A doubt regarding position representation of product of operators

[Kashmir]

We've two operators $\hat{a}$,$\hat{b}$. I know their position representation $\langle r|\hat{b} \mid \psi\rangle=b$
$\langle r|\hat{a}| \psi\rangle=a$

Is it generally true that the position representation of the combined operator $\hat{a}\hat{b}$ is $a b$ where $a, b$ are the individual representations of the operators ?

 

[PeroK]

We've two operators $\hat{a}$,$\hat{b}$. I know their position representation $\langle r|\hat{b} \mid \psi\rangle=b$
$\langle r|\hat{a}| \psi\rangle=a$

Is it generally true that the position representation of the combined operator $\hat{a}\hat{b}$ is $a b$ where $a, b$ are the individual representations of the operators ?

It doesn't look right, does it? Is there a counterexample?

 

[PeterDonis]

We've two operators $\hat{a}$,$\hat{b}$. I know their position representation $\langle r|\hat{b} \mid \psi\rangle=b$
$\langle r|\hat{a}| \psi\rangle=a$

What sort of mathematical objects are $a$ and $b$? Are they numbers? Functions? Something else?

 

[dextercioby]

Of course, this also begs the question "what is the combined operator"?

 

[Kashmir]

I'm sorry. I used wrong notation. My question is :

If two operators $\hat{a},\hat{b}$ have their position representations as $a, b$ ( both $a, b$ are operators but here they act on wavefunctions rather than kets) then is it true in general that the operator $\hat{a}\hat{b}$ , which is a product of the two operators has its position representation as the product of the position representation of the individual operators i.e is the position representation of $\hat{a}\hat{b}$ given as $a b$ ?

 

[PeterDonis]

If two operators $\hat{a},\hat{b}$ have their position representations as $a, b$ ( both $a, b$ are operators but here they act on wavefunctions rather than kets) then is it true in general that the operator $\hat{a}\hat{b}$ , which is a product of the two operators has its position representation as the product of the position representation of the individual operators i.e is the position representation of $\hat{a}\hat{b}$ given as $a b$ ?

You still haven't answered the question I asked in post #3. You should. Answering it should indicate to you that the question you are asking here is not even well-defined (because it's based on an implicit assumption that is not valid). Can you see why (i.e., what the invalid assumption is)?

 

[Kashmir]

What sort of mathematical objects are $a$ and $b$? Are they numbers? Functions? Something else?

They are objects that can act on a wavefunction $\psi(x)$, for example it can be $-i \hbar \frac{\partial}{\partial x}$, or just simply a multiplication by $x$.

 

[vanhees71]

That's a bit of a dilemma of notation. On the one hand you have the abstract operators in the abstract Hilbert space and on the other hand the same operators acting on "wave functions" in a given representation, i.e., wrt. a complete orthonormal basis of compatible observables.

E.g., for the momentum operator for motion in one dimension you have
$$\hat{p} \psi(x)=-\mathrm{i} \hbar \partial_x \psi(x)=\langle x|\hat{p}|\psi \rangle.$$
On the left-hand side it's the operator acting on the wave function, i.e., the position representation of the state vector. On the right-most side of the equation, it's the operator acting in the abstract Hilbert space.

 

[PeroK]

They are objects that can act on a wavefunction $\psi(x)$, for example it can be $-i \hbar \frac{\partial}{\partial x}$, or just simply a multiplication by $x$.

To put your question in mathematical language. We have a correspendence between states and position-space wave-functions:$$|\psi \rangle \leftrightarrow \psi$$where$$\forall x: \ \langle x|\psi \rangle = \psi(x)$$This implies a correspondence between operators on abstract states $|\psi \rangle$ and operators on square-integrable functions $\psi$:$$\hat A \leftrightarrow \bar A$$where:$$\forall x, \psi: \ \langle x|(\hat A\psi) \rangle = (\bar A \psi)(x)$$What you are asking is whether this correspondence preserves operator composition? I.e. do we have:$$\forall \hat A, \hat B: \ \hat A \hat B \leftrightarrow \bar A \bar B$$Can you prove that as a exercise in linear algebra?

 

[Kashmir]

To put your question in mathematical language. We have a correspendence between states and position-space wave-functions:$$|\psi \rangle \leftrightarrow \psi$$where$$\forall x: \ \langle x|\psi \rangle = \psi(x)$$This implies a correspondence between operators on abstract states $|\psi \rangle$ and operators on square-integrable functions $\psi$:$$\hat A \leftrightarrow \bar A$$where:$$\forall x, \psi: \ \langle x|(\hat A\psi) \rangle = (\bar A \psi)(x)$$What you are asking is whether this correspondence preserves operator composition? I.e. do we have:$$\forall \hat A, \hat B: \ \hat A \hat B \leftrightarrow \bar A \bar B$$Can you prove that as a exercise in linear algebra?

