# Operators acting on bras

1. Aug 8, 2012

### dEdt

In Quantum Mechanics, we have linear operators which can act on a ket to produce a new ket. However, we also allow the same operators to act on a bra vector to produce a new bra vector. That is, if $\langle\phi|$ is a bra and A is an operator, the action of A on $\langle\phi|$ is to produce a new bra denoted by $\langle\phi|A$. Furthermore, we demand that
$$\left(\langle\phi|A\right)|\psi\rangle=\langle\phi|\left(A|\psi\rangle\right)$$
for all kets $|\psi\rangle$.

This is how the action of an operator on a bra vector was (roughly) described in Dirac's Principles, as well as in other texts that I've seen. Next, Dirac asserts that this "uniquely determines" $\langle\phi|A$.

I was trying to prove, or at least justify this claim, but to no avail. Nor have I seen a proof anywhere else. Can anyone help?

2. Aug 8, 2012

### bhobba

The uniqueness follows from the assumed one to one correspondence between bras and kets. A acting on a ket produces another ket that when acted on by a bra defines a linear function on that ket (the one the linear operator produces). But you can also consider it a linear function of the original ket before it was acted on by the operator so defines another bra which is defined as the bra that results from A acting on the bra.

The thing to realize however in Dirac's presentation is it untrue. It is true, that the Bras and Kets can be put into one to one correspondence for a Hilbert space but untrue for what is called a Rigged Hilbert Space. However it is only Rigged Hilbert spaces that allow the introduction of the Dirac Delta function used liberally by Dirac.

Eventually its probably a good idea to learn about Rigged Hilbert Spaces, and Ballentine gives a good introduction, but for now simply think of the Dirac Delta function as an actual function that for all practical purposes is a Dirac Delta function but strictly speaking it isn't. This is similar to what you see sometimes in applied math where dx is considered a quantity of first order smallness rather than what it actually is a differential form.

Thanks
Bill

3. Aug 8, 2012

### dextercioby

This is technically not correct. It's NOT the same operator, but a shifted operator from the original Hilbert space into the topological dual of a dense subset of it. It's called the dual operator and acts linearly on linear functionals on the original Hilbert space producing other linear functionals.

4. Aug 8, 2012

### tom.stoer

... and for rigged Hilbert spaces you may run into big trouble ...

be careful with Dirac's notation: suppose you have the position operator x, the momentum operator p and momentum eigenstates |k>; now let's calculate

<k'|[x,p]|k> = <k'|xp-px|k> = <k'|xk-k'y|k> = (k-k') <k'|x-x|k> = 0

and

<k'|[x,p]|k> = <k'|i|k> = i δ(k'-k) ≠ 0

5. Aug 8, 2012

### dEdt

Sorry, but what does this mean?

6. Aug 8, 2012

### jmcelve

Let me give this a shot! (I'm stealing pretty much all of this from Ballentine - even the notation - since I'm reading most of it myself at the moment. This is my way of practicing to see if I understand it.)

For a given n-dimensional vector space $V$, a linear functional $F(\phi)$ in the corresponding dual space $V'$ is defined by a unique vector $f$ by the inner product $F(\phi) = (f, \phi)$ with $\phi$ being any arbitrary vector in $V$. For any set of orthonormal basis vectors $\{ \phi_n \}$ for $V$, we can uniquely write any vector in $V$ by expanding it in terms of its basis vectors: $\psi = \sum_n x_n \phi_n$, with $F(\psi) = \sum_n x_n F(\phi_n)$. To show the uniqueness of any $F(\psi)$, we first construct a vector $f = \sum_n [F(\phi_n)] ^* \phi_n$. When we take the inner product $(f, \psi)$, we obtain $\sum_n x_n F(\phi_n) = F(\psi)$. Thus, $F(\psi)$ is uniquely defined for each $\psi$.

7. Aug 8, 2012

### bhobba

Yea that's true - in what I said it defines a new operator acting on the dual but is in fact a different operator.

Thanks
Bill

8. Aug 8, 2012

### bhobba

The uniqueness of the operator thus defined acting on the space of duals (technically that's what bras are) - it would have been better for me to say the uniqueness and linearity of the operator thus defined which is the real issue ie a linear operator acting on Kets also defines a unique linear operator acting on Bra's - but your question was in terms of uniqueness.

Thanks
Bill

Last edited: Aug 8, 2012