Identity Operator: Vector Expressions in Basis A

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Discussion Overview

The discussion revolves around the application of the identity operator in different vector bases, specifically addressing how vectors expressed in one basis relate to another basis when applying the identity operator. The scope includes theoretical considerations and mathematical reasoning related to linear algebra and quantum mechanics.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants question whether the components of the identity operator should be expressed in their own basis or in terms of another given basis A when applied to a vector.
  • One participant suggests that as long as the inner products are calculated consistently, it does not matter which basis is used for the components of the identity operator.
  • Another participant emphasizes the importance of consistency in expressing vectors and applying the identity operator, noting that the inner products will match the bases appropriately.
  • It is mentioned that the completeness of the basis applies only to orthonormal bases, indicating a potential limitation when bases are not mutually orthogonal.

Areas of Agreement / Disagreement

Participants express differing views on the treatment of bases when applying the identity operator, with no consensus reached on the best approach. Some emphasize consistency in inner product calculations, while others raise concerns about the completeness of non-orthonormal bases.

Contextual Notes

Limitations include the assumption that the completeness of the basis applies only to orthonormal bases, and the discussion does not resolve how to handle non-orthonormal bases in this context.

aaaa202
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I was wondering about this: The identity operator writes a vector in the basis that is used to express the identity operator:
1 = Ʃlei><eil
But if you are to apply it to a vector in a given basis A should the lei> then be expressed in terms of their own basis or in terms of A?
 
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aaaa202 said:
I was wondering about this: The identity operator writes a vector in the basis that is used to express the identity operator:
1 = Ʃlei><eil
But if you are to apply it to a vector in a given basis A should the lei> then be expressed in terms of their own basis or in terms of A?

It doesn't really matter as long as you can calculate the inner products.
Maybe an example might help you. (Since I'm not a physicist, I don't know what the correct notation is for bra-ket, but I'll try).

Take the following basis
[tex]e_1=(1,0,0),~~ e_2=(1,1,0),~~ e_3= (1,1,1)[/tex]
Take the inner product
[tex]<(a,b,c)|(x,y,z)>=ax+by+cz[/tex]
This formulation of the inner product is going to be dependent on the basis of course.

Anyway, let's calculate
[itex](|e_2><e_2|)|a>[/itex]

where a is given by [itex](1,2,3)[/itex]. Then we get
[tex]<e2|a>=1*1+1*2+0*3=3[/tex]

So [itex]|e_2><e_2|a>= 3|e_2>=(3,3,0)[/itex].

Now what if a is given in terms of another basis? So what if

[tex]a=e_1+2e_2+3e_3[/tex]

One possibility is to write a in the normal basis, then we get

[tex]a=(1,0,0)+(2,2,0)+(3,3,3)=(6,5,3)[/tex]

and then calculate as usual:

[tex]|e_2><e_2|a>=(11,11,0)[/tex]

Another possibility is to write the inner product in terms of the basis [itex]\{e_1,e_2,e_3\}[/itex]. That is, find a formula for

[tex]<ae_1+be_2+ce_3|xe_1+ye_2+ze_3>[/tex]

and then you can calculate [itex]|e_2><e_2|a>[/itex] without converting to another basis.

Does this answer your question?
 
Not at all. You just have to be consistent about how you do your inner products. Let's say you have vector |b> expressed as linear combination of |ai> that are the basis of A. In other words, |b> = [itex]\displaystyle\small \sum_i[/itex]bi|ai>, where bi=<b|ai> are the components of |b> in basis A. Now you want to apply unit operator consisting of your |ei> vectors. Fine. You get I|b> = [itex]\displaystyle\small \sum_{ij}[/itex]|ei><ei|aj><aj|b>. The inner products <ei|aj> effectively take care of matching one basis to the other. In matrix form, these will form the transformation matrix from one basis to another.

Edit: micromass beat me to it.
 
aaaa202 said:
I was wondering about this: The identity operator writes a vector in the basis that is used to express the identity operator:
1 = Ʃlei><eil
But if you are to apply it to a vector in a given basis A should the lei> then be expressed in terms of their own basis or in terms of A?

There is one caveat, the completeness of basis only applies to orthonormal basis. If your basis are not mutually orthogonal, the sum of their outer product won't come to unity. But in QM, this is usually satisfied.
 

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