Why Does xp ≠ i * hbar? | Wolfram Alpha

  • Thread starter leonmate
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In summary: This is an important connection in physics, because it allows us to calculate things like the probability of finding a particle in a particular region.
  • #1
leonmate
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xp - px = x * (-i*hbar*d/dx) - (-i * hbar * d/dx (x)) = i * hbar

I can see that the px = i * hbar
But, why does xp = 0 ... ? I just get - i * hbar everytime I try, Wolfram Alpha agrees...

Can anyone show me why this isn't the case please?
 
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  • #2
leonmate said:
I can see that the px = i * hbar
Really? Show how you got that.
 
  • #3
The annoying thing about commutators is that, because they are operators, when you do it out explicitly you have to apply them to a function to make them work out, so apply [x,p] to f and see what happens.
 
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  • #4
Ok,

What I've been doing is just d/dx ( const * x ) = x
That works when we apply p to x. But not x to p I suppose.
 
  • #5
leonmate said:
Ok,

What I've been doing is just d/dx ( const * x ) = x
That works when we apply p to x. But not x to p I suppose.
These are operator relationships. To calculate what you call "px", you need to evaluate:
$$
\left( -i \hbar \frac{\partial}{\partial x} \right) x \psi$$
Note that you apply x first, then the derivative. So you'll need to use the product rule.
 
  • #6
Bit aside, how arises I in p and x?
 
  • #7
mac_alleb said:
Bit aside, how arises I in p and x?
It shows up when you apply the product rule for derivatives:
$$(\hat p\hat x f)(x)=(\hat p(\hat x f))(x) = -i(\hat x f)'(x) =-i\frac{d}{dx}(\hat x f(x)) =-i\frac{d}{dx}(xf(x))=\cdots.$$
 
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  • #8
mac_alleb said:
Bit aside, how arises I in p and x?

Have you heard about a Poisson bracket?
 
  • #9
Hi,

Roughtly speaking, the derivative operator "p" acts over everything to the right and the wave function it's always implicitly the last term, so you really have:

pxΨ= i * hbar ∂x(xΨ) = i * hbar * [x * ∂x(Ψ) + Ψ * ∂x(x) ]

Where Ψ is the wave function in the evaluated point, implicitly, Ψ(x).

x(x) = 1 so the conmutation is [x,p]Ψ = (xp - px)Ψ = i*h_bar Ψ.
Sergio
 
  • #11
Derivative is Ok, but what for I ? Moreover, how imaginary valued are connected with physical?
 
  • #12
mac_alleb said:
Derivative is Ok, but what for I ?
What do you get when you use the product rule to evaluate the last expression in post #7?

mac_alleb said:
Moreover, how imaginary valued are connected with physical?
There are many such connections. For example, ##|\psi(x)|^2\Delta x## is approximately equal to the probability that a particle detector near ##x## that's covering a region of size ##\Delta x## will detect a particle.
 
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Related to Why Does xp ≠ i * hbar? | Wolfram Alpha

1. Why does xp ≠ i * hbar?

The equation xp ≠ i * hbar is a fundamental concept in quantum mechanics known as the Heisenberg uncertainty principle. This principle states that it is impossible to know the precise position and momentum of a particle at the same time. This is because the more accurately we measure the position of a particle, the less accurately we can measure its momentum, and vice versa. Therefore, the product of the uncertainty in position (x) and momentum (p) cannot be equal to a constant, in this case, i * hbar.

2. What does xp ≠ i * hbar mean?

The equation xp ≠ i * hbar means that the measurement of a particle's position (x) and momentum (p) cannot both be known with absolute certainty. This is a fundamental principle in quantum mechanics and has been experimentally verified numerous times.

3. How does this equation relate to the Heisenberg uncertainty principle?

The equation xp ≠ i * hbar directly represents the Heisenberg uncertainty principle, which states that the product of the uncertainty in position and momentum of a particle cannot be less than or equal to a constant value, in this case, i * hbar. This principle is a fundamental aspect of quantum mechanics and has significant implications for our understanding of the behavior of particles at the subatomic level.

4. Can you provide an example to illustrate why xp ≠ i * hbar?

An example to illustrate the Heisenberg uncertainty principle and why xp ≠ i * hbar is the double-slit experiment. In this experiment, a beam of particles, such as photons, is directed towards a barrier with two narrow slits. As the particles pass through the slits, they create an interference pattern on a screen behind the barrier. This pattern can only be explained by the wave-like properties of particles, such as their momentum. However, when we try to measure the exact position of a particle, we disrupt its momentum and the interference pattern disappears. This shows that it is impossible to know both the position and momentum of a particle simultaneously, confirming the Heisenberg uncertainty principle.

5. Why is xp ≠ i * hbar important in quantum mechanics?

The equation xp ≠ i * hbar is crucial in quantum mechanics because it helps us understand the fundamental nature of particles at the subatomic level. It also has practical applications in fields such as quantum computing and cryptography. The Heisenberg uncertainty principle, represented by this equation, is one of the key principles that distinguish quantum mechanics from classical mechanics and has revolutionized our understanding of the universe.

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