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Vector Representations of Quantum States

  1. Jul 6, 2011 #1
    I happen to be studying the basics of quantum mechanics at the moment and have made acquaintance with the vector representation of quantum states, in particular the two states of electron spin.

    For this question let's just say the spin can be up or down. The state of the spin is represented by vector, and here's where my trouble in understanding shows up.

    I'll have to write the column vectors as row vectors, although the issues should still be clear.

    What is giving me trouble is that fact the the array/vector has two elements instead of just one. I would think a single digit, either "0" or "1" would be sufficient to describe a state as being either up or down.

    So my question is why are two element/components required to describe the two states of electron spin. There are only two states, so a single bit should be sufficient. Two bits allow for four possible states. That begs the question, what do the unused state vector definitions | 1 1 | and | 0 0 | then represent, and if they fail to represent anything meaningful why even allow for their expression, which is another way of asking my first question, why use two components in the vectors in the first place.
     
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  3. Jul 6, 2011 #2

    Fredrik

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    Superpositions. When the spin state is
    [tex]\begin{pmatrix}a\\ b\end{pmatrix}=a\begin{pmatrix}1\\ 0\end{pmatrix}+b\begin{pmatrix}0\\ 1\end{pmatrix},[/tex] the probability of measuring spin up is [tex]\frac{|a|^2}{|a|^2+|b|^2}.[/tex]

    You can certainly use two different symbols instead of these matrices to represent the "up" and "down" states, e.g. [itex]\chi_1[/itex] and [itex]\chi_2[/itex], but you would still have to make them basis vectors of a 2-dimensional vector space, to allow linear combinations like [itex]a\chi_1+b\chi_2[/itex].

    Note that [tex]\begin{pmatrix}a\\ b\end{pmatrix}[/tex] is just the matrix of components of [itex]a\chi_1+b\chi_2[/itex] in the ordered basis [itex](\chi_1,\chi_2)[/itex].

    By the way, I wouldn't say that the difference between wavefunctions for spin-0 particle and wavefunctions for spin-1/2 particles is that the latter are vectors. A "vector" is a member of a vector space, so wavefunctions for spin-0 particles are vectors too.
     
    Last edited: Jul 6, 2011
  4. Jul 6, 2011 #3
    I get the probability stuff and the matrix operations etc.

    It's just that first semantical step of representing/describing state up as | 1 0 | and state down as | 0 1 |.

    Those descriptions seem to include redundant information. For a coin toss, if heads is represented by | 1 |, that explicitly means heads is up and implicitly means tails is down.

    By representing heads with | 1 0 |, that is like literally saying, "The coin landed with heads up and tails down."

    I get the impression that the implicit information "and tails down" is made explicit so that the matrix operations can be done on the state descriptions, although it would be helpful if that was explicitly mentioned when the vector component format was first introduced.

    Maybe I'm just having a nit picking moment, although it's the kind of thing that gives me insomnia wondering why did they do it that way. :)
     
    Last edited: Jul 6, 2011
  5. Jul 6, 2011 #4

    dextercioby

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    It's much simpler than that: you need 2 numbers, not just one, to fully specify the spin state of a particle. For example, choose an electron. A possible spin state of it is given by the numbers 1/2 and 1/2. The first number fixes the particle species (electron, or system of spin 1/2), while the second one specifies the spin state by giving the projection of the spin angular momentum on an axis (Oz by convention) in multiples of hbar.

    And two numbers can be assembled into a matrix with 2 rows and 1 column => matriceal calculations are possible.
     
  6. Jul 6, 2011 #5
    Yes, in statements generalized that provide for arbitrary particles I can see that.

    However, I am talking specifically about discussions of electron spin states were one state is represented by | 1 0 | and the other by | 0 1 |,

    In this case I doubt a change of species is meant.
     
  7. Jul 6, 2011 #6

    SpectraCat

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    Think about the vector representation of a unit vector oriented along the x-axis. How many numbers do you need to uniquely represent that vector in 3D space? You need 3, one for each cartesian component: [itex]\hat{i}=\begin{pmatrix}1\\0\\0\end{pmatrix}[/itex]. Spins are represented in a 2-D space, so you need to specify the coefficients for both dimensions in order to completely specify the state.
     
