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From this link : [http://math.ucr.edu/home/baez/lie/node12.html][1], it is said that, from starting a quantum state, $$\vec{v}=a|\text{up} \rangle+b|\text{down} \rangle,$$ we can define the matrix projection by doing:

$$\text{Matrix}_\text{projection}=vv^{*} =\left[ \begin{array}{cc}a \\ b \end{array} \right] \left[ \begin{array}{cc}a^* & b^* \end{array} \right] = \left[ \begin{array}{cc} a a^* & a b^* \\ a^* b & b b^* \end{array} \right] \, .$$

From now, I will write " $v$ " instead of " $\vec{v}$ " for abreviation of vectors (except for basis vectors $\vec{v_i}$, see at the end).

The matrix above looks like a tensorial product between vector $v$ and its dual $v^{*}$.

Then, one introduces into this link the projection of an arbitrary vector $w$ along vector $v$ like this : $$\text{proj}_v(w) = \langle v, w \rangle v$$ I don't understand the following equation: $$v (v^* w) = (v v^*)w \, $$

Indeed, I am stuck on the transition

$$v(v^* w) = vv^* w$$

because "$vv^* w=\text{Matrix}_\text{projection}\,w$" does not make appear the vectors basis $\vec{v_i}$ unlike the term $$v(v^{*}w)=\sum_{i}(v_{i}^{*}w) \vec{v_{i}}$$

$vv^{*} w$ represents the action of projection matrix on $w$ vector but doesn't specify the basis vector $\vec{v_{i}}$.

I would like to get clarifications about this issue (maybe notations I use or used are bad since I prefer considering coordinates into basis vectors).

Elements of answer :

I try to give some elements for my original issue.

1)For the first expression : If I take the expression $(\vec{v}.\vec{v}^{*})$, I can assimilate it to a (1,1) tensor, so like a matrix equal to in my case :

$$\begin{pmatrix} v^{1}_{1} & v^{1}_{2}\\ v^{2}_{1} & v^{2}_{2}\\ \end{pmatrix}$$

So, I can write the $i-th$ component, I get :

$$\bigg[(\vec{v}\vec{v^*}) \vec{w}\bigg]_{i}=\sum_{j} v^{i}_{j} w^{j}$$

2) For the second expression : $\vec{v}(\vec{v^*} \vec{w})$, one can have with :

$$\vec{v}=\sum_{i}y^{i}\vec{e_{i}}$$

So by taking the $i-th$ component of this expression (i.e relatively to $\vec{e_{i}}$ basis vector) :

$$\bigg[\vec{v}(\vec{v^*} \vec{w})\bigg]_{i}=\sum_{j}x_{j}y^{i}\,w^{j}=\sum_{j}v^{i}_{j}\,w^{j}$$

with $x_{j}y^{i} = v^{i}_{j}$

3) Conclusion : Finally, I find the same expressions for the $i-th$ component of output vector (with first and second expression, i.e $$(\vec{v}\vec{v^*}) \vec{w}=(\vec{v^{*}}\vec{w}) \vec{v}$$

But my confusion comes from the case 1), especially the equality :

$$(\vec{v}\vec{v^*}) \vec{w}=\sum_{i} \sum_{j} v^{i}_{j} \sum_{k} w^{k}\vec{e_{k}}$$

I should rather introduce the $\vec{e_{i}}\otimes\vec{e^{j}}$ :

$$(\vec{v}\vec{v^*}) \vec{w}=\sum_{i} \sum_{j} v^{i}_{j} \vec{e_{i}}\otimes\vec{e^{j}} \sum_{k} w^{k}\vec{e_{k}}$$

but this introduction of term $\vec{e_{i}}\otimes\vec{e^{j}}$ doesn't seem to be useful, this even doesn't change nothing in my demo.

From the begining, I tried to have a matricial product point of view for both expression, so I would like to know if I can extend the product $(\vec{v}\vec{v^*})$ as a tensorial product of vector $\vec{v}$ and its dual such as I can manipulate it like a classical matrix operating on vector $\vec{w}$ ?

Anyone could confirm me this implicit extension in expression ($\vec{v}\vec{v^{*}})$ since this product doesn't look like naturally to a matrix ?

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1 Answer 1

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$$ v(v^*w) = (vv^*)w $$

is the associative property, so you're probably safe as long as you avoid octonions (though Baez loves the octonions).

Your final comment regarding "preferring basis vectors" is interesting. The beauty of the above formulation is that it is coordinate-free, as it should be: the projection of $w$ onto $v$ does not depend on the definition of a basis--so why introduce one?

As long as:

$$ P^2 = P $$

it's a projection.

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  • $\begingroup$ -@JEB . I would like to get the proof of the affirmation "the projection of w onto v does not depend on the definition of a basis" . The demo that I have found are pretty difficult to grasp. Do you know a simple proof ? Regards $\endgroup$
    – user87745
    Jul 28, 2018 at 23:46
  • $\begingroup$ en.wikipedia.org/wiki/Coordinate-free $\endgroup$
    – JEB
    Jul 29, 2018 at 0:20
  • $\begingroup$ -@JEB . When you say "The beauty of the above formulation is that it is coordinate-free", you surely want to say that the inner product "$(v^*w)$" with vectors v and w expressed in basis $B$, is equal to the inner product "$(v^*w)$" with v and w expressed in basis $B'$, don't you ? $\endgroup$
    – user87745
    Aug 4, 2018 at 13:11
  • $\begingroup$ @youpilat13 No. The general idea of coordinate free formulation is that a physical law can be expressed as a relationship between geometric objects (scalars vectors, tensor in R3, M4, Hilbert space...). These objects exist without a coordinate system, and in some cases, coordinates can obscure the underlying geometric relationship. $\endgroup$
    – JEB
    Aug 4, 2018 at 15:32
  • $\begingroup$ -@JEB . I just wanted to notice that inner product $(v^{*}w)$ is invariant under changing of basis but not the global term $v(v^*w)=\langle v,w \rangle v$ since $\langle v,w \rangle$ is invariant and vector $v$ depends of basis : do you agree ? $\endgroup$
    – user87745
    Aug 4, 2018 at 18:45

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