Momentum operator in "open" space

In summary, the conversation discusses the definition and properties of the operator P on the Hilbert space H=L^2([0,1],dx). It is shown that while P is Hermitian, it is not self-adjoint. However, by enlarging its domain to include functions that satisfy a certain condition, P can be made self-adjoint. The question then asks about the applicability of this procedure to H=L^2((-\infty,\infty),dx) or "open" space. The response clarifies that the requirement for functions to be absolutely continuous on any finite interval and belong to the space L^2(\mathbb{R}) ensures that the boundary terms can be eliminated when integrating by parts
  • #1
ShayanJ
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Consider the Hilbert space [itex]H=L^2([0,1],dx) [/itex]. Now we define the operator [itex] P=\frac \hbar i \frac{d}{dx} [/itex] on this Hilbert space with the domain of definition [itex] D(P)=\{ \psi \in H | \psi' \in H \ and \ \psi(0)=0=\psi(1) \} [/itex].
Then it can be shown that [itex] P^\dagger=\frac \hbar i \frac{d}{dx} [/itex] with [itex] D(P^\dagger)=\{ \varphi \in H | \varphi ' \in H \}[/itex].
So the operator P is Hermitian but no self-adjoint. But its possible to make P self-adjoint by enlarging its domain to the following:
[itex]
D(P)=\{ \psi \in H | \psi ' \in H \ and \ \psi(0)=e^{i \alpha} \psi(1) \} \ \ \ \ \ (\alpha \in \mathbb R)
[/itex].
My question is, how does this procedure work if the Hilbert space is [itex] H=L^2((-\infty,\infty),dx)?[/itex](In "open" space, which is of course not in its mathematically rigorous meaning.)
Thanks
 
  • #3
In <open space>, there's no restriction on the domain of the derivative operator other than the norm of the image be finite: ##D(P)=D(P^{\dagger})=\{φ∈H|φ′∈H\}##
 
  • #4
dextercioby said:
In <open space>, there's no restriction on the domain of the derivative operator other than the norm of the image be finite: ##D(P)=D(P^{\dagger})=\{φ∈H|φ′∈H\}##

I don't think that works. How should we kill the boundary terms when we do the integration by parts then?
 
  • #5
That ## \phi(-\infty) = 0## and ## \phi(\infty) = 0## follows from the requirements that the functions in the maximal domain of P be absolutely continuous on any finite interval of ##\mathbb{R}## and moreover, as I said, both ##\phi(x)## and ##\phi'(x)## belong to ##\mathcal{L}^2(\mathbb{R})##. This is firstly stated on page 106 of Akhiezer and Glazman's <Theory of Linear Operators in Hilbert Space> and later on page 111. Alternatively, check out page 18 of http://arxiv.org/abs/quant-ph/0103153v1.
 
  • #5
for your question. The procedure for making the momentum operator self-adjoint in an open space with Hilbert space H=L^2((-\infty,\infty),dx) is similar to the one described in the given content. First, we define the operator P=\frac \hbar i \frac{d}{dx} on H with the domain of definition D(P)=\{ \psi \in H | \psi' \in H \}. Then, we show that P^\dagger=\frac \hbar i \frac{d}{dx} with D(P^\dagger)=\{ \varphi \in H | \varphi' \in H \}. However, as you mentioned, this operator is not self-adjoint.

To make P self-adjoint, we need to enlarge its domain of definition to include functions that satisfy a certain boundary condition. In this case, the boundary condition is given by \psi(\infty)=e^{i \alpha} \psi(-\infty). This means that the function must approach the same value at both positive and negative infinity, but with a phase shift of e^{i \alpha}. This condition ensures that the operator is Hermitian and has a self-adjoint extension.

To summarize, the procedure for making the momentum operator self-adjoint in an open space with Hilbert space H=L^2((-\infty,\infty),dx) involves defining the operator with a certain domain of definition, showing that it is Hermitian but not self-adjoint, and then extending the domain to include functions that satisfy a specific boundary condition. This is done to ensure that the operator is self-adjoint and can be used in calculations and analysis in the given Hilbert space.
 

Related to Momentum operator in "open" space

1. What is the momentum operator in "open" space?

The momentum operator in "open" space is a mathematical operator used in quantum mechanics to describe the momentum of a particle in an unbounded or infinite space. It is represented by the symbol p and is defined as the operator that operates on a wavefunction to give the momentum of the particle.

2. How is the momentum operator related to the position operator?

The momentum operator and the position operator are related through the Heisenberg uncertainty principle. The position operator is represented by the symbol x and is defined as the operator that operates on a wavefunction to give the position of the particle. The two operators do not commute, meaning that their order of operation affects the outcome. This is mathematically represented as [x,p]=iħ, where i is the imaginary unit and ħ is the reduced Planck's constant.

3. What is the difference between the momentum operator in "open" space and in "closed" space?

The momentum operator in "open" space is defined for particles in an unbounded or infinite space, while the momentum operator in "closed" space is defined for particles in a bounded or finite space. In "closed" space, the momentum operator is limited by the boundaries of the space, whereas in "open" space, the momentum can have any value.

4. How is the momentum operator used in quantum mechanics?

In quantum mechanics, the momentum operator is used to calculate the expectation value of the momentum of a particle. It is also used to calculate the momentum of a particle at a specific point in time or to calculate the change in momentum over time.

5. Can the momentum operator be used for systems with multiple particles?

Yes, the momentum operator can be extended to systems with multiple particles by adding the momentum operators of each individual particle. This allows for the calculation of the total momentum of the system as well as the momentum of each individual particle.

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