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Operator works on a quantum state yields another state?

  1. Dec 1, 2015 #1
    The well-known eigen value expression A(a)=a(a) assuming the operator which represents a physical phenomena acts on a quantum state which is represented by an eigen vector, (a) corresponds to an observed value a.
    But I am wondering if the same operator A can act on (a) and produce another eigen vector in another space (b) with an observed value b.
    This is stemmed from singular value decomposition where a matrix A works on orthogonal vector V yielding σU where σ is the eigen value and U is another orthonormal space. So I am wondering if an application of this method is present in quantum theory.
     
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  3. Dec 1, 2015 #2

    BvU

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    Hello Adel,

    My first reaction when reading your post is "Yes, but it's not the same operator" and "No, the produced eigenvector is in the same space". And I think of ladder operators .

    I get the idea that what you call another orthonormal space is in fact the same space with another basis
     
  4. Dec 1, 2015 #3

    Mentz114

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    I'm not entirely sure what you mean. A projection operator can change the state the first time it operates. so ##|v_k\rangle \langle v_k| \psi\rangle=| \psi'\rangle##. Twice has a different effect, ##|v_k\rangle \langle v_k| v_k\rangle \langle v_k| \psi\rangle=c| \psi'\rangle## because ##|v_k\rangle (\langle v_k| v_k\rangle) \langle v_k| = c|v_k\rangle \langle v_k| ## where ##c## is a real number.
     
  5. Dec 1, 2015 #4
    I think by ##|v_k\rangle \langle v_k|##, you mean a projection matrix, correct? If so, it should be a symmetrical real matrix or a complex conjugate one. In this case the operator acts on a vector ##|\psi\rangle## to produce another vector ##|\psi'\rangle## in the same vector basis. But what I mean is the action of the same operator on the vector ##|\psi\rangle## will produce another vector in different basis. Would be that possible physically? Bear in mind that the operator matrix in the second case may not be a symmetrical matrix which made me think about SVD as a possible mathematical solution.
     
    Last edited: Dec 1, 2015
  6. Dec 1, 2015 #5

    kith

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    Could you give a tangible example of the mathematical situation you have in mind?
     
  7. Dec 1, 2015 #6
    I think you might have asked about the "physical situation" where the math is applicable.
    Any way let,
    ##|\psi\rangle=\sum |v_k\rangle\langle v_k|\psi\rangle##
    ##M|\psi\rangle=\sum M|v_k\rangle\langle v_k|\psi\rangle## where ##M## is a non-symmetric matrix and ##M|v_k\rangle=\sigma_k|u_k\rangle## Also ##|u_k\rangle_s ## are new base vectors.
    so, ##M|\psi\rangle=\sum \sigma_k|u_k\rangle\langle v_k|\psi\rangle##.
     
  8. Dec 1, 2015 #7

    bhobba

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    To understand where eigenvalues come from in QM you need to see an axiomatic treatment and what the foundational axiom is. See post 137:
    https://www.physicsforums.com/threads/the-born-rule-in-many-worlds.763139/page-7

    It relies on the important but advanced Gleason's theorem which beginning texts, correctly IMHO, shy away from. But once you have grasped the fundamentals as hopefully you have then you can see what is actually going on.

    Even that is not the most elegant way of viewing QM - the following, while not as axiomatically neat, is much more intuitive and gets to the heart of QM IMHO:
    http://arxiv.org/pdf/quant-ph/0101012.pdf

    After a while you start to feel, perhaps it's just hubris, but you feel that QM is not quite as weird as at first thought.

    Thanks
    Bill
     
    Last edited: Dec 1, 2015
  9. Dec 2, 2015 #8

    kith

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    No, that is your question. I ask you to explain the exact mathematical situation which you have in mind by giving an example. The exact mathematical situation wasn't clear to me from your previous posts and it still isn't clear to me from your recent post. As long as the math isn't clear it doesn't make sense to talk about its relation to the physics.

