# QM: Bra's and ket's

1. Jan 30, 2009

### Niles

1. The problem statement, all variables and given/known data
Hi all.

An operator is given by:

$$\widehat O = \sum\limits_i {\sum\limits_j {O_{ij} \left| {\psi _i } \right\rangle \left\langle {\psi _j } \right|} }$$

Am I allowed to move the bra and the ket around to achieve the following?

$$\widehat O = \sum\limits_i {\sum\limits_j {O_{ij} \left\langle {\psi _j } \right|\left| {\psi _i } \right\rangle } }.$$

Best regards,
Niles.

2. Jan 30, 2009

### Phrak

It looks a little odd. As long you know that
$$\left\langle {\psi _j } \right|\left| {\psi _i } \right\rangle$$
isn't an inner product.
Usually written
$$\widehat O = \left\langle {\psi _j } \right| {O_{ji} \left| {\psi _i } \right\rangle$$
or without indices
$$\widehat O = \left\langle {\psi} \right| O \left| {\psi} \right\rangle\ .$$

Last edited: Jan 30, 2009
3. Jan 30, 2009

### Niles

What do you mean by "contract"? And in general, is it allowed to interchange bra's and ket's as one wishes?

4. Jan 30, 2009

### Niles

Ok, I see that you changed your post. I thought that a bra and a ket gave an inner product? Why is it not like that in this case?

5. Jan 31, 2009

### Phrak

Yeah, bad terminology on my part, sorry.

I was a little uncertain, before, because I hadn't seen others use indexing on bras, kets, and operators before.

Without getting too technical, A ket is a column vector and a bra is a row vector in matrix notation.

In matrix notation, a row vector times column vector is a scalar.
Also, a column vector times a row vector is a matrix.

As soon as you add the subscripts, the notational meaning changes, and the associative properties of matrix multiplication no longer hold--because the elements are no longer maxrices. With the indices, the objects in question are really the elements of matrices, bras and kets.

For matrices, in general, $$\ AB \neq BA \$$, $$\ \ \ ABC \neq ACB \neq BAC \$$, etc.

$$\ C=AB \$$ in matrices is the same as $$\ C_{ik} \sum\limits_j {A_{ij} B_{jk} }$$

The columns of A are multiplied by the rows of B.

With indexing, the order doesn't matter, the indexing overrides the matrix multiplication rules. With indexing the result is the same, no matter the ordering. The summation takes care of that.

$$C_{ik} = \sum\limits_j {A_{ij} B_{jk} }$$

$$C_{ik} = \sum\limits_j {B_{jk} A_{ij} }$$

This holds, for matrices and kets.

$$\left| U \right> = A \left| V \right>$$

$$\left| U_{i} \right> = \sum\limits_j { A_{ij} \left| V_{j} \right> }$$

$$\left| U_{i} \right> = \sum\limits_j { \left| V_{j} \right> A_{ij} }$$

And matrices and bras.

$$\left< U \right| = \left< V \right| } A \$$

$$\left< U_{j} \right| = \sum\limits_i { \left< V_{i} \right| A_{ij} }$$

$$\left< U_{j} \right| = \sum\limits_i { A_{ij} \left< V_{i} \right| }$$

The inner product of a bra and a ket is a scalar.

$$S = \left< U \right| \left| V \right>$$

$$\left| N = \sum\limits_i { \left< U_{i} \right| \left| V_{i} \right> }$$

$$\left| N = \sum\limits_i { \left| V_{i} \right> \left< U_{i} \right|}$$

This last equation is not the same as $$\ W = \left| V \right> \left< U \right|$$ !

And finally the inner product of a ket and bra is a matrix of some sort--not a quantum mechanical operator--but a product space.

$$W = \left| U \right> \left< V \right| }$$

$$\left| W_{ij} = \left| U_{i} \right> \left< V_{j} \right|$$

$$\left| W_{ij} = \left< V_{j} \right| \left| U_{i} \right>$$

Hope this helps.

6. Jan 31, 2009

### Niles

I am a little uncertain of why the following is true:

$$\left| N = \sum\limits_i { \left< U_{i} \right| \left| V_{i} \right> } = \sum\limits_i { \left| V_{i} \right> \left< U_{i} \right|}.$$

The first part is the sum of some scalars, so N is a number. Then we sum over outer products, which you say is a scalar as well?

The rest is very good!

7. Jan 31, 2009

### brasidas

I strongly doubt whether you can do that...

Think about it purely in mathematical sense. Ket vector is a column vector (n x 1 vector) while bra vector is a row vector. (1 x n vector) Usually operators are n x n square matrix, so if you have a $$\left< U_{i} \right|\hat Q \left| V_{i} \right>$$, you end up with a scalar value while $$\left| V_{i} \right>| \hat Q |\left< U_{i} \right|$$ results in a matrix . The former is the inner product, and the latter is the outer product. They are different things, of course. In your case, you are given:

$$\left| V_{i} \right>\left< U_{i} \right|$$

which is, essentially a matrix, and your desired goal is:

$$\left< U_{i} \right|\left| V_{i} \right>$$

which is a scalar. So if you stick a test bra in front and test ket after the operator, you'd get a product of two inner products in the given situation, while the other one will give you a scalar multiple of the inner products between test vectors. They cannot be the same in general.

