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When I learned quantum mechanics, my professor frequently emphasized that a matrix is a representation of an operator, not the operator itself, and a column ($1\times M$ matrix) is not a vector, it's a representation of a vector. This was done to teach us to think of vectors as elements of an abstract vector space, so that we could freely change bases and not stick to e.g. position representation, as well as work with representation-free equations.

But now when I started reading about the formal definition of a vector space, it appeared that columns can actually themselves be elements of a vector space, and then the matrices would themselves be operators. Thus, as I understood, they would be representations of themselves.

I'm puzzled a bit: can we really talk about kets as being the same as column vectors, neglecting that columns are usually viewed as representations of abstract vectors, or should we really differentiate between a ket and a column, in context of quantum mechanics?

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  • $\begingroup$ Is that really different from saying that $(0, 1)^t$ is not that upwards unit vector, but just a representation of it? And if I turn my head 90 degrees it is $(-1, 0)^t$ but still the same vector. $\endgroup$ – mvw Jun 25 '14 at 12:26
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There is nothing to be puzzled with. But I admit this can be confusing at the beginning.

Let $V$ be a (abstact) vector space over some field $k$, with finite dimension $n\geq 0$. Then for any basis $f = (f_1 ,\dots, f_n)$ of $V$, there is an isomorphism $\varphi_f \colon k^n \to V$ mapping $(1,0,\dots,0)$ to $f_1$, $(0,1,\dots,0)$ to $f_2$, etc. until $(0,\dots,0,1)$ to $f_n$. What is called the coordinates of a vector $v \in V$ (or what you called the representation of $v$) in the basis $f$ is just the preimage of $v$ by $\varphi_f$. Ok, then what ?

Well then take $V$ to be the $k$-vector space $k^n$. Denote $e_i$ the vector $(0,\dots,0,1,0,\dots,0)$ where the $1$ is in the $i$th position. Then $e = (e_1,\dots,e_n)$ is a base of $V=k^n$ and then there is an isomorphism $\varphi_e \colon k^n \to k^n$ mapping $e_i$ to $e_i$. Well this is the identity, so the coordinates of a vector $v \in k^n$ is $\varphi_e^{-1}(v) = v$. And, as you said in your question, the vector is the same as its coordinates in the base $e$ : so we could just identify a vector with it coordinates...

But now take the family $f = (f_1=e_2,f_2=e_1,f_3=e_3,f_4=e_4,\dots,f_n=e_n)$ (that is the family $e$ where I just permuted the two first vectors). This family $f$ is also a basis of $k^n$. So there is an isomorphism $\varphi_f \colon k^n \to k^n$ mapping $e_i$ to $f_i$. In this basis $f$, the coordinates of a vector $v \in k^n$ is $\varphi_f^{-1}(v)$. For example, the vector $e_1$ has coordinates : $$ \varphi_f^{-1}(e_1) = (0,1,0,\dots,0). $$ So working in the basis $f$, you must not identify a vector with its coordinates, as they are different!

The bottom line is : if you fixed a basis once and for all, you can as well work directly with the coordinates and not bother with the difference between the coordinates and the actual vector ; but if you do some change of basis, then it is crucial to differentiate between a vector $v$ and all of its coordinates $\varphi_f^{-1}(v)$ which varie with the base $f$.

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