# Proof that all operators are linear

1. Feb 17, 2013

### dEdt

"Proof" that all operators are linear

I've "proven" that all operators acting on a Hilbert space are linear. Obviously this isn't true, so there must be a fault in my reasoning somewhere. I having trouble finding it though, and would appreciate input by someone who can.

Let $|\psi\rangle = \alpha |a\rangle+\beta|b\rangle$, $A$ be an arbitrary operator, and $|\phi\rangle$ be an arbitrary vector.
$$\langle\phi|A|\psi\rangle =(\langle\phi|A)|\psi\rangle=(\langle\phi|A) (\alpha |a\rangle+\beta|b\rangle)=\alpha (\langle\phi|A)|a\rangle+\beta(\langle\phi|A) |b\rangle= \alpha \langle\phi|(A|a\rangle)+\beta\langle\phi|(A |b\rangle)= \langle\phi| (\alpha A|a\rangle +\beta A|b\rangle).$$

So
$$\langle \phi |(A|\psi \rangle)=\langle\phi| (\alpha A|a\rangle +\beta A|b\rangle)$$
for all $\langle \phi |$. Therefore
$$A(\alpha |a\rangle+\beta|b\rangle)= \alpha A|a\rangle +\beta A|b\rangle.$$

2. Feb 17, 2013

### micromass

Staff Emeritus
Re: "Proof" that all operators are linear

Please define what you mean with operator.
Also, please define what you mean with $<\varphi|A|\psi>$, $<\varphi|A$and $A|\psi>$ and why they are linear.
And please explain why $(<\varphi|A|)|\psi>=<\varphi|(|A|\psi>)$.

Last edited: Feb 17, 2013
3. Feb 17, 2013

### dEdt

Re: "Proof" that all operators are linear

An operator is a map from the Hilbert space to itself.
$A|\psi\rangle$ is the vector produced by acting the operator $A$ on the element of the Hilbert space $|\psi\rangle$.

The action of the same operator on a dual vector (or bra vector) is denoted $\langle \phi| A$. It is defined so that $(\langle \phi| A)|\psi \rangle=\langle \psi| (A|\psi \rangle)$ for any bra and ket. This number is thus denoted $\langle \psi| A |\psi \rangle$.

4. Feb 17, 2013

### Fredrik

Staff Emeritus
Re: "Proof" that all operators are linear

At the third equality, you're using that $\langle\phi|A$ is linear. But that "product" is defined by $\left(\langle\phi|A\right)|\chi\rangle =\langle\phi|\left(A|\chi\rangle\right)$ for all $|\chi\rangle$ in the Hilbert space, and I don't see a reason to think that this makes $\langle\phi|A$ linear.

$\langle\phi|A$ is clearly linear if both $\langle\phi|$ and $A$ are, but we haven't assumed that A is linear.

5. Feb 17, 2013

### dEdt

Re: "Proof" that all operators are linear

The third equality holds by the linearity of the inner product, no?
$$\langle \chi | \psi \rangle = \langle \chi|( \alpha |a\rangle + \beta |b\rangle)=\alpha \langle \chi| a\rangle + \beta \langle \chi|b\rangle$$
Now just replace $\langle \chi|$ with $\langle \phi|A$.

6. Feb 17, 2013

### M@2

Re: "Proof" that all operators are linear

Try to consider A2

7. Feb 17, 2013

### Fredrik

Staff Emeritus
Re: "Proof" that all operators are linear

That's exactly what we can't do, because we haven't proved that $\langle\phi|A$ is a bra, i.e. that it's a bounded linear map from the Hilbert space into ℂ.

What you proved in post #1 is that for all $A:\mathcal H\to\mathcal H$, if $\left\langle\phi\right|A\in \mathcal H^*$, then $A$ is linear.

Last edited: Feb 17, 2013
8. Feb 18, 2013

### The_Duck

Re: "Proof" that all operators are linear

The inner product between a bra and a ket is linear because a bra is (defined to be) a linear map from the space H of kets to the complex numbers. $\langle \phi | A$ defines a map from H to the complex numbers but it is not necessarily a linear map unless A is linear. Accordingly the bra-ket notation becomes quite misleading as $(\langle \phi | A)$ is not a bra, even though it looks like one.

We are (or at least I am) accustomed to think of a bra $\langle \phi |$ as a row vector, while kets are column vectors. This is valid because any linear map from a vector space to the complex numbers can be written as a row vector where the linear map is implemented by multiplying the row vector by the column vector we are mapping. However, if A is not linear then the map from H to the complex numbers given by $(\langle \phi | A)$ cannot be thought of as a row vector. There is no row vector that implements this map from H to the complex numbers.

9. Feb 18, 2013

### stevendaryl

Staff Emeritus
Re: "Proof" that all operators are linear

I don't think that bra-ket notation can be used for nonlinear operators.

10. Feb 20, 2013

### Morgoth

Re: "Proof" that all operators are linear

Just to point out a mistake. A cannot act on <Χ|... it's A* that acts on <X|.
In your analysis you chose to say that
A=A* (or in other words that your operator is Hermitian)
However Hermitian operators are LINEAR.

Last edited: Feb 20, 2013
11. Feb 20, 2013

### Fredrik

Staff Emeritus
Re: "Proof" that all operators are linear

Any operator can "act" on a bra in the sense that we can define <X|A to be the (not necessarily linear) functional such that (<X|A)|Y> = <X|(A|Y>). A* is not fundamentally different from A. They're both operators.

12. Feb 20, 2013

### Morgoth

Re: "Proof" that all operators are linear

When A*=A, we are talking about hermitian operators.
If you really want A to act on BRAs you have to have the A* acting on KETs.
that is because the direct product of bra and ket is a complex number:
<g|f>=<f|g>*
that is the reason if A acts on the kets you get the A[del] acting on bras, or vice versa.
The action of the same operator on both sides holds true only for hermitian operatos (which are linear).
In fact he proved that hermitian operators are linear....

((P.S. SOMEONE INFORM ME HOW TO USE BRA-KET IN LATEX XD))

Last edited: Feb 20, 2013
13. Feb 20, 2013

### Fredrik

Staff Emeritus
Re: "Proof" that all operators are linear

Right, but no one mentioned A* above, did they?

You don't have to have an A* at all. The OP considered an arbitrary map from a Hilbert space into the same Hilbert space, and A* is usually only defined for linear A. I don't think it can be defined for arbitrary A.

Start by clicking the quote button next to one of the posts above, e.g. my post #4, to see how we're doing it. The codes you need to know are \langle and \rangle. See this FAQ post for general information about LaTeX at PF.