
     
  16. Sep 29, 2016 #15

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    This is what's called a group or Lie-algebra representation. The rotations are in this case represented by a two-dimensional complex Hilbert space with the rotations represented by unitary ##\mathbb{C}^{2 \times 2}## matrix. Now in quantum theory the vector representing a pure state is not unique, but you can always multiply it with a phase factor. Thus the rotation group is here represented by its socalled covering group, which is a twofold covering of the rotation group. This means that a rotation with an angle of ##2 \pi## maps the spin-state vector not to itself as expected but multiplies it by -1. That's however, a minor detail.

    The rotation group is a socalled Lie group, i.e., any group element can be built from infinitesimal transformations, and the corresponding generators are angular-momentum operators ##\hat{J}_k## that obey the commutator relations
    $$[\hat{J}_k,\hat{J}_l]=\mathrm{i} \epsilon_{klm} \hat{J}_m.$$
    There's one socalled Casimir operator for this Lie algebra, namely the modulus squared of the angular momentum operator ##\hat{\vec{J}}^2=\hat{J}_1^2+\hat{J}_2^2+\hat{J}_3^2##. You can easily check that the three components ##\hat{J}_k## all commute with ##\hat{\vec{J}}^2##, and such an operator is called a Casimir operator of the Lie algebra.

    Now it can be shown that all representations of the Lie algebra are uniquely determined by the eigenvalues of ##\hat{\vec{J}}^2##, and the possible eigenvalues are ##J \in \{0,1/2,1,3/2,\ldots \}##. Then the standard basis of each of these representations is chosen as the eigenbasis of ##\hat{J}_z##, and the eigenvalues are ##M_3 \in \{-J,-J+1,\ldots,J-1,J \}##. For ##J=1/2## you get a 2D vector space spanned by the two orthonormal vectors ##|M_3 =\pm 1/2 \rangle##. Each vector in the representation is thus given as a linear combination
    $$|\psi \rangle=\psi_{1/2} |M_3=1/2 \rangle + \psi_{-1/2} |M_3=-1/2 \rangle,$$
    and the usual convention is to map the vectors to the realization of the 2D vector space to ##\mathbb{C}^2## and to write the components as a column vector (which in this case is called a spinor, because this represents the spin of a spin-1/2 particle):
    $$\psi=\begin{pmatrix} \psi_{1/2} \\ \psi_{-1/2} \end{pmatrix}.$$
    Now you can also evaluate, using the commutator algebra, how ##\hat{J}_1## and ##\hat{J}_2## act on these basis vectors (you find the derivation in any good textbook on quantum mechanics, e.g., J.J. Sakurai, Modern Quantum Mechanics, Addison Wesley). For convenience one write for the corresponding matrices wrt. to the above basis
    $$\hat{J}_k=\frac{1}{2} \hat{\sigma}_k,$$
    where the ##\hat{\sigma}_k## are the socalled Pauli matrices. They turn out to be
    $$\sigma_1=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma_2=\begin{pmatrix} 0 & -\mathrm{i} \\ \mathrm{i} & 0 \end{pmatrix}, \sigma_3=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$
    Of course ##\hat{\sigma}_3## is diagonal, because we've chosen the eigenvectors of the 3-component of the spin to determine the basis we want to work with.

    Now you can calculate how a rotation acts on the spinors by composing it by infinitesimal transformations, which leads to a matrix exponential. Take the rotation axis (in 3D space!) to be given by the unit vector ##\vec{n}## and the rotation angle ##\varphi##. Then the rotation matrix is given by
    $$\hat{R}(\vec{n},\varphi)=\exp \left(-\frac{\mathrm{i}}{2} \varphi \vec{n} \cdot \vec{\sigma} \right ).$$
    Using the power series of the exponential function you can show that
    $$\hat{R}(\vec{n},\varphi)=\cos (\varphi/2) \hat{1} -\mathrm{i} \vec{n} \cdot \hat{\vec{\sigma}}.$$
    As you see for ##\varphi=2\pi## you get ##\hat{R}=-\hat{1}##. To get back to the unit matrix you must set ##\varphi=4 \pi##. That's the above mentioned double cover of the rotation group. You can easily check that the rotation matrices are unitary and that their determinant is 1. Thus we represent the usual rotation grou SO(3) by its covering group SU(2).

