^, the operators in quantum mecħanics

In summary, the conversation discusses the doubt about the validity of the equation pΨ = (ħ/i) dΨ/dx and its derivation from Schrodinger's equation. The question also addresses an issue regarding the operation of p on a wavefunction that satisfies the Schrodinger's equation, and the uncertainty principle. In the end, the conversation concludes that the value for the momentum operator was not derived from the Schrodinger's equation, but rather from the classical case using the fact that it is the generator of infinitesimal translations.
  • #1
shanu_bhaiya
64
0
The doubt:
It's not a problem, but a doubt. We know that in general quantum physics at undergraduate level, we write pΨ = (ħ/i) dΨ/dx. My doubt is that if we derived this equation from Schrodinger's equation only, so we must operate p on a wave-function only, which satisfies Schrodinger's equation.

But I went further, and I saw a problem in the book - "Concepts of Modern Physics - A. Beiser", Chapter 5, Problem 9. It was asked to find the value of <xp>-<px>. Now to solve the problem, p is operated on xΨ (for <px>) and Ψ (for <xp>) both simultaneously. But I can prove that if Ψ satisfies Schrodinger's equation, xΨ cannot. So how can we operate p on xΨ, when it doesn't satisfy the Schrodinger's equation.
Or may be p can operate on anything, then we need a proof, which I may have not yet studied, is it true that there exists such a proof?

Please someone help.
 
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  • #2
Good question: [itex] \hat{p} [/itex] is meant to operate on any wavefunction with suitable properties (such as being square integrable and differentiable), but being a solution to the S.E. is not one of those properties! Otherwise it would even be hard to define [itex] \hat{p}^2 [/itex] acting on the wavefunction, since [itex] \hat{p}\psi [/itex] is also, in general, no longer a solution to the S.E.
 
  • #3
Thanks a lot for reply.
borgwal said:
Good question: [itex] \hat{p} [/itex] is meant to operate on any wavefunction with suitable properties (such as being square integrable and differentiable), but being a solution to the S.E. is not one of those properties!
But we derived the meaning of operator p from SE:

Ψ = Aexp((i/ħ)xp)
Differentiate w.r.t. x
dΨ/dx = ((i/ħ )p)Aexp((i/ħ)xp)
put value of Ψ from first equation
dΨ/dx = ((i/ħ )p)Ψ
pΨ = (ħ/i) (dΨ/dx)


Now, 1st equation is nothing but SE, as it's differential form is SE after eliminating A.

borgwal said:
Otherwise it would even be hard to define [itex] \hat{p}^2 [/itex] acting on the wavefunction, since [itex] \hat{p}\psi [/itex] is also, in general, no longer a solution to the S.E.
Actually, if Ψ is a solution, dΨ/dx is also a solution of SE as you can check. Further, [itex] \hat{p}^2 [/itex] can still be derived as follows:

From the last equation above
pΨ = (ħ/i) (dΨ/dx)
Differentiaite it again w.r.t. x
p(dΨ/dx) = (ħ/i) (d2Ψ/dx2)
Replace the value of dΨ/dx from the equation of p operator
p(pΨ/(ħ/i)) = (ħ/i) (d2Ψ/dx2)
p2Ψ = (ħ/i)2 (d2Ψ/dx2)


And hence we can get the value of [itex] \hat{p}^2 [/itex]. Waiting for your reply...
 
  • #4
No, [itex] \hat{p}\psi [/itex] is *not* a solution for the S.E. with a spatially dependent potential.

The meaning of the operator d/dx can be made plausible by looking at a particular wavefunction exp(ikx), but that doesn't mean it is derived from there. In fact, what one requires is the commutation relation [itex] [\hat{x},\hat{p}]=i\hbar [/itex], and from that relation it follows what [itex] \hat{p} [/itex] is.

In any case, you seem to think that the S.E. does not contain a potential...but it does!
 
  • #5
shanu_bhaiya said:
But we derived the meaning of operator p from SE:
No, that's not true at all. In fact, the momentum operator used in the SE can be different in different circumstances (see, for example, charged particle in a magnetic field).

The QM momentum operator is "derived" by extension of the classical case, using the fact that momentum is the generator of infinitesimal translations.

See:
1. Goldstein, Classical Dynamics
2. Sakurai, QM
 
  • #6
Thanks to both of you for clearing the doubt.
 
