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Bra-ket notation

  1. Jul 5, 2009 #1

    I have been working out of the beginning of Shankar lately, and I wanted to address some confusion I have had with regard to Dirac notation.

    I know that physicists tend to love the notation, but to me, so far, it is confusing, inconsistent, and even occasionally contradictory. Here is a short list of some of what bothers me about it. I guess I am hoping that some of you can explain to me why some of these problems are not as bad as they seem, or even why they aren't actually problems.

    1. Labels:
    Kets are used to denote vectors, and a label is placed on the ket to name it. This is okay for me except when it comes to performing operations on vectors. For instance, let V be a vector space over C, and take |A> and |B> to be vectors in V. Also, let a and b be complex numbers. Then, Shankar uses the notation

    a |A> + b |B> = |a A + b B>.

    Therefore, a A + b B becomes a label for the new ket. This is okay with me except for cases like this: Let |1>, |2>, ..., |n> be a basis for V. Then, (3+2i) |1> - 4 |2> = |(3+2i) 1 - 4 2>.

    I know that one shouldn't get hung up on notation, but that just kills me.

    2. Inner products:
    This is basically the same thing as the first, but inner product labeling gets even worse. We denote the inner product of |V> with |W> by <V|W>. This seems to imply that the inner product of |LABEL 1> with |LABEL 2> is <LABEL 1|LABEL 2>. However, Shankar also writes things like
    <V| (aW + bZ) which is decent except that it has the same labeling issues.

    3. Dual space:
    Thanks to the Riez Representation Theorem, every element of the dual space can be written as a bra. Because of this, I think writing things like <V| is very good notation. However, Shankar feels that ANYTHING acting on or acted on by a bra should go after the bra.

    For instance, instead of 2 <V|, Shankar writes <V| 2 which is okay I guess if we know ahead of time that 2 is in the field and not the vector space. But what about this: let T be a linear transformation acting on bras. Then shankar writes <V| T instead of T <V| or T(<V|). Then he awful things like this:
    <V|aT =<V|Ta when what he means is T(a<V|) = a T<V|

    What about <V| T L where T and L are linear operators? Does it mean LT(<V|) or TL(<V|)?

    Maybe I'm overreacting to all of this, but this really seems to be a big mess!

    Anyway, thanks for reading! I would appreciate any discussion!
  2. jcsd
  3. Jul 5, 2009 #2


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    This is not strictly part of dirac notation. If you find this way of labelling confusing, you can just call the new vector |C>.
  4. Jul 6, 2009 #3


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    I have never seen that notation before. I absolutely hate it.

    This is standard. In other posts where I have tried to help people understand bra-ket notation, I've been writing the inner product of two vectors x and y as (x,y) to avoid confusion with the bra and ket symbols. With this notation, what you just said takes the form

    [tex]\big(|V\rangle,|W\rangle\big)=\langle V|W\rangle[/tex]

    Of course, that's probably only useful in discussions like this one.

    This is standard too. <V|T, where T is an operator (acting on kets), is defined as the functional that takes an arbitrary ket |U> to <V|(T|U>).

    There's no need to ever mention operators on the dual space. Are you sure that's what he does? I think he's just trying to define expressions like <V|T, where T is an operator on the space of kets. It needs to be defined the way I did it above, if we want everything that looks like a multiplication to be "associative".

    You might find this post useful. Note that equation 5 is an example of why we should allow the scalar to appear on the right of the bra. We want everything to look like we're just multiplying things together, using an associative multiplication operation. You should also check out this thread (read the last post first).
    Last edited: Jul 6, 2009
  5. Jul 11, 2009 #4
    Dirac notation is very good for large sequence, where you can see OK [tex]|...>[/tex] this is vector in direct space; [tex]<...|[/tex] this is vector in dual space; [tex]<...>[/tex] this is scalar; [tex]|...|[/tex] this is an operator. And of course this notation is great when you whant to change the representation.
    For me operators are [tex]\hat{A},\hat{B},\hat{C}...[/tex] with hat. Vectors are [tex]|\psi>,|\varphi>,|\phi>...[/tex]
  6. Jul 11, 2009 #5


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    3. Dual space

    Imagine we have N-dimensional space, with hermitian operator A, let the states |a> be represented by column vectors, then A is a NxN matrix

    clearly A|a> = a |a> is in linear algebra a NxN matrix mutliplying a column vector with N rows yielding a coloumn vector with N. Things are ok?

    Now the the ket should be represented by a row vector, with N columns, clearly A<a| is not defined as a matrix-vector multiplication in linear algebra, thus we must have <a|A to get "row vector times matrix = row vector"

    1. Labels:
    You might want to try the first chapter in the text by Sakura, modern quantum mechanics, there he writes a |A> + b |B> = c|C>
  7. Jul 12, 2009 #6
    Or look all that in for exaple [tex]\mathcal{R}^3[/tex] space. Dual of [tex]\mathcal{R}^3[/tex] is [tex]\mathcal{R}^3[/tex]. [tex]\mathcal{R}^3[/tex] is Hilbert space. This is of course particular case of [tex]\mathcal{R}^{N}[/tex].
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