What is a "shift" in a separable Hilbert space? I thought I knew what a  (left) "shift" is. If $(e_n)$ is the standard unit vector basis in $\ell_2$, then $T:\ell_2\to \ell_2$  defined by 
$$
Te_1=0\\ Te_j=w_j e_{j-1}, \forall j>1
$$
is a weighted left shift on $\ell_2$. 
That suggested the definition that an operator $T$ on a separable Hilbert space $H$ is a left shift if there exist an orthonormal sequence $(e_n)$ such that in that orthonormal sequence $T$ looks like above. 
Now consider $\bigoplus \ell_2  :=\{((x_1, x_2, \dots) : \sum\|x_i\|^2<\infty \}$, the direct sum of countable copies of $\ell_2$, and the operator $S$ defined as:
$$
S(x_1, x_2, x_2, x_3\dots)=(x_2, x_3, \dots)
$$
This looks like a left shift, but does it satisfy my ad-hoc definition above? Is my definition of a "shift" faulty, and if so what is a "shift" on an abstract separable Hilbert space?   
Edit: In view of the comments below, the second "shift" does not satisfy my, obviously faulty definition. Clearly, it is natural to consider $S$ above as being a shift. Consider now $X$ an isomorphism between $\ell_2$ and $\bigoplus \ell_2$. Then $X^{-1}SX$ is a shift on $\ell_2$ that does not look at all like the $T$ above. How do I recognize directly such an operator as being a shift, without noticing (how?) that it has a "shift" like form on another representation of a separable Hilbert space? 
 A: I suggest for you the following definitions :
Let $H$ be a separable Hilbert space. A bounded operator $A$ on $H$
is a right shift if and only if for some orthogonal basis $\left(
e_{n}\right) _{n\in 
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\mathbb{N}
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}$ for $H$ we have $A\left( e_{n}\right) \in span\left\{ e_{n+1}\right\} $
for every $n\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
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$.
Let $H$ be a separable Hilbert space. A bounded operator $A$ on $H$ is a
left shift if and only if for some orthogonal basis $\left( e_{n}\right)
_{n\in 
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\mathbb{N}
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}$ for $H$ we have $A\left( e_{0}\right) =0$ and $A\left( e_{n+1}\right) \in
span\left\{ e_{n}\right\} $ for every $n\in 
%TCIMACRO{\U{2115} }%
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\mathbb{N}
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$.
But if your Hilbert space is not necessarily separable you can use
alternatively the two following definitions :
Let $H$ be a Hilbert space. A bounded operator $A$ on $H$ is a generalized
right shift if and only if there exists a sequence of closed non-trivial orthogonal
subspaces $\left( H_{n}\right) _{n\in 
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\mathbb{N}
%EndExpansion
}$ such that $\underset{n\in 
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\mathbb{N}
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}{\bigoplus }H_{n}=H$ and $A\left( H_{n}\right) \subset H_{n+1}$ for every $%
n\in 
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%BeginExpansion
\mathbb{N}
%EndExpansion
$.
Let $H$ be a Hilbert space. A bounded operator $A$ on $H$ is a generalized
left shift if and only if there exists a sequence of closed non-trivial orthogonal
subspaces $\left( H_{n}\right) _{n\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
}$ such that $\underset{n\in 
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\mathbb{N}
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}{\bigoplus }H_{n}=H$, $A\left( H_{n+1}\right) =\left\{ 0\right\} $ and $%
A\left( H_{n+1}\right) \subset H_{n}$ for every $n\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
$.
Then we can say that according to all above definitions, your first example is for a left shift and the second one is for a generalized left shift but not a left shift.
