Position Operator Action on Wave Function: $\psi(x)$

[davidge]

Would the action of the position operator on a wave function ##\psi(x)## look like this?

$$\psi(x) \ =\ <x|\psi>$$ $${\bf \hat x}<x|\psi>$$

Question 2: the position operator can act only on the wave function?

 

[hilbert2]

A vector ##|\psi>## is an abstract state and the ##<x|\psi >## is a wave function. An operator like ##\mathbf{x}## acts on the state ##|\psi >##, and the wavefunction of the state that has been acted on by the operator is ##<x|\mathbf{x}|\psi >##.

 

[davidge]

Thanks Hilbert2. So we have the freedom to move the operator ##\hat x## from ##\hat x <x|\psi>## to ##<x|\hat x|\psi>##?

 

[hilbert2]

If you're talking about the abstract ##\mathbf{x}## operator, it can only act on a vector ##|\psi >##, not on the wavefunction ##<x|\psi >##. But if you mean the position space representation of the operator ##\mathbf{x}##, then it is something that acts on the wavefunction, not an abstract state.

 

[davidge]

Ah, ok.
What is the motivation for defining such a operator? In Quantum Mechanics one can define a time-evolution operator which gives you a state vector at a later time, one can define a rotation operator which gives you a rotated state vector, etc... What about the position operator?

 

[hilbert2]

The matrix element ##<\psi |\mathbf{x}|\psi >## gives the expectation value of the position of a particle that is in state ##|\psi >##. That's one motivation. The wavefunctions describing eigenstates of ##\mathbf{x}## are not proper functions like those of a momentum operator or hamiltonian operators, which makes them a bit difficult to handle.

 

[davidge]

I see
In what does the position operator differs from the translation operator?

 

[PeroK]

I see
In what does the position operator differs from the translation operator?

Where and how are you learning QM? Your questions are seemingly random.

 

[davidge]

Where and how are you learning QM?

Most from Sakurai's book and McIntyre (on introductory QM)

Your questions are seemingly random.

It's just that one question leads me to another

 

[PeroK]

Most from Sakurai's book and McIntyre (on introductory QM)

It's just that one question leads me to another

How much linear algebra do you know? Sakurai, in my opinion, assumes a good grasp of the relevant undergraduate maths - especially linear algebra.

 

[PeroK]

I see
In what does the position operator differs from the translation operator?

I have the "revised" edition of Sakurai. Page 42 has the relevant section on "Position Eigenkets and Position Measurements". Page 44 describes "Translation" and the Translation operator.

 

[hilbert2]

I see
In what does the position operator differs from the translation operator?

There's not much in common, because the translation operator is not hermitian. The momentum operator is the generator of translation, and I guess that the position operator is a generator of a galilean frame change, which is a translation in momentum space.

 

[davidge]

How much linear algebra do you know?

Well, I attended Linear Algebra classes last semester at university. So I would say I know the basics of it.

I have the "revised" edition of Sakurai. Page 42 has the relevant section on "Position Eigenkets and Position Measurements". Page 44 describes "Translation" and the Translation operator.

I think this is not avaiable in older editions of the book.

There's not much in common, because the translation operator is not hermitian. The momentum operator is the generator of translation, and I guess that the position operator is a generator of a galilean frame change, which is a translation in momentum space.

How could we see this? Can you give me an example where we can see that the position operator is the generator of a galilean frame change?

 

[hilbert2]

How could we see this? Can you give me an example where we can see that the position operator is the generator of a galilean frame change?

If you have a momentum eigenstate ##\psi (x)=e^{ipx/\hbar}## and multiply it with ##e^{i(\Delta p) x/\hbar}##, you get a state with a momentum that's increased by ##\Delta p##. Just like operation with ##e^{i(\Delta x) \mathbf{p}/\hbar}## where ##\mathbf{p}## is the position space momentum operator is equivalent to space translation by ##\Delta x##.

 

[davidge]

If you have a momentum eigenstate ##\psi (x)=e^{ipx/\hbar}## and multiply it with ##e^{i(\Delta p) x/\hbar}##, you get a state with a momentum that's increased by ##\Delta p##. Just like operation with ##e^{i(\Delta x) \mathbf{p}/\hbar}## where ##\mathbf{p}## is the position space momentum operator is equivalent to space translation by ##\Delta x##.

Thank you