Raising and lowering operators of the Hamiltonian

In summary, the operators ##a## and ##a^{\dagger}## satisfy the commutation relation ##[a,a^{\dagger}] = 1## and we are asked to find the normalization of the state ##|\psi \rangle = C (a^{\dagger} )^2 |0\rangle##, where the vacuum state ##|0\rangle## is such that ##a|0\rangle = 0##. In order to do so, we first write out the commutator explicitly and then try various manipulations to simplify it, such as multiplying by ##(a^{\dagger})^2## on the left and ##|0\rangle## on the right. However,
  • #1
CAF123
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Homework Statement


a) The operators ##a## and ##a^{\dagger}## satisfy the commutation relation ##[a,a^{\dagger}] = 1##. Find the normalization of the state ##|\psi \rangle = C (a^{\dagger} )^2 |0\rangle##, where the vacuum state ##|0\rangle## is such that ##a|0\rangle = 0##

b)A one dimensional simple harmonic oscillator has the Hamiltonian $$\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}mw^2 \hat{x}^2.$$ The operator ##\hat{a}## is defined as $$\hat{a} = \left(\frac{mw}{2\hbar}\right)^{1/2}\hat{x} + \frac{i}{\sqrt{2mw\hbar}} \hat{p}$$

Suppose that the oscillator frequency is changed slightly to ##w \rightarrow \sqrt{1+\epsilon} w## What are the exact new energy eigenvalues? Now calculate the first order perturbation in the energy and compare with the exact answer.

Homework Equations


In a), perhaps that ##a|n\rangle = \sqrt{n}|n-1\rangle## and that ##a^{\dagger} |n\rangle = \sqrt{n+1} |n+1\rangle##

The Attempt at a Solution


In a), write out the commutator explicitly ##aa^{\dagger} - a^{\dagger}a = 1##. Now I tried various manipulations of this equation to get something that resembled the equation given. Multiply by ##(a^{\dagger})^2## on the left and ##|0\rangle## on the right would give ##(a^{\dagger})^2 a a^{\dagger} |0\rangle - (a^{\dagger})^3 a|0\rangle = (a^{\dagger})^2 |0\rangle.## But this simplifies to LHS=RHS, so it is of no use.

b)New energy eigenvalues would be ##E_n = \hbar (\sqrt{1+\epsilon} w) (n+1/2)## and the shift is ##E' = \langle n | \hat{H'} | n\rangle## where ##\hat{H'}## is the Hamiltonian above but with ##w## replaced with ##\sqrt{1+\epsilon}w## Do we treat ##\epsilon## as a constant here?

Many thanks.
 
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  • #2
For a), just compute the vector itself, a†|0> = √0+1 |1> = |1>, etc.
 
  • #3
dextercioby said:
For a), just compute the vector itself, a†|0> = √0+1 |1> = |1>, etc.
Applying a† again would give √2 |2>, not some constant premultiplying the state |0>. So I think somehow we have to incorporate the commutation relation given, which I tried to do in the OP.
 
  • #4
Try using ##\|C(a^\dagger)^2\vert 0 \rangle\|^2=1##.
 
  • #5


I would first commend the student for their attempt at solving the problems and for recognizing and utilizing relevant equations. I would then provide guidance on how to approach the problems and provide some additional information that may be helpful.

For part a), it may be helpful to use the fact that the operator ##a^{\dagger}## can be expressed in terms of the creation and annihilation operators for the harmonic oscillator, i.e. ##a^{\dagger} = \sqrt{2m\omega\hbar}\hat{a}^{\dagger}/\sqrt{\hbar}##. This will allow for the explicit calculation of the normalization constant ##C##.

For part b), it is important to note that the perturbation in the energy is given by the expectation value of the perturbation Hamiltonian, which is proportional to ##\epsilon##. Therefore, it is reasonable to treat ##\epsilon## as a small constant in this case. Additionally, it may be helpful to use the known expression for the perturbation in the energy of a harmonic oscillator when the frequency is changed slightly.

I would also suggest that the student review the concept of raising and lowering operators and their properties, as well as the calculation of energy eigenvalues and eigenstates for the harmonic oscillator.
 

Related to Raising and lowering operators of the Hamiltonian

What are raising and lowering operators of the Hamiltonian?

Raising and lowering operators are mathematical operators used in quantum mechanics to describe the behavior of particles in a quantum system. They are used to determine the energy levels and transitions of a system by acting on the wave function.

What is the significance of raising and lowering operators?

Raising and lowering operators are important because they allow us to calculate the energy spectrum of a quantum system and predict the probability of transitions between energy levels. They also provide a mathematical framework for understanding the behavior of particles in a quantum system.

How do raising and lowering operators work?

Raising and lowering operators work by acting on the wave function of a quantum system. The raising operator increases the energy of the system by one quantum level, while the lowering operator decreases the energy by one quantum level. By applying these operators repeatedly, we can determine the energy levels and transitions of the system.

What are the commutation relations for raising and lowering operators?

The commutation relations for raising and lowering operators are important because they determine the behavior of these operators when they are applied to a wave function. The commutation relation between the raising and lowering operator is given by [a,a†] = 1, where a is the lowering operator, a† is the raising operator, and [a,a†] is the commutator.

How are raising and lowering operators related to the Hamiltonian?

Raising and lowering operators are related to the Hamiltonian through the energy eigenvalue equation, which states that the Hamiltonian acting on the wave function is equal to the energy of the system times the wave function. By using raising and lowering operators, we can determine the energy levels and transitions of a quantum system, which are described by the Hamiltonian.

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