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I What does the state vector mean?

  1. Mar 31, 2017 #1
    To me the state vector represents the following....

    1) The number of elements in the state vector is the number of possible outcomes. Call that number n.
    2) The value of each element in the state vector is the probability amplitude associated with that outcome.

    If that is true, then it seems to me, that a state vector which is a linear combination of the eigenstates, cannot be interpreted as creating a new possible outcome. It can only be interpreted as a way to calculate the amplitude of one of the n, apriori, possible outcomes.
     
    Last edited: Mar 31, 2017
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  3. Mar 31, 2017 #2

    DrClaude

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    This doesn't make sense to me.
     
  4. Mar 31, 2017 #3
    Thanks. I am trying to say that a state which is a linear combination of eigenstates cannot be interpreted as creating a new possible outcome, where the new possible outcome, for instance, would be a coin in the head and tail state at the same time. That state would represent a new possible outcome. But it is more logical to interpret it as a way to calculate the probability amplitudes of the two existing possible outcomes/states.

    There cannot be more than n possible outcomes.
     
    Last edited: Mar 31, 2017
  5. Apr 1, 2017 #4

    stevendaryl

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    The number of possible outcomes depends on what it is that you're measuring. If you measure the component of the spin of an electron along the z-axis, there are two possible outcomes. If you measure momentum, there are infinitely many possible outcomes.
     
  6. Apr 1, 2017 #5
    I thought you would have said there were 3 possible outcomes because of this superposition state ##S_z = \frac{1}{\sqrt{2}}\big(\uparrow+\downarrow\big)##
     
  7. Apr 1, 2017 #6

    Nugatory

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    No matter which spin measurement you make, there are two possible outcomes: up and down. That's another way of saying that the Hilbert space containing the spin states is two-dimensional. However, that Hilbert space is only a subspace of the infinite-dimensional Hilbert space that contains the state vectors of the particle in question.
     
  8. Apr 1, 2017 #7

    stevendaryl

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    No, the number of outcomes, as I said, is determined by the observable that you are measuring. Every observable corresponds to an operator, and the number of possible outcomes is equal to the number of eigenvalues for the operator.

    For the observable of "z-component of spin", there are two eigenvalues, +1/2 and -1/2, which means that those are the only two possible outcomes of a measurement.

    There are infinitely many different state vectors, though (although every state vector is a linear combination of two possibilities: [itex]|u\rangle[/itex] and [itex]|d\rangle[/itex])
     
  9. Apr 1, 2017 #8
    I guess you mean a state ##\left| \psi \right> =\frac { \left| \uparrow \right> +\left| \downarrow \right> }{ \sqrt { 2 } }## of a spin-1/2 particle. What you have written is the spin state using a certain representation (The z-direction spin basis), and there are two possible outcomes when measuring that spin in the z-direction, those being up or down, with a 50% probability of either. The same is true for measuring the spin in the y-direction, but since that state is an eigenstate of the x-direction of spin operator with eigenvalue ħ/2, then measuring the spin in the x-direction will always yield an 'up' result.
     
  10. Apr 1, 2017 #9

    PeterDonis

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    You're confusing the state vector with its representation in a particular basis. It seems like we already went over this in a previous thread.
     
  11. Apr 1, 2017 #10
    No, I am not confusing the state vector with its representation in a particular basis.
     
  12. Apr 1, 2017 #11
    I really like this response. Specially the last sentence.
     
  13. Apr 2, 2017 #12

    PeterDonis

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    Yes, you are, in the OP, when you say this:

    These two statements are statements about the representation of the state vector in a particular basis--the one corresponding to the measurement whose outcomes you are describing.
     
  14. Apr 2, 2017 #13

    As others have said, the number of possible outcomes is equal to the number of eigenvalues of the particular operator. That was actually what I was thinking when I wrote the original post. Unless you are saying that changing the basis changes the number of eigenvalues of a particular operator (which I do not think is the case when you apply a unitary operation to an operator.)
     