    I hope this explains a little how the abstract angular-momentum (in this case spin-1/2) eigenvectors in Hilbert space relate to the usual rotations in 3D Euclidean space.
     
  17. Sep 29, 2016 #16

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You can't apply a 3D rotation to a 2D vector. There is an analogy with classical spin. If you know the total angular momentum, you need only specify the axis of rotation. But, this axis can be specified by two parameters, normally the spherical angles ##\theta## and ##\phi##. Every spin state in classical mechanics can therefore be specified by the total spin AM and a "vector" ##(\theta, \psi)##.

    In this case we have a 2D vector, which represents a direction in 3D space. It's very important when dealing with this vector to distinguish between the 2D "vector" itself and the 3D vector it represents.

    In a similar way a vector in your complex Hilbert space represents the spin state of a 3D object with 3 axes of rotation. With some added complexities over the classical case. What the spin state represents here is very different from an axis of rotation. It's doubly
    important, therefore, not to confuse this spin state with a classical axis of rotation in 3D. What the spin state represents is something much more abstract.
     
    Last edited: Sep 29, 2016
  18. Sep 29, 2016 #17

    dyn

    User Avatar

    Hi. Thanks for your replies. I think I can follow the maths but let me try to be more precise with my question. I have the ket | j = 1/2 , mj = 1/2 > which can be represented by ( 1 0 )T. This is a ket in a 2-D complex Hilbert space
    From this reply I take it that the ket has no direction in 3-D Euclidean space. Is this correct ?
    I have read for example that this state can be rotated by angle α about the x-axis and the rotated state is given by e- i α σx/2 ! 1 0 )T where σx is the relevant Pauli matrix . Does this represent a rotation in 3-D Euclidean space by angle α ?
    If so , how can a ket/vector/state with no direction in 3-D space be rotated by an angle in 3-D space ?
    Thanks
     
  19. Sep 30, 2016 #18

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Read my #15 carefully again!
     
  20. Sep 30, 2016 #19

    dyn

    User Avatar

    I have re-read #15 and #16 again. This is my understanding of it. Hopefully someone can tell me if I am correct or not.
    If I have the state | j = 1/2 , mj = 1/2 > , ie. ( 1 0 )T. This is a vector in 2-D Hilbert space. It has no direction in 3-D space but represents a vector in 3-D. But using the rotation operator expressed in terms of Pauli matrices the 3-D representation can be rotated around any of the 3 orthogonal axes x , y , z. So the 3-D representation can be rotated by any angle but it did not have a starting direction and so has no finishing direction. All we know is that it has been rotated by a certain angle.
    Am I correct ? Thanks.
     
  21. Oct 1, 2016 #20

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    I don't understand what you want to say.

    In quantum theory you have states and observables. A bit simplified you can look at the special case of pure states as represented by vectors in a Hilbert space. The observables are represented by self-adjoint operators. That's the mathematical framework. Now comes the interpretation (I use the minimal interpretation and no philosophical additions, and I hope this thread won't again be hijacked by some philosophical gibberish later):

    The possible outcome of measuring an observable ##A## which is represented by the self-adjoint operator ##\hat{A}## is an eigenvalue of this self-adjoint operator. If ##|a,\beta \rangle## are a complete set of orthonormal eigenvectors of ##\hat{A}## to the eigenvalue ##a## and the state of the system is represented by the normalized vector ##|\psi \rangle##, then the probability to find the value ##a## when measuring ##A## is
    $$P(a|\psi)=\sum_{\alpha} |\langle a,\alpha|\psi \rangle|^2.$$
    That's the meaning of the vectors in Hilbert space in physical terms.

    Now symmetry transformations like rotations in space are represented by unitary operators which build a representation of the corresponding symmetry group, which means that you have for two rotations ##R_1,R_2 \in \mathrm{SO}(3)##
    $$\hat{U}(R_2) \hat{U}(R_1)=\hat{U}(R_2 R_1),$$
    where ##\hat{U}(R)## is the unitary representation of the rotations in the Hilbert space of quantum theory.

    Now you can ask for the socalled irreducible representations, which are those representations of the group where there are no proper subspaces of the Hilbert space which is left invariant by all group operations. One example of such an irreducible representation is the spin-1/2 representation we are discussing here, and the unitary representation in this case in terms of the eigenbasis of the spin-z component is given with help of the Pauli matrices as indicated above.
     
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: Spin eigenvectors
  1. High Spin (Replies: 1)

Loading...