  8. Jul 6, 2011 #7
    Maybe that's where I'm getting off track. I was thinking the spin representation as one dimensional, something like calling heads (+1) and tails (-1). Basically that is the point. I was presuming that the spin state can be represented by using only one dimension.

    So, the question is still there, why use two dimensions?

    For reference say we're talking about the eigen vector for the sigma 3 matrix.

    ( I'm also trying to figure out how to use the latex feature. Is there a tutorial on that?)

    With the sigma 3 we have eigen vectors of | 1 0 | and | 0 1 |. It seems to me each of these vectors conveys information that could be conveyed with just one element in each, either a 1 or a 0.

    So my question is still and this is worded awkwardly are these vectors literally saying in the positive axis direction the spin is pointing positive up and in the negative axis direction the spin is pointing negative down, and for the other in the positive axis direction the spin is pointing negative up and in the negative axis direction the spin is pointing positive down?

    Why is it that you say electron spin is represented in 2-D?

    It seems to me each of the sigma matrices with its own eigen vectors handles one dimension of a 3-D representation of the electron's spin.
     
  9. Jul 6, 2011 #8

    Fredrik

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    That's exactly what post #2 answers. Maybe you're not familiar with the fact that all vectors in the same 1-dimensional subspace represent the same state? There are no superpositions in a 1-dimensional vector space.

    The best way to learn is to just use the quote button to see how others are doing it. Check out my post #2 and note how itex and tex tags produce different results. This document will be useful too.

    One of the problems with that is of course that (0) is the zero vector. If you try to use that as a state vector, you'd end up assigning probability 0 to all measurement results.

    (It doesn't look like you've really understood vector spaces and matrices. For example, how could anything but a 2×1 matrix be an eigenvector of a 2×2 matrix?).

    No. I don't really understand what you're saying here, but this looks very different from any accurate description I've seen. :smile:
     
    Last edited: Jul 6, 2011
  10. Jul 6, 2011 #9

    Fredrik

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    I don't often disagree with you, but unless I have misunderstood what you're saying, this is wrong. The number that identifies the particle species is the j in the eigenvalue equations [tex]\begin{align}\vec S^2|jm\rangle &=j(j+1)|jm\rangle\\ S_z|jm\rangle &= m|jm\rangle\end{align},[/tex] while the number of rows in these column matrices is equal to the number of values m can take.
     
  11. Jul 6, 2011 #10

    What I get out of post #2, now that I've reread it a few times, is that the state vector needs 2 dimensions, so that is can be combined in linear form to allow the superposition of the two states can be represented.

    I did ask if this was done for mathematical convenience.

    That bit in the other post was worded badly. How about this. The representation of | 1 0 | is like literally saying and visually representing that the the north end of the electron is pointing in the positive axis direction and the south end of the electron is pointing in the negative axis direction.

    I guess its the abstractness of the representation that I'm getting caught up in. With the 1 above the 0 in a column vector it's literally a picture with the electrons north end pointing up and south end pointing down. The matrix operations, due to the actual spacial layout of the matrix elements, are like physical manipulations of the visual representation of the spin orientation.

    I've done some coding, and from that perspective a construction such as the state vector would be called a one dimensional array with two elements. I'm beginning to think describing the state vector as "two dimensional" is misleading
     
    Last edited: Jul 6, 2011
  12. Jul 6, 2011 #11

    SpectraCat

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    You can probably find one on the web .. I just learned it by copying and pasting and messing around with the source from other posts .. the Tex source code is shown when you use the PF QUOTE feature.

    I think the source of your problem is that you are taking the colloquial terminology associated with spins too literally. When we measure the projection of a spin on a given axis of quantization, there are two possible results, [itex]\pm\frac{\hbar}{2}[/itex]. Colloquially, we call the positive value "spin up", and the negative value "spin down". It is important to understand this does NOT mean the spin angular momentum is actually aligned with the axis is the up or down direction ... that is explicitly forbidden by QM, since then you would know the projection of the spin on any axis perpendicular to the axis of quantization (it would be zero). Since the operators corresponding to those projections don't commute, that is impossible. Note also that the magnitude of the spin is given by [itex]|s|=\hbar\sqrt{s(s+1)}=\hbar\frac{\sqrt{3}}{2}[/itex], which is larger than the magnitude of the projection, indicating that the "vectors" are not parallel.

