Kinetic energy in quantum mechanics

  • #1
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Main Question or Discussion Point

Given psi as function of x^2, and the potential energy as function of x, find the kinetic energy.

My reasoning:
KE=P^2/2m and use the momentum operator.

My professor's reasoning:
Calculate the hamiltonian operator and subtract the potential energy then divide by psi.

Note:
I talked to my professor about the part where he divided by psi since then kinetic energy will be a function of x which supposedly implies that momentum is a function of x which goes against the uncertainty principle. He didnt give me a statisfactory reply, and i feel that i might be having a concept hole.
 

Answers and Replies

  • #2
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  • #3
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You really need to have a look at Chapter 3 of Ballentine:
https://www.amazon.com/dp/9814578584/?tag=pfamazon01-20

What you did is correct - but why - that's the rub.

Thanks
Bill
So it is wrong to divide by psi... Phew life makes so much more sense now... Thank you you recommended Ballentine a couple of times for me now, and i will surely check this "holy grail of QM" :D thank you
 
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  • #4
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  • #5
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I went through the book, but its so much above my level, so could you please explain it to me (atleast intuitively)
Basically symmetry considerations mean the energy and momentum operator must be the form it is and a free particle must be as p^2/2m.

Thanks
Bill
 
  • #6
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Basically symmetry considerations mean the energy and momentum operator must be the form it is and a free particle must be as p^2/2m.

Thanks
Bill
I just need to make sure... Do we have to divide by psi? H(psi)=E(psi) so does E equal H(psi)/psi ?
 
  • #7
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Or even if KE(psi)=KE(psi) so KE=KE(psi)/psi ?
 
  • #8
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Its p^2/2m where P in the momentum operator - you had it correct from the start.

I don't know where this divide by psi comes from.

Thanks
Bill
 
  • #9
vanhees71
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I'm not sure, whether I understand the question you ask right. I guess, you study nonrelativistic quantum mechanics in the position representation, i.e., Schrödinger wave mechanics. For a particle moving in one spatial dimension, the usual Hamilton operator then looks
$$\hat{H}=-\frac{\hbar^2}{2m} \partial_x^2+V(x),$$
where ##V## is some potential (or 0 for free particles).

Now given a wave function, ##\psi(x)##, representing a state, i.e., a square-integrable function, which can be (by convention to make life easier) normalized to 1,
$$\int_{\mathbb{R}} \mathrm{d} x |\psi(x)|^2=1,$$
it doesn't make too much sense to ask about what the kinetic energy might be. One "translation" of this vague idea, which makes sense, to ask for the expectation value of the kinetic energy. The operator for kinetic energy is the first term in the Hamiltonian, i.e.,
$$\hat{T}=-\frac{\hbar}{2m} \partial_x^2.$$
The expectation value is then given by
$$\langle T \rangle_{\psi} = \int_{\mathbb{R}} \mathrm{d} x \psi^*(x) \hat{T} \psi(x)= \int_{\mathbb{R}} \mathrm{d} x \psi^*(x) \left [-\frac{\hbar}{2m} \partial_x^2 \psi \right ].$$
Nowhere you have to divide by the wave function. Why should one?

Enough of speculation: Do you have a completely formulated question by your professor to answer?
 
  • #10
atyy
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Note:
I talked to my professor about the part where he divided by psi since then kinetic energy will be a function of x which supposedly implies that momentum is a function of x which goes against the uncertainty principle. He didnt give me a statisfactory reply, and i feel that i might be having a concept hole.
Did he divide by psi or the "square" of psi? If the latter, then it is just a normalization, as vanhees71 is writing in post #9. Usually the wave function is normalized in the first step, so that dividing by the "square" of psi is just dividing by one. It is a matter of convention whether to use a normalized or unnormalized state.
 
  • #11
vanhees71
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Hm, but then you have to divide by the integral over the (modulus) squared wave function and not simply through the square...
 
  • #12
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kishki.jpg


The whole question was about finding the kinetic energy, the two solutions (mine and the professor's differed) i got very confused due to what he did.
 
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  • #13
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The dividing by psi is not valid. What you did is.

Thanks
Bill
 
  • #14
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The dividing by psi is not valid. What you did is.

Thanks
Bill
I will have to talk to my professor about that, so i have to argue against dividing by psi, but i just cant find the right argument as to why he is wrong. I could say that dividing by psi to find the kinetic energy makes the schrodinger equation meaningless as a statement of the conservation of energy? But i dont know why i feel that the whole question is meaningless when examined under the light of the uncertainty principle - how could one even determine the kinetic energy as a function of x?
 
  • #15
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Or mathematically speaking, you can not remove the function that the operator acts on, and since E and U are operators it is mathematically erroneous to divide by psi as my professor did?
 
  • #16
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Or mathematically speaking, you can not remove the function that the operator acts on, and since E and U are operators it is mathematically erroneous to divide by psi as my professor did?
Of course.

Think about it. You have converted an equation equating vectors to one equating operators. It makes no sense.

Thanks
Bill
 
  • #17
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Of course.

Think about it. You have converted an equation equating vectors to one equating operators. It makes no sense.

Thanks
Bill
Thank you very much.. I hope he is convinced, specially since this was a question in our mid term
 
  • #18
atyy
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Hm, but then you have to divide by the integral over the (modulus) squared wave function and not simply through the square...
Yes, I was just sketching what I thought the OP's professor might have been thinking of. I think the OP should write the exact question, otherwise it's hard to know what is being asked.
 
  • #19
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Yes, I was just sketching what I thought the OP's professor might have been thinking of. I think the OP should write the exact question, otherwise it's hard to know what is being asked.
The question was exactly as mentioned in the post, except it had the psi and potential energy as functions. All it asked was to find the kinetic energy in terms of h bar and mass. The function of psi and u was a little bit messy.. I actually have another question now, what does it even mean to calculate the kinetic energy in quantum mechanics?
 
  • #20
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It means p^2/2m where p is the momentum operator applied to the wavefunctions allowable from the Schroedinger equation.

Thanks
Bill
 

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