Also... if you have:

$$\left|U\right> = \sum\limits_i {A_{i} \left|U_{i}\right> }$$

Then...

$$\left<U\right| = (\left|U\right>)^+ = \sum\limits_i {A_{i}^*\left<U_{i}\right|}$$

Last edited: Jan 31, 2009
8. Jan 31, 2009

### Ben Niehoff

No, you are not allowed to do that, and I think some of the earlier replies gave you bad advice. You are never allowed to assume any bra or ket is an ordinary number. They must always be considered vectors in Hilbert space, and the notation is a way of denoting abstract matrix algebra, where multiplication is NOT commutative. The products

$$\langle \psi | \phi \rangle$$

and

$$| \psi \rangle \langle \phi |$$

give completely different objects! The first is a scalar, while the second is a matrix. They cannot be used interchangeably.

The expression

$$\hat O = \sum_{m,n} O_{mn} |m \rangle \langle n|$$

is just the expansion of O in some basis $|m \rangle$. When plugged into an inner product, it means this:

$$\langle \phi | \hat O | \psi \rangle = \sum_{m,n} \langle \phi | O_{mn} |m \rangle \langle n| \psi \rangle$$

$$= \sum_{m,n} O_{mn} \langle m | \phi \rangle^* \langle n| \psi \rangle$$

which is now finally the sum of products of ordinary scalars.

9. Jan 31, 2009

### Niles

Great, I guess that is the definitive answer to my question. Thanks for replying.

And thanks to all for participating. I really appreciate your help.

10. Jan 31, 2009

### Phrak

That was, of course, one of the points I was making. But you've missed the entire issue of the ambiguity that arises with an explicit summation, which nullifies the advantage of using bra ket notation in the first place.

Of course, doing this

$$\sum\limits_i { \left| v_{i} \right> \left< u_{i} \right|} \ ,$$

should be avoided. Either it results in a scalar, or it's meaningless.

11. Jan 31, 2009

### Hurkyl

Staff Emeritus
Huh? What's ambiguous? That looks like a perfectly meaningful expression to me -- the summand expresses a linear operator, a type of object for which addition (and thus finite summation) is well-defined. Even infinite-summation can be defined if you include a notion of convergence.

12. Jan 31, 2009

### Phrak

Which is it then? Do you want to operate the ket on the bra and then take the summation, or the other way around?

13. Jan 31, 2009

### Hurkyl

Staff Emeritus
The one that is usually specified by such an expression. If the u's and v's represent real numbers, there is no ambiguity in $\sum_i v_i u_i$, is there? The semantics here is exactly the same: for each i, you multiply the ket and bra to get a linear operator, and then sum the resulting operators.

I don't think any other reading could make sense, because both the ket and bra involve the index of summation....

14. Jan 31, 2009

### Phrak

Yes, yes. I should have said confusing rather than ambiguous. So what's your answer to Niles' question?

15. Jan 31, 2009

### Hurkyl

Staff Emeritus
My answer would have been "no", since {bra}*{ket} is (AFAIK) always used to represent the 'inner product', so the two expressions are of different types, so they couldn't possibly be equal, even by chance. I've never seen {bra}*{ket} used to represent a linear functional like you suggested, but that is yet another type of object.

Such a thing is quite awkward to write, but I would expect some other notation to be used. If someone insisted on keeping the notation as similar as possible, I would expect something like $\langle \psi_i \mid \_\_ \mid \psi_j \rangle$ with a underscore or dash to emphasize that there is a place to plug in something, rather than just expressing a product of the bra with the ket.

Ah, now that I think of it, the notation I would expect to actually see is the trace. What you express as the result of plugging an operator T into the expression $\langle \psi_i \mid \mid \psi_j \rangle$, I would expect to see as $\mathrm{Tr}\left( \mid \psi_j \rangle \langle \psi_i \mid T \right)$.

Last edited: Jan 31, 2009
16. Jan 31, 2009

### brasidas

... That's not to be avoided...

It's called "Projection Operator", with

$$\sum\limits_n { \left| n \right> \left< n \right|} \ = \hat 1,$$

where $\{ \left | n \right > | n = 1, 2, 3... \}$ represents orthonormal basis, and $\hat 1$ represents an identity matrix...

Last edited: Jan 31, 2009
17. Feb 1, 2009

### philip041

I know this is bad banter; don't use apostrophes to pluralise!!!

18. Oct 16, 2010

### WallOfShame

is there always a "bra" for a "ket"? Can u plz justify?
thanking you.