    Anyway, let's say we chose the z-axis for quantization. The two eigenvalues [itex]\pm\frac{\hbar}{2}[/itex] correspond to the basis states that we will annotate |+> and |->. Now when we write these in vector notation, we have:
    [itex]|+>=\begin{pmatrix}1\\0\end{pmatrix}, |->=\begin{pmatrix}0\\1\end{pmatrix}[/itex].
    As you say, this is somewhat redundant, and doesn't convey anymore info than we already knew from the labels for the eigenstates. However, what if there is another spin in our system, that has been measured using the x-axis as the quantization axis. How can we represent those results, which we'll call [itex]|s_x^+>[/itex] and [itex]|s_x^->[/itex] for clarity, in our initial |+> and |-> basis for z-axis quantization? That is where the sigma matrices come in .. if you do the appropriate transformations, you find that:
    [itex]|s_x^+>=\begin{pmatrix}\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{pmatrix}[/itex] and [itex]|s_x^->=\begin{pmatrix}\frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}\end{pmatrix}[/itex] in the |+>,|-> basis.

    So, that is why you need to specify both components of a spin in the vector representation.
    Because choosing an axis of quantization is equivalent to choosing a two-dimensional basis. Once you choose an axis of quantization, ANY spin variable can be represented by specifying just two components.

    No, the sigma matrices define transformations between the 2D bases representing cases where each of the cardinal axes is chosen as the quantization axis. Since the projection of spin on different axes don't commute, you can only ever know the projection on a single axis.

    [EDIT] Note that when I have been discussing "spins" above, I have been assuming we are talking about single fermions with a spin of [itex]\frac{1}{2}[/itex]. As Frederik points out, if you are talking about a boson, or a total spin S formed from the coupling of multiple fermions, then you need to use a more general form, where the dimension of the basis is 2S+1.
     
    Last edited: Jul 6, 2011
  13. Jul 6, 2011 #12
    Well I thought I followed that. Then I realized the first sigma 1 state you mentioned | 1/2^(1/2) 1/2^(1/2) |. That would seem to conflict with my two points on a line analogy.

    ***

    I think I got it now, at least for my own satisfaction. The state vector is a representation of the spin, and I did mention that this was about an electron's spin in the my first post, perpendicular to the axis of the spin rather than parallel to it.

    That is like looking at a coin in the plane of the coin and putting a 1 on the heads side of the plane and a 0 on the tails side of the plane, rather then looking directly down on it and seeing which side is facing up and using just one digit either a 1 or 0 to represent the visible side.

    I'm interpreting this to be a one dimensional representation of the spin orientation that need two elements to express. This is exactly like using two points on a line, point N represents the north end of the electron and point S represents the south end. Both points are needed to indicate the orientation, with point S being either to the right or left of point N depending of course on the orientation.

    I understand the algebra of how the three sigmas work together, and how the unit vector can be used to determine the spin probability in any arbitrary direction. What I was having trouble with was understanding exactly how the state vector was representing the spin orientation.
     
    Last edited: Jul 6, 2011
  14. Jul 6, 2011 #13
    Uh oh, some of what you said is starting to make sense. I've been following a lecture series about this, and the sigmas were presented as is without any elaboration about transforming between one and another. So I lack a clue as to how to transform [itex]\sigma[/itex]3 to [itex]\sigma[/itex]1 etc.

    I just realized what you said there, spin up and spin down are not parallel, Doh! they're orthogonal, as shown by the inner product of their eigen states being zero. That was actually mentioned in the lectures. I failed to notice that and was erroneously presuming them to be parallel. There goes my two points on line analogy...


    After mulling over you post and the others as well as my notes from the lectures i realize I may have to get a book on the subject. In any case the lectures presented the eigen values of all the sigmas as +1 and -1, and avoided any mention of the magnitude of the spin.

    Also the measurement process was described as observing if the system either emitted a photon when it was aligned to the quantization axis or if ti aligned without emitting a photon. There was an absence of any elaboration on the amount of the energy carried off by the photon if emitted. The lecture series is a basic introduction.

    I do realize that once a system is prepared, then all that is known is the probability of the spin state in any given direction until it's measured in that direction. Any recommendation on good printed material covering the subject, something available in PDF form would be most convenient at the moment.

