# Homework Help: Bra - ket?

1. Oct 26, 2012

### rubertoda

bra - ket??

Hi, maybe a stupid question, but i would like to know if, if We have a real number, but we are i a vector space, and the operator is hermitian, is |a> is equal to < a |*?

i assume this, because if a is the vector (1,0) (spin up), and only real entries.

im trying to make ((Sy|a>) |b>) to ((<a|Sy)|b>) somehow..if both a and b are real..plz solve this mess

I am talking about Sx, Sy, Sz is hermitian operators.. and these are operating..
are these 3 hermitian conjugates to each other?
i might add, that Sy can ONLY operate on the 'a' state ( <a| or |a>)

and, finally, how would you write out <a|Sy?, is it a vector?

Last edited: Oct 26, 2012
2. Oct 26, 2012

### Chopin

Re: bra - ket??

Can you give a little more information about what you mean when you talk about real numbers? Bras and kets are members of a Hilbert space, so the concept of "real number" doesn't mean anything when talking about them.

3. Oct 26, 2012

### Jesssa

Re: bra - ket??

Do you mean that,

Sy|a> = a |a> where a is real?

<a| Sy = row vector times matrix = row vector

4. Oct 27, 2012

### rubertoda

Re: bra - ket??

yes, i mean row vector times matrix, i would like to show that Sy|a> = <a|Sy, if Sy is hermitian..

5. Oct 27, 2012

### Fredrik

Staff Emeritus
Re: bra - ket??

If |a> is a member of a Hilbert space H, then so is Sy|a>. But <a| is a member of the dual space H*, defined as the set of continuous linear functions from H into ℂ. <a|Sy is another member of H*, so it can't be equal to any member of H, like e.g. Sy|a>.

It would however make sense to ask if they have the same components, with respect to some basis {|i>} and its dual basis {<i|} (the latter being defined by <i|j>=δij). The ith component of <a|Sy is by definition <a|Sy|i>. (See this post for more about components of linear maps with respect to a basis). The ith component of Sy|a> is just its projection onto the ith basis vector, <i|Sy|a>.

6. Oct 27, 2012

### Jesssa

Re: bra - ket??

another way to think of it,

Sy|a> = <a|Sy

Sy|a> = matrix times column vector = column vector

<a|Sy = row vector times matrix = row vector

it is not possible that Sy|a> = <a|Sy

7. Oct 27, 2012

### Fredrik

Staff Emeritus
Re: bra - ket??

This argument is fine when the vector space is finite.