Talk:Schmidt decomposition

message: I haven't changed it because I'm not sure, but I guess that in the following paragraph e_j 
should be f_j. So:
The Schmidt decomposition is essentially a restatement of the singular value decomposition in a different context.
Fix orthonormal bases \{ e_1, \cdots, e_n \} \subset H_1 and \{ f_1, \cdots, f_m \} \subset H_2. 
We can identify an elementary tensor e_i \otimes e_j with the matrix e_i e_j ^T,  
where e_j ^T is the transpose of e_j. A general element of the tensor product
:v = \sum _{1 \leq i \leq n, 1 \leq j \leq m} \beta _{ij} e_i \otimes e_j 
Should be:
The Schmidt decomposition is essentially a restatement of the singular value decomposition in a different context. 
Fix orthonormal bases \{ e_1, \cdots, e_n \} \subset H_1 and \{ f_1, \cdots, f_m \} \subset H_2. 
We can identify an elementary tensor e_i \otimes f_j with the matrix e_i f_j ^T, 
where f_j ^T is the transpose of f_j. A general element of the tensor product
:v = \sum _{1 \leq i \leq n, 1 \leq j \leq m} \beta _{ij} e_i \otimes f_j
Please have a look and change it if I'm rigth or correct me here if I'm wrong.
130.102.2.60 03:34, 12 April 2007 (UTC)
replies: text: yes, you're right. thanks for pointing it out. Mct mht 07:14, 12 April 2007 (UTC)

message: Why call it a statement and not a theorem, lemma, or proposition?
In the statement you use \alpha_i, while at the end of the proof you use \sigma_i.  It might not be clear to all readers that the \alpha_i in the statement are equal to the \sigma_i in the proof.  It may be worthwhile to keep the separate variables names to maintain the distinction between singular values and Schmidt coefficients, however, at some point it should be made explicit that \alpha_i=\sigma_i.
It seems to me that the Schmidt rank should be defined as the number of nonzero Schmidt coefficients, otherwise the Schmidt rank is always m, in which case the definition is unnecessary.  Furthermore, it would imply that every v is an entangled state, which is clearly not true.
In the last paragraph in the section on Schmidt rank and entanglement, the symbol v is used in two different ways, once as an element of H_1\otimes H_2 and once in an expression involving the tensor product of two vectors.
Mathmoose 14:23, 21 July 2007 (UTC)
replies: text: well, all valid points (modulo "come on, man..." :-) ). wanna go ahead and make the changes you propose? Mct mht 03:59, 25 July 2007 (UTC)

text: on the article, do you see the "edit this page" button on top? double click that and edit away. :-) Mct mht 03:07, 1 August 2007 (UTC)

text: hm, that can't be right. you sure you're on this page when you click it? Mct mht 22:39, 5 August 2007 (UTC)
thread_title:  quibbles 

message: The entire section on plasticity seems out of place.  It is poorly explained in the context of the Schmidt decomposition and seems in need of a strong overhaul, removal, or to be moved to a more appropriate page. JoseyS (talk) 08:11, 27 December 2017 (UTC)
thread_title:  Section on plasticity  

message: I'm learning about this for the first time, and the statement of the theorem feels out of order to me. Maybe I am misinterpreting it.
It currently says
Let H_1 and H_2 be Hilbert spaces of dimensions n and m respectively. Assume n \geq m. For any vector w in the tensor product H_1 \otimes H_2, there exist orthonormal sets \{ u_1, \ldots, u_m \} \subset H_1 and \{ v_1, \ldots, v_m \} \subset H_2 such that w= \sum_{i =1} ^m \alpha _i u_i \otimes v_i, where the scalars \alpha_i are real, non-negative, and, as a (multi-)set, uniquely determined by w.
Again, I'm learning this for the first time, but this statement seems weird. Because it starts with "for any vector w", it sounds like the orthonormal sets depend on w. But then this would be trivial, since one could just write w = u \otimes v for u \in H_1, and v \in H_2, complete u (resp. v) to a basis of H_1 (resp. an m dimensional subspace of H_2) and perform Gram-Schmidt starting from u and v separately to fill out an orthonormal set. Then you would have \alpha_1 = 1 ,u_1 = u, v_i = v. The rest of the \alpha_i are then zero. Of course this isn't constructive.
I expect what is meant is that given H_1 and H_2, there exists these orthonormal sets such that any w can be written in this way for some set of \alpha_i. So I suggest the following edit. Basically replaces "for all [], there exists ()" with "there exists () such that for all []".  Is it reasonable, or am I missing something?
Let H_1 and H_2 be Hilbert spaces of dimensions n and m respectively. Assume n \geq m. There exist orthonormal sets \{ u_1, \ldots, u_m \} \subset H_1 and \{ v_1, \ldots, v_m \} \subset H_2 such that any vector w in the tensor product H_1 \otimes H_2 can be written w= \sum_{i =1} ^m \alpha _i u_i \otimes v_i, where the scalars \alpha_i are real, non-negative, and, as a (multi-)set, uniquely determined by w.
Thanks. Fish sounds (talk) 20:51, 30 July 2020 (UTC)
thread_title:  Statement of theorem seems awkward