Harmonic Oscillator Negative Energy(Quantum)

In summary, the state u(x) = Bxe^[(x^2)/2] is an energy eigenstate of the system with V(x) = 1/2*K*X^2.
  • #1
Relativeee
3
0

Homework Statement


Ok so the question is, is the state u(x) = Bxe^[(x^2)/2] an energy eigenstate of the system with V(x) = 1/2*K*X^2 and what is the probability per unit length of this state.

Homework Equations


The Attempt at a Solution

So the way i did this was, to find if the state is an energy eigenstate of the system i checked to see if it satisfied the time independent Schrodinger equation. After all of the Calculations , came up with, 3- 2E/hw = 0, and the definition of E is E = (n+1/2)hw, so the only way for this solution to be satisfied is if the energy is negative, -2 to be exact. So with this I am pretty sure that this cannot be an energy eigenstate of the system since the harmonic oscillator has to have a positive energy(unless there is a specific exception i do not know about) and therefore we cannot find the probability density because the wave-function is not normalizable(blows up at inifinity) and that's what is bugging me, is it as simple as that or am i missing something, like for instance is there a way to normalize this wave-function so as to find the probability density?
 
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  • #2
Relativeee said:

The Attempt at a Solution

So the way i did this was, to find if the state is an energy eigenstate of the system i checked to see if it satisfied the time independent Schrodinger equation. After all of the Calculations , came up with, 3- 2E/hw = 0, and the definition of E is E = (n+1/2)hw, so the only way for this solution to be satisfied is if the energy is negative, -2 to be exact.
No it isn't. Try that again.
 
  • #3
Relativeee said:
After all of the Calculations , came up with, 3- 2E/hw = 0, and the definition of E is E = (n+1/2)hw, so the only way for this solution to be satisfied is if the energy is negative, -2 to be exact.
That's not very exact. If 3 - 2 E/hbar*ω = 0 and you solve for E, what do you get? Hint: Move -2 E/hbar*ω over to the right side and change its sign.
 
  • #4
Oh my goodness i apologize, i meant to put 3 + 2E/hw
 
  • #5
So, given your definition of E, n must equal to -2. However, n has to be 0 or positive integer.
Does that mean this is the energy eigenstate of the system?
 
  • #6
Relativeee said:
Oh my goodness i apologize, i meant to put 3 + 2E/hw
OK, in that case, we can perhaps help you if you show the work that led you to that expression.
 

What is a harmonic oscillator?

A harmonic oscillator is a physical system that exhibits a repetitive pattern of motion around a stable equilibrium point. It is typically characterized by a restoring force that is proportional to the displacement from the equilibrium point.

What is negative energy in the context of a harmonic oscillator in quantum mechanics?

In quantum mechanics, negative energy refers to the energy levels below the ground state of a system. In the case of a harmonic oscillator, these negative energy levels represent a state of energy lower than the lowest possible energy state, known as the zero-point energy.

How does a harmonic oscillator with negative energy behave?

A harmonic oscillator with negative energy behaves differently than one with positive energy. It exhibits oscillatory motion around the equilibrium point, but with a decreasing amplitude as the energy decreases. This is because the system is constantly losing energy due to quantum mechanical effects.

What is the significance of negative energy in quantum mechanics?

Negative energy plays a crucial role in quantum mechanics as it allows for an accurate description of physical phenomena such as the Casimir effect and the Lamb shift. It also has implications for the stability of matter and the behavior of particles at the quantum level.

Can negative energy be observed in the real world?

While negative energy is a concept used in quantum mechanics, it has not yet been directly observed in the real world. However, its effects have been observed through experimental studies of quantum systems, providing strong evidence for its existence.

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