Identity Operator: Vector Expressions in Basis A

In summary, the identity operator can be expressed in terms of a given basis, but it is more complete to do it in terms of the basis that the vector resides in.
  • #1
aaaa202
1,169
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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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  • #2
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?
 
  • #3
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.
 
  • #4
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.
 

FAQ: Identity Operator: Vector Expressions in Basis A

What is an identity operator in vector expressions?

An identity operator in vector expressions refers to a vector that, when multiplied by a given vector, yields the same vector as a result. In other words, it is a vector that does not change the original vector's direction or magnitude.

What is the basis of a vector expression?

The basis of a vector expression refers to the set of vectors used to express a given vector. These vectors are typically linearly independent and form the basis for all other vectors in the expression.

How do you determine if two vector expressions are equal using identity operators?

To determine if two vector expressions are equal using identity operators, you must check if the basis vectors of the two expressions are equal. If the basis vectors are equal, then the two expressions are considered equal.

Can an identity operator be used in different bases?

Yes, an identity operator can be used in different bases as long as the basis vectors are still linearly independent. However, the resulting vector expression may look different in each basis.

What is the significance of identity operators in linear algebra?

In linear algebra, identity operators play a crucial role in defining vector spaces and determining linear independence. They also allow for the simplification of vector expressions and help in solving various mathematical problems.

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