    Been doing a bit more reading. So for 1/2 spin particles there are 2s+1 = 2 spin states(-1/2, +1/2), for spin 1 particles there are 2s + 1 = 3 spin states( -1, 0, +1) and for spin 3/2 particles there are four spin states generated by the combinations of the three 1/2 spin elementary spin carriers(-3/2, -1/2, +1/2 and +3/2).

    It seems like the relations between the sigmas [itex]\sigma[/itex]1[itex]\sigma[/itex]2 = i[itex]\sigma[/itex]3 etc are describing the properties of the fabric of space.

    ***

    This is really beginning to bugger me. From a bit more reading is seems that the vector space is called two dimensional because it's composed of, wait for it, two orthogonal vectors.

    However, that is a completely separate issue from the dimensionality of the individual vectors that comprise that 2-D vector space. Those individual vectors have two components/elements rather than dimensions.

    So I'm back to my original question although is has become a bit more specific now and I guess rhetorical as well, just how is it justified/rationalized that the parallel spin descriptions of up or down as measured on an axis of quantization are mathematically described by orthogonal vectors? I guess this involves the conflict between the colloquial verses the mathematical meanings of spin, and I guess further study of the math will reveal just what it is that to two component vectors are representing.

    ***

    What about this: it seems like the sigma number order was taken from quaternion notation, meaning that [itex]\sigma[/itex]3 is the system preparation axis where the probability of measuring the spin orientation is 1 and 0 depending which way the measurement is taken on the axis of preparation.

    The state vectors of [itex]\sigma[/itex]1 and [itex]\sigma[/itex]2 are the probabilities of measuring the spin orientation orthogonal to the axis of preparation. From this perspective I could still use my two points on a line analogy to convey the spin orientation information for the only applicable sigma, which would be [itex]\sigma[/itex]3 the axis of preparation.
     
    Last edited: Jul 7, 2011
  15. Jul 7, 2011 #14
    The pieces are falling into place. I see that the components of the eigenstates simply represent probabilities, the two points on a line analogy for [itex]\sigma[/itex]3 is quite a poor one, and the orthogonality of the eigenstates for each sigma simple implies mutual exclusivity of those states rather than physical orthogonality of the spin orientation.

    I believe I've found my way into Hilbert Space!

    Now to learn the lay of the land...
     
    Last edited: Jul 7, 2011
  16. Jul 7, 2011 #15

    Fredrik

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    You're definitely making progress. I couldn't really make sense of the things you were saying yesterday, but now you're starting to sound like one of us. :smile:

    The commutation relations certainly do. They arise naturally from the assumption that space is isotropic, unfortunately through some very heavy mathematics.

    Those column vectors are 2×1 because the vector space is 2-dimensional, but it's not hard to think of examples of an n-dimensional vector space where the members are m-tuples with n≠m. See e.g. what I said about the spin matrices in this thread.

    1. Eigenvectors corresponding to different eigenvalues are always orthogonal.
    2. If you prepare an "up" state and then measure the spin component along that axis, the result will always be "up". If the vectors had not been orthogonal, QM would have assigned a non-zero probability to the "down" result.
     
  17. Jul 9, 2011 #16
    Now my mind is spinning. N-dimensional mutually orthogonal basis vectors, linear hermitian Hamiltonian operators over the outer tensor product of abstract n-dimensional vector spaces etc. It all makes me wish I had actually completed a formal study of calculus and classical mechanic instead of just the basics of both.

    I found the Shankar book. It seems to be well detailed, although he's one of my least favorite Professors. I think Yale corrupted his soul.

    In any case I think I've found the beginning of the path. I've already seen that a number of people are saying that a good understanding of the matrix and vector math is fundamental to exploring the rest QM so I suppose that my first goal. I might have to do some related prerequisite study, although that's what I do, I'm constantly studying one thing or another, and I have been wanting to finish my calculus study so I can start playing with all those partial derivatives and tensors of general relativity.

    I am looking forward to being able at some point to model a simple complex system. Hmm, did I just express an oxymoron there? Anyway, to be able to explore what is meant by the statement that some of the properties of the complex system lack supervenience upon the properties of the individual systems that make up the complex.
     
    Last edited: Jul 9, 2011
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