    If I take the vector
    \begin{pmatrix}
    1\\
    1
    \end{pmatrix}
    and throw it into the matrix
    \begin{pmatrix}
    1 & 2\\
    0 & 0
    \end{pmatrix}
    I get
    \begin{pmatrix}
    1\\
    2
    \end{pmatrix}
    What do you mean by saying that the resulting vector is "in [a] different basis"?
     
  10. Dec 2, 2015 #9
    I think you meant:
    \begin{pmatrix}
    1 & 0\\
    2 & 0
    \end{pmatrix}
    Yes this is an example of how a matrix can act on a vector producing another vector. Calculating eigen values and eigen vectors of this matrix, you will find one non-zero eigen value which is ##\sigma_1=1## corresponds to an eigen vector,
    \begin{pmatrix}
    1\\
    2
    \end{pmatrix}
    The other eigen value is zero.
    So,
    $$Mv_1=\sigma_1 v_1$$
    which can be written as
    M=v1σ1 v1T
    with v1 forms one of base of the vector space. The other base, v2 does not appear in the above equation because its corresponding eigen value =0.
    This is an example of a diagonalizable matrix with two eigen values, ##\sigma_1=1 and \sigma_2=0##.

    Now consider a non-diagonalizable matrix,
    \begin{pmatrix}
    1 & 1\\
    0 & 1
    \end{pmatrix}
    There is no set of eigen basis for this matrix to act upon because the algebraic multiplicity is not equal to the geometrical multiplicity. But there is a way out; a set of eigen vector of vector base ##V## can be found so that the action of M on it yields eigen values in another set of eigen vectors ##U##.
    So, ##Mv_k=\sigma_k u_k##, here, k=1,2
    ##v_1## is;
    \begin{pmatrix}
    \frac{2}{1+\sqrt5}\\
    1
    \end{pmatrix}
    and ##v_2## is
    \begin{pmatrix}
    -1\\
    \frac{\sqrt5 -1}{2}
    \end{pmatrix}
    ##u_1## is;
    \begin{pmatrix}
    1\\
    \frac{\sqrt5 -1}{2}
    \end{pmatrix}
    and ##u_2## is;
    \begin{pmatrix}
    \frac{-2}{\sqrt5 +1}\\
    1
    \end{pmatrix}

    Also,
    $$\sigma_1=\sqrt\frac{3+\sqrt5}{2}$$
    and
    $$\sigma_2=\sqrt\frac{3-\sqrt5}{2}$$
    So,
    $$Mv_1=\sigma_1 u_1$$
    and
    $$Mv_2=\sigma_2 u_2$$
    So if there is a vector ##x## which is a superposition of eigen vectors in the base ##V##,
    $$x=\alpha_1 v_1+\alpha_2 v_2$$
    Acting on it by M yields;
    $$Mx=M\alpha_1 v_1+M\alpha_2 v_2$$
    $$Mx=\alpha_1 M v_1+\alpha_2 M v_2$$
    $$Mx=\alpha_1 \sigma_1 u_1+\alpha_2 \sigma_2 u_2$$
    $$Mx=\omega_1 u_1+\omega_2 u_2$$
    Which means transforming the vector x from vector basis ##V## into another eigen basis ##U##.
     
    Last edited: Dec 2, 2015
  11. Dec 2, 2015 #10

    kith

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    Yes, thanks for correcting this.

    Also thanks for your description, I think I get the idea. So you are looking for matrices which are not diagonalizable and want to know whether the vectors ##u_k## have a significance in QM?

    The two most important classes of operators in QM are observables and symmetry operators. Observables correspond to physical quantities and are self-adjoint (this is one of the postulates of QM). Symmetry operators are unitary. So this way you won't get non-diagonalizable matrices.

    As BvU mentioned, a third class of operators are ladder operators. These are used to model the creation or annihilation of a quantum of energy, angular momentum, etc. The corresponding matrices in 2D are
    \begin{pmatrix}
    0 & 1\\
    0 & 0
    \end{pmatrix}
    and
    \begin{pmatrix}
    0 & 0\\
    1 & 0
    \end{pmatrix}
     
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