The classically forbidden region is the x values outside of where the turning points would be for a classical harmonic oscillator. For a classical harmonic oscillator the total energy (E) is composed of KE + PE. So if the KE = 0, then the PE is the same as the total E; it can never be any larger. So the "classically forbidden" region would be those positions (x values) where the PE is larger than the total E. Example: in the ground state (n=0) we have E = ½ hν. When x = √(hν/k), then the (classical) PE is ½ k x² = ½ k (hν/k) = ½ hν = E. So, classically, we would never find x > √(hν/k) or x < −√(hν/k). Those regions are forbidden for the classical harmonic oscillator. But the quantum mechanical harmonic oscillator CAN be found in those regions that would be forbidden for the classical harmonic oscillator. These are the outermost tails of the probability distribution. It is an interesting question to ask how often the quantum harmonic oscillator is found in regions where the classical harmonic oscillator could not access. In fact, this is probably the homework problem that led to your question.
Thank you for these lectures! Intuitively we think of the two atoms "vabrating" back and forth like two masses connected with a spring, continuously moving away then back together again. But when there are nodes, like at E1 (x=0), is the better interpretation that if the molecule were in energy level E1 and we tried to measure its bond length (somehow), we would never find its separation to be the resting bond length (x=0)? This is even weirder for particle in a box, does it mean that if we knew there were one particle in the box at a specific energy level with nodes at specific locations, that we would never measure the particle the particle at the node locations (near the locations...)? Thank you
Yes, that's exactly correct. It's difficult to break with our macroscopic intuition built from experience with balls and springs and boxes. But at the microscopic level, things don't behave the same way, and it can be very counterintuitive.
At some energy level, i, if we take the expected value of the potential energy (Int_x p_i(x)*(1/2)*k*x^2 dx), where p_i(x) is the square of the wave function at energy level i will we get E_i, or are we not counting the kinetic energy?
That would only be the average potential energy. You're right that it neglects the kinetic energy. To get that, you'd have to include the KE operator (-h²/8π²m d²/dx²) in the integral. You'd also have to be careful to sandwich it with the wavefunction, instead of combining them in your p(x) term.
@@markpwoodward Sure you could! That would be a perfectly fair homework problem at this point in the course. (Although assigning it in ASCII text would perhaps be cruel and unusual punishment.)
@@PhysicalChemistry I did it! I'm now a quantum mechanic!! Thank you for the encouragement. The Hermit polynomials were tricky. I used the relation "d/dx H_n(x) = 2n H_{n-1}(x)". The kinetic energy operator, making use of that relation, produces 5 terms. One term cancels with the potential energy. Using the Hermite recurrence relation presented in class, the other 4 terms can be merged into a constant times the wave function. The integral of the two wave functions is 1 and, after a lot of algebra, the constant becomes "(n+1/2) h/2pi sqrt(k/mu)", as we want. Super satisfying!
he's so calm. that's good. thanks for the video
I have question! Where are classical forbbiden and classical allowed areas for Qantum harmonic oscillator? What detemines it?
The classically forbidden region is the x values outside of where the turning points would be for a classical harmonic oscillator.
For a classical harmonic oscillator the total energy (E) is composed of KE + PE. So if the KE = 0, then the PE is the same as the total E; it can never be any larger.
So the "classically forbidden" region would be those positions (x values) where the PE is larger than the total E.
Example: in the ground state (n=0) we have E = ½ hν. When x = √(hν/k), then the (classical) PE is ½ k x² = ½ k (hν/k) = ½ hν = E. So, classically, we would never find x > √(hν/k) or x < −√(hν/k). Those regions are forbidden for the classical harmonic oscillator.
But the quantum mechanical harmonic oscillator CAN be found in those regions that would be forbidden for the classical harmonic oscillator. These are the outermost tails of the probability distribution.
It is an interesting question to ask how often the quantum harmonic oscillator is found in regions where the classical harmonic oscillator could not access. In fact, this is probably the homework problem that led to your question.
Thank you for these lectures! Intuitively we think of the two atoms "vabrating" back and forth like two masses connected with a spring, continuously moving away then back together again. But when there are nodes, like at E1 (x=0), is the better interpretation that if the molecule were in energy level E1 and we tried to measure its bond length (somehow), we would never find its separation to be the resting bond length (x=0)? This is even weirder for particle in a box, does it mean that if we knew there were one particle in the box at a specific energy level with nodes at specific locations, that we would never measure the particle the particle at the node locations (near the locations...)? Thank you
Yes, that's exactly correct.
It's difficult to break with our macroscopic intuition built from experience with balls and springs and boxes. But at the microscopic level, things don't behave the same way, and it can be very counterintuitive.
@@PhysicalChemistry Makes sense. Thank you. Thank you.
At some energy level, i, if we take the expected value of the potential energy (Int_x p_i(x)*(1/2)*k*x^2 dx), where p_i(x) is the square of the wave function at energy level i will we get E_i, or are we not counting the kinetic energy?
That would only be the average potential energy. You're right that it neglects the kinetic energy. To get that, you'd have to include the KE operator (-h²/8π²m d²/dx²) in the integral. You'd also have to be careful to sandwich it with the wavefunction, instead of combining them in your p(x) term.
@@PhysicalChemistry Got it, not that I could do it, but I conceptually. Thank you
@@markpwoodward Sure you could! That would be a perfectly fair homework problem at this point in the course. (Although assigning it in ASCII text would perhaps be cruel and unusual punishment.)
@@PhysicalChemistry I did it! I'm now a quantum mechanic!! Thank you for the encouragement.
The Hermit polynomials were tricky. I used the relation "d/dx H_n(x) = 2n H_{n-1}(x)". The kinetic energy operator, making use of that relation, produces 5 terms. One term cancels with the potential energy. Using the Hermite recurrence relation presented in class, the other 4 terms can be merged into a constant times the wave function. The integral of the two wave functions is 1 and, after a lot of algebra, the constant becomes "(n+1/2) h/2pi sqrt(k/mu)", as we want. Super satisfying!