- #1

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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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- Thread starter leonmate
- Start date

- #1

- 84

- 1

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?

- #2

Doc Al

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Really? Show how you got that.I can see that the px = i * hbar

- #3

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- #4

- 84

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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

Doc Al

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These are

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.

$$

\left( -i \hbar \frac{\partial}{\partial x} \right) x \psi$$

Note that you apply x first,

- #6

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Bit aside, how arises I in p and x?

- #7

Fredrik

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It shows up when you apply the product rule for derivatives:Bit aside, how arises I in p and x?

$$(\hat p\hat x f)(x)=(\hat p(\hat x f))(x) = -i(\hat x f)'(x) =-i\frac{d}{dx}(\hat x f(x)) =-i\frac{d}{dx}(xf(x))=\cdots.$$

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- #8

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Bit aside, how arises I in p and x?

Have you heard about a Poisson bracket?

- #9

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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

pxΨ= i * hbar ∂

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

∂

Best regards,

Sergio

- #10

bhobba

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Have you heard about a Poisson bracket?

To the OP. Just on that point you may find the following helpful:

http://bolvan.ph.utexas.edu/~vadim/classes/2013s/brackets.pdf

Thanks

Bill

- #11

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Derivative is Ok, but what for I ? Moreover, how imaginary valued are connected with physical?

- #12

Fredrik

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What do you get when you use the product rule to evaluate the last expression in post #7?Derivative is Ok, but what for I ?

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.Moreover, how imaginary valued are connected with physical?

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