The Student Room Group

Quantum mechanics -harmonic oscillators

Say we have a 1D harmonic oscillator with potential V=12mωc2x2V = \frac{1}{2} m \omega_c^{2} x^2 (classical frequency). Let the wavefunction of a particle be described by ψ0=Aexp(bx2)\psi_0 = A \exp(-bx^2).

How do I find the probability of finding the particle outside of the 'classically allowed region'? I figured the classical limit is basically the amplitude a of the particle's SHM, so I integrate the wavefunction squared from a to infinite (and multiply by 2 for symmetry). Not quite sure how to find a, though.

Thanks for any help !
Original post by trm90
Say we have a 1D harmonic oscillator with potential V=12mωc2x2V = \frac{1}{2} m \omega_c^{2} x^2 (classical frequency). Let the wavefunction of a particle be described by ψ0=Aexp(bx2)\psi_0 = A \exp(-bx^2).

How do I find the probability of finding the particle outside of the 'classically allowed region'? I figured the classical limit is basically the amplitude a of the particle's SHM, so I integrate the wavefunction squared from a to infinite (and multiply by 2 for symmetry). Not quite sure how to find a, though.

Thanks for any help !


Find A using the normalisation condition (probability of particle existing somewhere between -infty and +infty sums to 1), i.e.

<ψ0ψ0>=1 <\psi_0|\psi_0^{*}> = 1

Otherwise I think the method you suggest is correct!
Reply 2
Original post by Prime Suspect
Find A using the normalisation condition (probability of particle existing somewhere between -infty and +infty sums to 1), i.e.

<ψ0ψ0>=1 <\psi_0|\psi_0^{*}> = 1

Otherwise I think the method you suggest is correct!


Oh, so the wavefunction amplitude is pretty much equivalent to the classical amplitude? In that case that would make sense. Thanks very muc h!
Original post by trm90
Oh, so the wavefunction amplitude is pretty much equivalent to the classical amplitude? In that case that would make sense. Thanks very muc h!


Ah sorry I didn't read your first post properly, got the a's confused - thought you wanted to find A not a...

Unfortunately its not true that A = a; instead I think you want to say that the expected quantum value for energy (the average value, or observed value) should equal the observed energy of the classical harmonic oscillator, i.e.

Eav=<ψ0H^ψ0> E_{av} = <\psi_0|\hat{H}|\psi_0^{*}>

The classical harmonic oscillator has a constant total energy, which is given by

Etot=mwc2a22 E_{tot} = \dfrac{mw_c^2a^2}{2}

hence I think this total energy can be equated to the expected energy of the quantum system. Therefore I think that the equation

mwc2a22=<ψ0H^ψ0> \dfrac{mw_c^2a^2}{2} = <\psi_0|\hat{H}|\psi_0^{*}> should hold. From this I think you should be able to determine a and hence solve the problem... I hope!
(edited 13 years ago)
Reply 4
Original post by Prime Suspect
Ah sorry I didn't read your first post properly, got the a's confused - thought you wanted to find A not a...

Unfortunately its not true that A = a; instead I think you want to say that the expected quantum value for energy (the average value, or observed value) should equal the observed energy of the classical harmonic oscillator, i.e.

Eav=<ψ0H^ψ0> E_{av} = <\psi_0|\hat{H}|\psi_0^{*}>

The classical harmonic oscillator has a constant total energy, which is given by

Etot=mwc2a22 E_{tot} = \dfrac{mw_c^2a^2}{2}

hence I think this total energy can be equated to the expected energy of the quantum system. Therefore I think that the equation

mwc2a22=<ψ0H^ψ0> \dfrac{mw_c^2a^2}{2} = <\psi_0|\hat{H}|\psi_0^{*}> should hold. From this I think you should be able to determine a and hence solve the problem... I hope!


Thanks very much, and the method seems more than viable. I'm going to give this a go on the train home and will let you know if I work it out later :-)
Reply 5
Original post by Prime Suspect
Ah sorry I didn't read your first post properly, got the a's confused - thought you wanted to find A not a...

Unfortunately its not true that A = a; instead I think you want to say that the expected quantum value for energy (the average value, or observed value) should equal the observed energy of the classical harmonic oscillator, i.e.

Eav=<ψ0H^ψ0> E_{av} = <\psi_0|\hat{H}|\psi_0^{*}>

The classical harmonic oscillator has a constant total energy, which is given by

Etot=mwc2a22 E_{tot} = \dfrac{mw_c^2a^2}{2}

hence I think this total energy can be equated to the expected energy of the quantum system. Therefore I think that the equation

mwc2a22=<ψ0H^ψ0> \dfrac{mw_c^2a^2}{2} = <\psi_0|\hat{H}|\psi_0^{*}> should hold. From this I think you should be able to determine a and hence solve the problem... I hope!

EDIT: Argh, messed things up again!

