New answers tagged notation
1
Q1/2. In Dirac notation, one does not usually write expressions like $|\psi(x,t)\rangle$ because the ket symbol denotes an element of a Hilbert space, not its corresponding representation in a particular basis. One does, however write expressions like $|\psi(t)\rangle$ to denote the state of the system at time $t$. If you wanted to write such a state in ...
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In the comments Alfred raises the notion of the dual space. In fact, if you try to read Dirac's principles of QM, you will find that he starts with dual space.
In Dirac notation $|z\rangle$ is an element of an abstract vector space $\mathcal{H}$. Then, there is a notion of dual space: the dual space $\mathcal{H}^*$ is the space of all (continious) linear ...
1
My question is why do we distinguish between both kinds of vectors
Take an arbitrary vector space. Then, the set of scalar-valued linear functions on that space inherits a linear structure in an obvious way and becomes a vector space in its own right.
We distinguish between these spaces because they are distinct.
Now, in the finite dimensional case ...
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Short answer: we keep track of covariant and contravariant vectors because it gives us the freedom to use different quantities with respect to different bases and to find dot products of covariant and contravariant vectors without having to move everything into the tangent or cotangent basis and use the metric every time we want to find a physically ...
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From Wiki:
For a finite-dimensional vector space, using a fixed orthonormal
basis, the inner product can be written as a matrix multiplication of
a row vector with a column vector:
$ \langle A | B \rangle = A_1^* B_1 + A_2^* B_2 + \cdots + A_N^* B_N = \begin{pmatrix} A_1^* & A_2^* & \cdots & A_N^* \end{pmatrix} \begin{pmatrix} B_1 \\ ...
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To formalize the comments as an answer:
The difference between requiring
$$(\alpha u,v)=\alpha(u,v)\quad\text{ (mathematician's definition)}$$
and
$$\langle u, \alpha v\rangle=\alpha\langle u,v\rangle\qquad\quad\,\,\text{ (physicist's definition)}$$
is purely one of convention, and the two definitions are equivalent as $(u,v)=\langle v,u\rangle$. There's no ...
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There are two aspects. One is sort of trivial and comprehensible; the other is a bit technical.
The trivial reason is that $\tilde t \bar{\tilde t}$ has two "accents" on top of each other and the symbol therefore occupies too much vertical space which is undesirable because we may get overlapping characters and/or non-uniform spacing between lines. The ...
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