Xjyuk

What is the difference between these two?

$langle x|xrangle$ and $|xranglelangle x|$

Are they the same? If they're the same, why are they used in these two different forms?

They are not the same. One is the outer product and one is the inner product.

In the finite-dimensional real-number case for example, if $$|xrangle = beginbmatrix1\2\3endbmatrix,$$ then $$|xranglelangle x| = beginbmatrix1\2\3endbmatrix beginbmatrix1&2&3endbmatrix = beginbmatrix1&2&3\2&4&6\3&6&9endbmatrix,$$ while $$langle x | xrangle = beginbmatrix1&2&3endbmatrix beginbmatrix1\2\3endbmatrix = 1 + 4 + 9 = 14.$$

In Bra-Ket notation $| x rangle$ is a vector and $langle x |$ is a covector. Formally, a covector is a map that goes from $V rightarrow mathbbR$, i.e. it “eats a vector” and gives you a number. When we write
$$
langle x | x rangle
$$
we are saying that the covector $langle x |$ is acting on the vector $|xrangle$, so $langle x | xrangle$ is a real number.

On the other hand,
$$
| x rangle langle x|
$$
isn’t acting on anything. If we toss some vector at it $|arangle$, we would have
$$
|xrangle langle x | a rangle
$$
which is just a real number times $|x rangle$. We have given the above a vector and we got a vector back out so it is a map between vectors. To summarize then, $|xrangle langle x|$ and $langle x | x rangle $ represent different types of objects: the first is a map from $V rightarrow V$, wheras the second is an element $b in mathbbR$.

To be concrete, let
$$
|xrangle =
beginpmatrix
1\
0
endpmatrix
$$
. Then we would write
$$
langle x | = beginpmatrix 1&0endpmatrix
$$
which is the Hermitian conjugate of $|xrangle$, in this case just the transpose. So
$$
langle x | x rangle = beginpmatrix 1&0 endpmatrix
beginpmatrix
1\
0
endpmatrix = 1
$$
And
$$
| x rangle langle x | =
beginpmatrix
1\
0
endpmatrix beginpmatrix1&0endpmatrix =
beginpmatrix
1&0\
0&0
endpmatrix
$$
A number and a map, as expected.

In general $langle Avert Brangle$ is a scalar product whereas $vert Aranglelangle Bvert$ is an operator. To this this last suppose
$vert psirangle$ is an arbitrary vector in your space. Then
$$
vert Aranglelangle Bvert psirangle
$$
is a vector proportional to $vert Arangle$ since $langle Bvert psirangle in mathbbC$ is a (complex) number, so that $vert Aranglelangle Bvert$ maps a vector to another vector.