If my Hamiltonian operator is of the form h22md2dx2+V(x)\dfrac{-h'^2}{2m} \dfrac{d^2}{dx^2} + V(x) then I'm not sure how I'm supposed to about any of the integrals, as I'll get two integrals with x^2 exp(...x^2) terms and I'm not told how to evaluate those (my lecturer specifically said that all standard integrals will be provided and that we won't have to spend even 10 seconds trying to evaluate them, and only the answer to an integral of the form exp(-2bx^2) is provided).
(edited 12 years ago)
Original post by trm90
EDIT: Argh, messed things up again!

If my Hamiltonian operator is of the form h22md2dx2+V(x)\dfrac{-h'^2}{2m} \dfrac{d^2}{dx^2} + V(x) then I'm not sure how I'm supposed to about any of the integrals, as I'll get two integrals with x^2 exp(...x^2) terms and I'm not told how to evaluate those (my lecturer specifically said that all standard integrals will be provided and that we won't have to spend even 10 seconds trying to evaluate them, and only the answer to an integral of the form exp(-2bx^2) is provided).


Use integration by parts and use that

xe2bx2dx=14be2bx2 \displaystyle \int xe^{-2bx^2} dx = \dfrac{-1}{4b}e^{-2bx^2}

(presence of the derivative)

I think that should work?

EDIT

Perhaps my method isn't correct then if you're not supposed to need to evaluate any integrals? Maybe instead you're supposed to consider energies, as for any energy Emwc2a22 E \neq \dfrac{mw_c^2a^2}{2} the quantum oscillator is in a region that's classically forbidden... not sure how to use that info though!
(edited 12 years ago)
Reply 7
Original post by trm90
EDIT: Argh, messed things up again!

If my Hamiltonian operator is of the form h22md2dx2+V(x)\dfrac{-h'^2}{2m} \dfrac{d^2}{dx^2} + V(x) then I'm not sure how I'm supposed to about any of the integrals, as I'll get two integrals with x^2 exp(...x^2) terms and I'm not told how to evaluate those (my lecturer specifically said that all standard integrals will be provided and that we won't have to spend even 10 seconds trying to evaluate them, and only the answer to an integral of the form exp(-2bx^2) is provided).


Surely you can use the fact that ψ0> |\psi_0> is an eigenstate of the Hamiltonian so you don't need to do any integrals.
Reply 8
EDIT: Okay, I evaluated the Hamiltonian and got:

E=h2bm4b2x2+12mωc2x2E = \dfrac{h'^2 b}{m} - 4b^2 x^2 + \dfrac{1}{2}m \omega_c^{2} x^2

I equated this with the energy 12mωc2x2\dfrac{1}{2} m \omega_c^{2} x^2 and got:

x=h24mbx = \sqrt{\dfrac{h'^2}{4mb}}

where b = mω2h\dfrac{m \omega}{2h'}

does that make sense?
(edited 12 years ago)
Original post by trm90
EDIT: Okay, I evaluated the Hamiltonian and got:

E=h2bm4b2x2+12mωc2x2E = \dfrac{h'^2 b}{m} - 4b^2 x^2 + \dfrac{1}{2}m \omega_c^{2} x^2

I equated this with the energy 12mωc2x2\dfrac{1}{2} m \omega_c^{2} x^2 and got:

x=h24mbx = \sqrt{\dfrac{h'^2}{4mb}}

where b = mω2h\dfrac{m \omega}{2h'}

does that make sense?


Sunelir is right - you know the ground state energy of the quantum oscillator is given by

<ψ0H^ψ0>=hˉwc2 <\psi_0|\hat{H}|\psi_0> = \dfrac{\bar{h}w_c}{2}

because ψ0 \psi_0 is an eigenstate of the hamiltonian this must hold. So you equate this to the classical energy to find a - sorry for making you evaluate the integrals, I am definitely a bit rusty on the old quantum mechanics!
Reply 10
Original post by Prime Suspect
Sunelir is right - you know the ground state energy of the quantum oscillator is given by

<ψ0H^ψ0>=hˉwc2 <\psi_0|\hat{H}|\psi_0> = \dfrac{\bar{h}w_c}{2}

because ψ0 \psi_0 is an eigenstate of the hamiltonian this must hold. So you equate this to the classical energy to find a - sorry for making you evaluate the integrals, I am definitely a bit rusty on the old quantum mechanics!


I see... this is a silly question, but I was supposed to know that psi was the ground state wavefunction right :o: ?

But I think I can take it from here, thanks guys!
Reply 11
Original post by trm90
I see... this is a silly question, but I was supposed to know that psi was the ground state wavefunction right :o: ?


Well the label on psi was 0 and not n.
Original post by trm90
I see... this is a silly question, but I was supposed to know that psi was the ground state wavefunction right :o: ?

But I think I can take it from here, thanks guys!


Yeah you can safely assume it is given the 0 subscript... I totally missed this at first as well!

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