  15. Apr 2, 2017 #14
    State vector |A> = c1 |A1> + c2 |A2> + c3 |A3> + …. + Cn |An>

    ie n terms which I think is what you are saying.

    If all the base vectors (|Am> ) are normalised ,that is their magnitude is equal to 1, then the coefficients (cm ) are equal to the amplitudes, so that the square of | cm | are the probabilities.

    If the system is definitely in state 3 then the probability for this is | c3 | - squared . However, if the system is in a state which is a superposition of two or more of the base states then several coefficients will need to be added to find this probability.

    At measurement there is only one outcome, despite the various probabilities for different possible states prior to measurement.

    I hope that helps.
     
  16. Apr 2, 2017 #15
    First I am going to express your equation for ##|A\rangle## .$$C_n = \langle n|C\rangle$$ Where ##\langle n## is the orthonormal basis(##A##) and ##|C\rangle## is the state vector.

    I think I can then write

    \begin{equation}C=\begin{bmatrix}
    \\{C_1}\\
    \\{C_2}\\
    \\{\vdots}\\
    \\{C_n}
    \end{bmatrix}\end{equation}

    and would each ##C_n## represent the probability amplitude of measuring the nth possible outcome?
     
    Last edited: Apr 2, 2017
  17. Apr 2, 2017 #16

    vanhees71

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    Again: An observable in quantum theory is represented by a self-adjoint operator on the Hilbert space. The possible values of the observable are the eigenvalues (more precisely the spectral values) of this self-adjoint operator. So let ##A## be the observable you measure and ##\hat{A}## the corresponding self-adjoint operator and ##|a,\beta \rangle## an orthonormalized set of eigenvalues of ##\hat{A}## to the eigenvalue ##a##. For simplicity let's assume that all eigenvalues and the labels ##\beta## are all discrete (the generalization to continuous parts of the spectrum are straight forward; everywhere where I write sums you have to add integrals, and where I write Kronecker-##\delta##'s write Dirac-##\delta##-distributions):
    $$\langle a,\beta|a',\beta' \rangle=\delta_{a,a'} \delta_{\beta,\beta'}, \quad \sum_{a,\beta} |a,\beta \rangle \langle a,\beta|=\hat{1}.$$
    This means that the eigenvectors of ##\hat{A}## can be chosen as a complete orthonormalized set, i.e., an orthonormal basis of the Hilbert space (which except for academic examples is the separable infinite-dimensional Hilbert space).

    Pure states are represented by rays in Hilbert space, which are themselves again represented by normalized state vectors. If the system is in the pure state, represented in this sense by the state vector ##|\psi \rangle##, then the probality to measure the value ##a## of the observable ##A## is given by Born's rule,
    $$P(a)=\sum_{\beta} |\langle a,\beta|\psi \rangle|^2.$$
     
  18. Apr 2, 2017 #17
    That doesn't look right. If I was writing it that way I would have the C's as a row vector or even as a diagonal of a square matrix with all other coefficients zero and then the base | A > 's as the column vector. Dirac introduced the ket vector as a column vector and the operator as a matrix. A bra vector is represented as a row vector to the left, if required.
     
  19. Apr 2, 2017 #18

    vanhees71

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    No, Dirac introduced the ket vector as an abstract vector. There are no columns and rows. It's just an abstract vector! You get columns and rows, if you choose a descrete basis of the Hilbert space and calulate the corresponding components of the vectors.
     
  20. Apr 2, 2017 #19
    Ok. Please, just simply, what is the name of a vector that just contains the probability amplitudes? And how do I express that vector in Dirac notation? Is that vector a state vector?
     
    Last edited: Apr 2, 2017
  21. Apr 2, 2017 #20
    Maybe I am using the wrong word. Maybe the word vector is throwing me off. Let me call it the n-tuple that represents the probability amplitudes. What do we call the n-tuple which contains the probability amplitudes? How do we represent that n-tuple using Dirac notation? Can I think of it as a column vector?
     
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