# I Question(s) about Dirac notation

1. Apr 3, 2017

### mike1000

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?

2. Apr 3, 2017

### Orodruin

Staff Emeritus
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.

3. Apr 3, 2017

### Khashishi

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.

4. Apr 3, 2017

### vanhees71

Forget about matrix notation. It's almost always more confusing than helpful, except in cases where it's a clever calculational tool.

5. Apr 3, 2017

### mike1000

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 $$u\otimes v = uv^T=\begin{pmatrix} \\ u_1\\ u_2 \\ u_3 \\ . \\ . \\ . \end{pmatrix}\begin{pmatrix}v_1 & v_2\end{pmatrix}=\begin{pmatrix}u_1v_1 & u_1v_2 & u_1v_3 & \dots \\ u_2v_1 & u_2v_2 & u_2v_3 & \dots \\ u_3v_1 & u_3v_2 & u_3v_3 & \dots\\ \dots & \dots & \dots &\ddots\end{pmatrix}$$
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
$$|A\rangle = |u\rangle |U\rangle$$

Last edited: Apr 3, 2017
6. Apr 3, 2017

### Khashishi

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.

7. Apr 3, 2017

### mike1000

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?

Last edited: Apr 3, 2017
8. Apr 3, 2017

### Orodruin

Staff Emeritus
No, it is a matrix representation of a tensor.

A tensor is not a data type, it is a mathematical object.

9. Apr 3, 2017

### Staff: Mentor

$|U\rangle$ 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 $a$".