  • #7
Let’s make it simple. If I am right, your question is pointing toward something which says,
“will a wave function, that is satisfying the SE (say Ψ) will remain a perfect wave
function when multiplied by something say ‘x’, and thus making it ‘xΨ’ ”? This question is
however a complete analog of something which says, “if we have a continuous,
differentiable and single valued function f(x), will it remain pursuing these properties
even if is multiplied by some other function say g(x)? ”
K(x) = f(x) * g(x)
In case of our analogy, it depends on the continuity and differentiability of g(x), if it is,
then K(x) would be, or otherwise. Fine, but in case of our wave function the thing goes
quite differently, there are so many propert ies like normalization, single valued etc etc etc,
that a function has to satisfy in order to be a perfect wave function . ‘p’, however should
only to be applied to a perfect wave func tion (as it also have to satisfy SE, which is only
possible for a perfect wave function).
Now, taking equation <px>–<xp> which is nothing but the uncertainty principle can be
written as
<ħ (∂x/∂x)/i> – <xħ(∂/∂x)/i>
=> <ħ/i> – <0>
=> <ħ/i> = ħ/i
This is a completely safe, in the sen se that we don’t need to operate the whole equation
on to a wave function kind of stuff (Ψ). Which is simply because the expectation values
of <px> or of <xp> yield no operator, and it ’s operator which operates.
<px>Ψ or <xp>Ψ is wrong, and so as ‘p*x Ψ’.
I think, perhaps you’ll get it now.
 
  • #8
String_man said:
“will a wave function, that is satisfying the SE (say Ψ) will remain a perfect wave
function when multiplied by something say ‘x’, and thus making it ‘xΨ’ ”?
It is sure that xΨ won't obey SE, you can confirm it by putting in the SE. You'll get some extra terms apart from SE.

String_man said:
Now, taking equation <px>–<xp> which is nothing but the uncertainty principle can be
written as
<ħ (∂x/∂x)/i> – <xħ(∂/∂x)/i>
=> <ħ/i> – <0>
=> <ħ/i> = ħ/i
This is a completely safe, in the sense that we don’t need to operate the whole equation
on to a wave function kind of stuff (Ψ). Which is simply because the expectation values
of <px> or of <xp> yield no operator, and it ’s operator which operates.
<px>Ψ or <xp>Ψ is wrong, and so as ‘p*x Ψ’.
I think, perhaps you’ll get it now.

Well, whatever... the main point was that the value for p-operator wasn't derived from SE, as Gokul43201 explained that p-operator has been derived by extension of the classical case using the fact that it is the generator of the infiniteimal translations. Although I need more insight in this.
 
  • #9
For any general linear operator A, to say that it is the "generator" of some group action G, means this:

[tex]e^A = G[/tex]

where the operator exponential is defined by the infinite series

[tex]e^A = 1 + A + \frac{1}{2} A^2 + \frac{1}{3!} A^3 + ...[/tex]

if you put the derivative operator in place of A in the above series, you can check that it is, indeed, the generator of translations.

Note: The factor [itex]-i \hbar[/itex] is used to make the momentum operator Hermitian (such that its inverse is its conjugate transpose). In this case, the infinitesimal translation operator is

[tex]T \psi(x) = \psi(x - \frac{i}{\hbar} \epsilon)[/tex]

rather than

[tex]T \psi(x) = \psi(x - \epsilon)[/tex]

This also makes the translation operator Hermitian. See this page: http://en.wikipedia.org/wiki/Momentum_operator
 
Last edited:

What is the ^ operator in quantum mechanics?

The ^ (hat or caret) operator in quantum mechanics is known as the position operator. It is used to describe the position of a particle in space and is represented as a vector with three components (x, y, z). It is a fundamental operator used in many equations and calculations in quantum mechanics.

How is the ^ operator used in quantum mechanics?

The ^ operator is used to calculate the expectation value of a particle's position in space. It is also used to describe the wave function of a particle and is a key element in the Schrödinger equation, which is used to determine the evolution of a quantum system over time.

What are the properties of the ^ operator in quantum mechanics?

The ^ operator in quantum mechanics is Hermitian, meaning that its eigenvalues are real and its eigenvectors are orthogonal. It is also unitary, meaning that its adjoint is equal to its inverse. These properties make it a powerful tool in calculating the behavior of quantum systems.

Can the ^ operator be used to measure the position of a particle in quantum mechanics?

No, the ^ operator is a mathematical representation of a particle's position and cannot be used to directly measure it. In quantum mechanics, the act of measurement can change the state of a system, so the ^ operator is used to calculate the probability of finding a particle at a certain position.

Are there other operators in quantum mechanics similar to the ^ operator?

Yes, there are other operators in quantum mechanics that represent different physical properties of particles, such as momentum, spin, and energy. These operators also follow similar mathematical rules and are integral to describing the behavior of quantum systems.

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