No. Any hints please?

 

[Kashmir]

I think I've done it. I'll post my attempt

 

[Kashmir]

Let the wavefunction of $|\psi\rangle$ be $\psi (x)$

Let $a, b$ be the position representation of $A, B$.

Then we've $b \psi (x)= \langle x | B|\psi\rangle$.

Let $f(x)=\langle x | B|\psi\rangle$ be wavefunction of $|f\rangle$

Then we've $a f (x)= \langle x | A|f\rangle$

Then $\begin{aligned} a b \psi(x) &=a\langle x|\hat{B}| \psi\rangle=a f(x) \\ &=\langle x|\hat{A}| f\rangle \\=\langle x|\hat{A}| \hat{B} \mid \psi\rangle \end{aligned}$

 

[PeroK]

No. Any hints please?

These things often fall out if you have the discipline to work through the definitions. The trick is not to wave your arms and say this step is obvious, but use what you know to be true.

Let's take two operators and their composition:$$\hat C = \hat A \hat B$$What we want to show is that$$\bar C = \bar A \bar B$$By definition of $\bar C$ we have$$\forall x, \psi: \langle x |\hat A \hat B|\psi \rangle = \langle x |\hat C|\psi \rangle = (\bar C \psi)(x)$$And what we need to show is that: $$\forall x, \psi: (\bar C \psi)(x) = (\bar A \bar B \psi)(x)$$A good idea now is to let $$|\phi \rangle = \hat B | \psi \rangle$$So that:$$\forall x: \phi(x) = \langle x |\phi \rangle = \langle x | \hat B | \psi \rangle = (\bar B \psi)(x)$$Which gives us the functional equation $$\phi = \bar B \psi$$And, again by definition we have:$$\forall x, \psi: \langle x|\hat A |\phi \rangle = (\bar A \phi)(x)$$Putting these last two equations together we have:
$$\forall x, \psi: \langle x|\hat A \hat B| \psi \rangle = \langle x|\hat A |\phi \rangle = (\bar A \phi)(x) = (\bar A \bar B \psi)(x)$$And that's it.

 

[PeroK]

Let the wavefunction of $|\psi\rangle$ be $\psi (x)$

Let $a, b$ be the position representation of $A, B$.

Then we've $b \psi (x)= \langle x | B|\psi\rangle$.

Let $f(x)=\langle x | B|\psi\rangle$ be wavefunction of $|f\rangle$

Then we've $a f (x)= \langle x | A|f\rangle$

Then $\begin{aligned} a b \psi(x) &=a\langle x|\hat{B}| \psi\rangle=a f(x) \\ &=\langle x|\hat{A}| f\rangle \\=\langle x|\hat{A}| \hat{B} \mid \psi\rangle \end{aligned}$

This is the right idea, but it lacks the discipline of a formal proof! It takes practice to develop formal proof-writing techniques.

 

[Kashmir]

This is the right idea, but it lacks the discipline of a formal proof! It takes practice to develop formal proof-writing techniques.

Thank you. I will work harder.

 

[PeterDonis]

They are objects that can act on a wavefunction $\psi(x)$, for example it can be $-i \hbar \frac{\partial}{\partial x}$, or just simply a multiplication by $x$.

So what does the "product" of two such objects mean? For example, say the two objects are $- i \hbar \partial_x$ and $- i \hbar \partial_y$. How would you multiply them?

(I see that @PeroK already gave an answer to the above, but was that the answer you had in mind in the OP of this thread?)

 

[Kashmir]

So what does the "product" of two such objects mean? For example, say the two objects are $- i \hbar \partial_x$ and $- i \hbar \partial_y$. How would you multiply them?

(I see that @PeroK already gave an answer to the above, but was that the answer you had in mind in the OP of this thread?)

Their multiplication means that I would apply them consecutively. I know that I should have written it as perok wrote but I wasn't able to gather it in mathematical language.

Yes what PeroK wrote is what I meant.

 

[PeterDonis]

what PeroK wrote is what I meant.

Ok, good.

 

[vanhees71]

In math that's a quite general notion: The composition of two operators acting in some way on "objects" often follows rules similar to a (usually not commutative) product. For linear operators mapping a vector space into itself (automorphisms) they form together with the addition a ring. For QT even more important us the fact that self-adjoint operators with the commurators (multiplied by $\mathrm{i}$) form a Lie algebra. These "Lie brackets" in addition build also a derivative structure. In this way the self-adjoint operators define one-parameter Lie groups, e.g., the Hamiltonian generates "time translations" etc.