9
$\begingroup$

I'm studying QM from Shankar. He introduces linear operators and says that an operator is an instruction for transforming one ket into another. But then a few lines below he says operators can also act on bras.

So does the complete specification of an operator include its action on bras? Or does its action on the kets determine its action on the bras?

$\endgroup$
  • 1
    $\begingroup$ Do you want something like this? Also check the Hermitian conjugate operator section, I guess the equations there will fully answer your question. $\endgroup$ – acarturk Sep 13 at 9:38
  • $\begingroup$ No I've read that. Can you point to where clarification might be needed in my question? $\endgroup$ – peterwright Sep 13 at 10:00
  • $\begingroup$ Let A be an operator. Suppose A sends ket V to ket W. I thought it might send bra V to bra W, but this is not the case (the adjoint operator does this). So I'm wondering what A sends bra V to. $\endgroup$ – peterwright Sep 13 at 10:02
  • 13
    $\begingroup$ This question now appears in the HNQ – expect many visitors who (like me) understand virtually nothing of neither question nor answers, but clicked because they were bemused by the notion of linear operators (whatever exactly those are) having a particular effect on brassieres. $\endgroup$ – Janus Bahs Jacquet Sep 14 at 8:50
  • 1
    $\begingroup$ @JanusBahsJacquet See Bra–ket notation, aka Dirac notation. $\endgroup$ – PM 2Ring Sep 15 at 20:55
10
$\begingroup$

If $A:V\to V$ is a linear operator on a vector space $V$ with basis $|m\rangle$, Dirac defined the action $\langle n|A$ of $A$ on the bra vector $\langle n|$ (an element of the dual space $V^*$)as a new bra vector that is evaluated on a ket $|m\rangle$ to give the same answer as the evaluation of $\langle n|$ on $A|m\rangle$. In other words he set $$ (\langle n|A)|m\rangle = \langle n|(A|m\rangle). $$ Since it does not matter where one puts the brackets Dirac just writes $$ \langle n|A|m\rangle $$ for both things. Note that the dagger operation $\dagger: |m\rangle\mapsto \langle m|= (|m\rangle)^\dagger$ that provides an antilinear map between $V$ and $V^*$ (and so defines the inner product on $V$) is not needed here. So $A$ acting on bras has nothing to do with the hermitian conjugate $A^\dagger$ which needs the inner product. As $A$ acting to the left on bra vectors is really an operator $V^*\to V^*$ rather than an operator $V\to V$ it usual in linear algebra to regard it as different operator $A^*:V^*\to V^*$ that is called the "conjugate," or the "transpose." The latter name is probably best as no complex conjugation is involved, and in the dual basis $A^*$ is representated by the transpose (but not the complex conjugate transpose) of the matrix representing $A$.

Note added: I just saw Qmechanic's answer. What he says: $A|\phi\rangle = |\chi\rangle \Leftrightarrow \langle \phi|A^\dagger =\langle \chi|$ is quite correct, but you really don't need to use the dagger to define left action of $A$.

$\endgroup$
  • 1
    $\begingroup$ To hammer home your last sentence: even if you have an inner product, not only do you not need to use it to define $\langle \phi | A$, but in fact it's extremely unnatural to do so. $\endgroup$ – tparker Sep 13 at 13:11
  • $\begingroup$ Hm, it looks to me that you can either specify the matrix elements of the operator or specify the dagger operation in combination with its action on the kets. Then the action of the operator is fixed and defined. Only the action on the ket's alone is not sufficient to decide what the action of $\langle v| \hat O$ is. Given only $ \hat O |v\rangle$ does not suffice to determine $\langle v|\hat O$, or am i wrong ? $\endgroup$ – Hans Wurst Sep 14 at 11:32
5
$\begingroup$
  1. Using Shankar's notation from his eq. (1.6.14), an operator acting on a bra$^1$ $$\langle \phi | ~=~ \langle \psi |\hat{A} $$ is equivalent to the Hermitian adjoint operator acting the corresponding ket $$|\phi \rangle ~=~ \hat{A}^{\dagger}|\psi \rangle . $$

  2. The Hermitian adjoint operator fulfills the relation $$ \langle w | \hat{A}v \rangle~=~\langle \hat{A}^{\dagger}w | v \rangle. $$

--

$^{1}$We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.

$\endgroup$
  • 2
    $\begingroup$ They're equivalent statements, but definitely not the same statement. Also, this just pushes back the question to the completely parallel question "How does the adjoint of a linear operator act on a ket?", which isn't manifestly obvious from the definition either. $\endgroup$ – tparker Sep 13 at 12:37
  • 2
    $\begingroup$ The OP was asking for a definition, not a statement of equivalence to a more complicated quantity whose own existence requires much more mathematical structure. This does not answer their question. $\endgroup$ – tparker Sep 13 at 13:15
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Sep 13 at 13:54
3
$\begingroup$

There is a standard definition for the action of a linear operator on a bra, although many authors (including Shankar) sloppily skip defining it. $\langle \phi| A$ is defined to be the linear functional such that $\big( \langle \phi| A \big) |\psi\rangle \equiv \langle \phi| \big(A |\psi\rangle\big)$. In other worlds, you first act the linear operator on the ket that's getting "eaten" and then act the bra $\langle \phi |$ on the output of that.

Similarly, we can define the action of the adjoint of a linear operator on a ket. $$A^\dagger | \psi \rangle$$ is defined by how an arbitrary bra $\langle \phi |$ acts on it: $$\langle \phi | \big( A^\dagger | \psi \rangle \big) := \big( \langle \phi | A^\dagger \big) | \psi \rangle = \langle A \phi | \psi \rangle.$$ Note that unlike the former case, this requires an inner product structure so that we can define the adjoint.

The inner product of a ket $A^\dagger | \psi \rangle$ with all possible kets $|\phi\rangle$ completely specifies the ket, because the positive-definiteness of the inner product implies that it's nondegenerate. It's not often appreciated that the adjoint of an operator only has a unique action on kets because of the positive-definiteness requirement on the inner product.

$\endgroup$
2
$\begingroup$

The actual statement holds in a much more general setting.

Let $\Phi:V\to V$ be any mapping, and let $f:V\to K$ be a function on $V$. Then we get a new function $\Phi^\ast f:V\to K$ defined by $v\mapsto f(\Phi(v))$.

In your case, $V$ is the space of kets, and $\Phi$ is a linear operator on it. A linear map $f:V\to\mathbb C$ is a bra. (Let's stay in the finite dimensional case to not have to worry about continuity and so.) Since $\Phi$ is linear, it is not hard to see that if $f$ is linear, then so is $\Phi^\ast f$. That is all there really is about how $\Phi$ acts on bra's.

If you take this further, then you see how the matrix of $\Phi$ on kets transforms when expressed in the dual basis of bra's. This is elaborated in the other answers.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.