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Change of Basis (Matrices)

  1. Jul 7, 2012 #1
    1. The problem statement, all variables and given/known data
    https://dl.dropbox.com/u/4788304/Screen%20shot%202012-07-08%20at%2002.53.44.JPG [Broken]

    This is the solution of Problem A.15 in Griffiths' Quantum Mechanics. Tx is the rotation matrix about x-axis for theta degrees; while Ty is the rotation matrix about y-axis for theta degrees.

    The thing I do not understand is the apparent mismatch between the components and the matrices. For example look at the last row; applying S to the original bases (i, j, k), you will get (-j, i, k), hence i' = -j, j' = i and k' = k, which was not what was written on the left.

    This seems like a really stupid question. Am I missing out something?

    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jul 7, 2012 #2
    Looks like a typo. Why don't you try writing a textbook, see how hard it is to get everything perfect. I'll bet you don't turn in perfect homeworks and tests.

    Maybe if you were smarter you would not have gotten so stuck on such a small error.

    Just kidding!
     
    Last edited by a moderator: May 6, 2017
  4. Jul 8, 2012 #3
    Thanks for the reply. I got it now, and it's not a typo (and I don't think it was a typo in the first place, that's why I asked). In a matrix the basis vectors are written as columns, not rows. It's indeed very silly of me to read off the rows as the new basis vectors and arrived my previous conclusion that something is wrong.

    And I was not criticising the textbook, so I have no idea why you were being rather cynical about it. Perhaps it's because I said "This seems like a really stupid question." and you thought I was referring to the questions in the book, but what I intended to mean is "This seems like a really stupid question for me to ask." :smile:

    But anyway, case closed.
     
  5. Jul 8, 2012 #4
    Aha, you're right. j'=Sj, i don't know why that escaped me. Must be getting cross-eyed.

    And why did i say that? Kidding primarily, maybe a little projecting of my own reflections on school. I remember noticing that I didn't get screwed up by a professor's mistake in lecture if I understood the material (perfectly). It's when you're barely getting the material that a teacher's typo will really trip you up. I'm not putting myself above you, I think I find mistakes in texts all the time, and at least half the time, after staring at it for a while, I realize it's not a typo. This problem case and point. And I did it too!
     
  6. Sep 5, 2012 #5
    Actually, this still doesn't make sense to me. As per the book (eq A.62) :

    [itex]a^{new} = S a^{old}[/itex]

    which says that the components of the vector [itex]a[/itex] expressed in the old basis, when multiplied by [itex]S[/itex] will yield the components of [itex]a[/itex] expressed in the new basis.

    Now multiply the S he gives in the solution, by i [which has (1,0,0) as components in the old basis]. The result is (0,1,0) which, according to above, represents the components of i in the new basis. But (0,1,0) in the new basis (which is just a rotation by 90 degrees counterclockwise around the k axis) is clearly not the original i vector!

    You can see it numerically : The change of basis in the problem is i' = j, j' = -i, k'=k. Since, the original basis vectors are i=(1,0,0), j=(0,1,0), k = (0,0,1) the new basis vectors are i' = (0,1,0) , j' = (-1, 0, 0), k = (0,0,1).

    So, if S was the correct matrix, (0,1,0) in this basis is supposed to equal (1,0,0). But 0*i' + 1*j' + 0*k' = 0 * (0,1,0) + 1 * (-1,0,0) + 0*(0,0,1) = (-1,0,0).
     
  7. Sep 5, 2012 #6
    That's right.

    That's wrong. j'= -i, as stated in the question, so matrix S will transform the basis vector (0,1,0) to (-1,0,0).
     
