I Question(s) about Dirac notation

mike1000
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I promise that anytime I have question about Dirac notation I will ask it in this thread.

I do not know how to parse the following Dirac notation.

|\Psi'\rangle = |u\rangle |U\rangle

Can someone please convert the Dirac notation to matrix notation?
 
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It is most likely to be interpreted as the tensor product of the two states. How you would write it as a matrix depends on what the original Hilbert spaces are and on the tensor product basis you pick.
 
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There's a lot of freedom in how you write it. If we take u to mean spin up of particle 1 and U to mean spin up of particle 2, then you could perhaps write it like
##\Psi' = \begin{pmatrix}
1 \\
0
\end{pmatrix}_1
\times
\begin{pmatrix}
1 \\
0\end{pmatrix}_2 =

\begin{pmatrix}
1 \\
0 \\
0 \\
0\end{pmatrix}
\begin{matrix}
uU \\
dU \\
uD \\
dD\end{matrix}##

The uU dU uD dD are just labels for the basis states in the column vector and aren't really part of the notation.
 
Forget about matrix notation. It's almost always more confusing than helpful, except in cases where it's a clever calculational tool.
 
Orodruin said:
It is most likely to be interpreted as the tensor product of the two states. How you would write it as a matrix depends on what the original Hilbert spaces are and on the tensor product basis you pick.
vanhees71 said:
Forget about matrix notation. It's almost always more confusing than helpful, except in cases where it's a clever calculational tool.

I will forget about it but first I have to get comfortable with Dirac notation, which, to me, is confusing.

I looked up a tensor product of two vectors and found in linear algeba that is the outer product of two vectors.( I use to call that the cartesian product)

Here is what I found. "In linear algebra, the outer product is the tensor product of two vectors" For example the outer product of two vectors is \begin{equation}u\otimes v = uv^T=\begin{pmatrix} \\ u_1\\ u_2 \\ u_3 \\ . \\ . \\ . \end{pmatrix}\begin{pmatrix}v_1 & v_2\end{pmatrix}=\begin{pmatrix}u_1v_1 & u_1v_2 & u_1v_3 & \dots \\ u_2v_1 & u_2v_2 & u_2v_3 & \dots \\ u_3v_1 & u_3v_2 & u_3v_3 & \dots\\ \dots & \dots & \dots &\ddots\end{pmatrix}\end{equation}
The tensor product (outer product) of two vectors enumerates all the possible combinations of the elements of each vector.

However, in Dirac notation the outer product is given by
$$A = |u\rangle\langle U|$$
which is not the same as this
$$|A\rangle = |u\rangle |U\rangle$$
 
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mike1000 said:
In linear algebra, the outer product is the tensor product of two vectors" For example the outer product of two vectors is
You can't just throw in a transpose like that. You should enumerate the possible combinations and then represent each combination as a term in a column vector, which is what I did in post 3. Matrices are kind of clumsy for treating tensors since you have to play around with the dimensions to get things right.
 
Khashishi said:
You can't just throw in a transpose like that. You should enumerate the possible combinations and then represent each combination as a term in a column vector, which is what I did in post 3. Matrices are kind of clumsy for treating tensors since you have to play around with the dimensions to get things right.

Thanks. I understand what you just said. But let me ask you this, is the matrix I show in my previous post considered a tensor? In other words, does the outer product of two vectors result in a tensor data type?
 
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mike1000 said:
is the matrix I show in my previous post considered a tensor?
No, it is a matrix representation of a tensor.

mike1000 said:
does the outer product of two vectors result in a tensor data type?
A tensor is not a data type, it is a mathematical object.
 
mike1000 said:
I bet I know what the problem is...##|U\rangle## does not represent a vector does it? Is it an entire vector space? If it is how does someone know that just by looking at the notation? Is it because is it capitalized? It seems that Dirac notation is very context sensitive and it is not easy finding a good reference on Dirac notation.
##|U\rangle## is a ket, and all kets are vectors in some vector space, so of course it is a vector. And of course ##|u\rangle|U\rangle## is also a ket, albeit one from a different vector space than either ##|U\rangle## or ##|u\rangle##, namely the product space of the two spaces (which may be the same) to which ##|U\rangle## and ##|u\rangle## belong. The following notations are all roughly equivalent, except in their capacity to annoy people who are annoyed by sloppy notation:
##|uU\rangle##
##|u\rangle|U\rangle##
##|u\rangle\otimes|U\rangle##

Do remember that the symbols inside the ket are just labels, chosen for convenience in whatever problem you're working with. If I write something like ##A|a\rangle=a|a\rangle##, the only significance of the ##a## inside the ##|a\rangle## is that I've decided that for whatever I happen to be working on at the moment the letter "a" will be a really convenient label to use for "that ket which is an eigenket of operator ##A## with eigenvalue ##a##".
 
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