  8. Sep 6, 2012 #7

    HallsofIvy

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    Some physics texts premultiply matrices. That is to apply a rotation about the x-axis to a vector [itex]a\vec{i}+ b\vec{j}+ c\vec{k}[/itex], they do the m)atrix multiplication
    [tex]\begin{bmatrix}a & b & c \end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & cos(\theta) & sin(\theta) \\ 0 & -sin(\theta) & cos(\theta)\end{bmatrix}[/tex]
    rather than
    [tex]\begin{bmatrix}1 & 0 & 0 \\ 0 & cos(\theta) & -sin(\theta) \\ 0 & sin(\theta) & cos(\theta)\end{bmatrix}\begin{bmatrix}a \\ b \\ c\end{bmatrix}[/tex]
    which is more standard in mathematics.

    That appears to be what is being done here.
     
  9. Sep 6, 2012 #8
    I think you misinterpreted what I said. I meant that if S is the correct matrix, then (0,1,0) is the representation of i in the new basis (i',j','k'). But if you multiply out (0,1,0) by (i',j',k'), you get (-1,0,0) when you are supposed to get (0,1,0).

    Let me try to formalize my reasoning more :

    The objects in our vector space are triples <x,y,z> where x,y,z [itex]\in[/itex] ℝ. Multiplication by a scalar and the addition of two triples are defined component wise.

    Our original basis vectors i,j and k are the following three triples : <1,0,0>, <0,1,0>, <0,0,1> respectively. The new basis vectors (i',j',k') are [following the change of basis stated in the problem] i' = j, j' = -i, k'=k which means i' = 1*<0,1,0>, j' = -1 * <1,0,0>, k' = 1 * <0,0,1>

    or i' = <0,1,0> , j' = <-1,0,0>, k' = <0,0,1>

    Now, any vector v in our space can be expressed as

    v = a*i + b*j + c*k in the old basis

    and also as

    v = a'*i' + b'*j' + c'*k' in the new basis

    for some scalars a,b,c,a',b',c'.

    Equation A.62 in the book says that components_new = S components_old, or in other words that expressed in matrix form, (a',b',c') = S (a,b,c).

    If we multiply any components (a,b,c) by the matrix S given in the solution to the problem, we get : (-b,a,c).

    So (a',b',c') = (-b, a, c) and therefore

    v = a'*i' + b'*j' + c'*k' = -b*i' + a*j' + c*k'
    = -b * <0,1,0> + a * <-1,0,0> + c * <0,0,1> = <0,-b,0> + <-a,0,0> + <0,0,c> = <-a,-b,c>

    This is a contradiction, because v was taken to be the triple <a,b,c>, and here we have v = <-a,-b, c>
     
  10. Sep 8, 2012 #9
    Wait hang on. If you multiply the components (a,b,c) by S, you will get (b, -a, c). i' = j, j' = -i, k'=k.
     
  11. Sep 9, 2012 #10
    By "multiply any components (a,b,c) by the matrix S", I meant regular matrix-vector multiplication :

    [itex]\begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix}
    [/itex],

    which is equal to [itex]\begin{bmatrix} -b \\ a \\ c \end{bmatrix}
    [/itex].

    I'm just using equation A.62 in the book :

    [itex]a^{new}_{i}[/itex] = [itex]\sum^{n}_{j=1}S_{ij} a^{old}_{j}[/itex]

    which is regular matrix-vector multiplication.
     
  12. Sep 9, 2012 #11

    vela

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    The solution has S and S-1 swapped. For the matrices in the solution, you have
    \begin{align*}
    S\begin{pmatrix} 1 \\ 0 \\ 0\end{pmatrix} &= \begin{pmatrix} 0 \\ 1 \\ 0\end{pmatrix} \\
    S^{-1}\begin{pmatrix} 1 \\ 0 \\ 0\end{pmatrix} &= \begin{pmatrix} 0 \\ -1 \\ 0\end{pmatrix}
    \end{align*} Now the vector ##\hat{j}## is represented by ##\begin{pmatrix} 0 \\ 1 \\ 0\end{pmatrix}## in the unprimed basis and by ##\begin{pmatrix} 1 \\ 0 \\ 0\end{pmatrix}## in the primed basis. You can see multiplication by S takes the representation for ##\hat{j}## in the primed basis to the representation in the unprimed basis whereas S-1 does not.
     
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