Finding <x> using raising and lowering operators and orthonormality

  • #1
milkism
118
15
Homework Statement
Finding P(x, t>0) and <x>.
Relevant Equations
See under.
I have this 1D LHO problem.
4cd913d9da3a743443ef7dc2d1c2ab1e.png

https://gyazo.com/4cd913d9da3a743443ef7dc2d1c2ab1e
For ##\psi_n (x)## I get
$$\left( \frac{\alpha}{\sqrt{\pi} 2^n n!} \right) ^{\frac{1}{2}} e^{\frac{- \alpha^2 x^2}{2}} H_n(x)$$
with ##E_n = (n+ \frac{1}{2}) \hat{h} \omega##. where ##\hat{h}## is hbar.
For ##\psi_{n+1}(x)## I get
$$ \left( \frac{ \alpha }{ \sqrt{\pi} 2^{n+1} (n+1)!} \right) ^{\frac{1}{2}} e^{ \frac{- \alpha^2 x^2}{2}} H_{n+1}(x)$$
with ##E_{n+1} = ((n+1)+ \frac{1}{2}) \hat{h} \omega##
We can find ##\Psi(x,t>0)## by multiplying the eigenfunctions with their corresponding factors and eigenenergies in the form of ##\e^{-\frac{i}{\hat{h} E_n t}}##, to find ##P(x, t>0)## we basically take ##|\Psi(x, t>0)|^2## which I think will be a long expression.
But how can we find <x>, if we don't know the actual expressions for the Hermite polynomials? How can we compute the integral?
 
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  • #2
How come nothing got latexed :(
 
  • #3
milkism said:
How come nothing got latexed :(
Use “$$”, not “$” for stand alone (centered) and for inline equations “##”.
 
  • #4
Still doesn't work xd
 
  • #5
milkism said:
Still doesn't work xd
You probably have an opened bracket somewhere.
 
  • #6
Almost haha, I hope the unlatexed are understandable.
 
  • #7
This one still is not quite right in your post:

$$ \left( \frac{ \alpha }{ \sqrt{\pi} 2^{n+1} (n+1)!} \right) ^{\frac{1}{2}} e^{ \frac{- \alpha^2 x^2}{2}} H_{n+1}(x)$$

As for the Physics... it's out of my league, but someone will be along.
 
  • #8
Thanks!
 
  • #9
milkism said:
Thanks!
you had some extra brackets in the exponent
 
  • #10
Use \hbar for ##\hbar##.

Express ##\hat x## in terms of the raising and lowering operators and use the orthogonality of the energy eigenstates.
 
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  • #11
vela said:
Use \hbar for ##\hbar##.

Express ##\hat x## in terms of the raising and lowering operators and use the orthogonality of the energy eigenstates.
Do I have to change the x in the $$\psi(x)$$ terms also in terms of raising-ladder operators? Or only $\hat{x}$ in $$< \Psi | \hat{x} | \Psi >$$
 
  • #12
Also I can't write anyting right now so I am using my head, don't we need to use two orthogonality expressions? One for ##\psi (x)## and one for the Hermite polynomials? Which do not have the same form of orthogonality. And also norm for Hermite squared is different. So I first solve the integral w.r.t x for ##\psi (x)## then an another integral for the hermites w.r.t alpha x?
 
  • #13
Nevermind, I'm lost, I don't know how can I use orthogonality in finding expectation value of x, if it was taking integral of probability, no problem.
 
  • #14
Oh! Wait I can rewrite x as $$\frac{\chi}{\alpha}$$ then my integral will look something like
1703875963797.png
 
  • #15
milkism said:
Also I can't write anyting right now so I am using my head, don't we need to use two orthogonality expressions? One for ##\psi (x)## and one for the Hermite polynomials? Which do not have the same form of orthogonality.
The orthogonality of the wave functions is the same as the orthogonality of the Hermite polynomials. You're probably confused because your expression for the wave function is slightly incorrect. ##H(x)## should be ##H(\alpha x)##. The argument of the Hermite polynomial should be dimensionless, so it can't just be a plain ##x## which has dimensions of length.
 
  • #16
milkism said:
Oh! Wait I can rewrite x as $$\frac{\chi}{\alpha}$$ then my integral will look something like
View attachment 337863
Oh wait nevermind we would get $$e^\frac{-\chi ^4}{4}$$
 
  • #17
OMG am so dumb, i forgot exponent rules!
 
  • #18
Is the solution
$$ < x > = \frac{1}{\alpha} \left( \gamma ^2 \left( \frac{\alpha}{\sqrt{\pi}2^n n!} \right) + (1- \gamma^2) \left( \frac{\alpha}{\sqrt{\pi}2^{n+1} (n+1)!} \right) \right)$$?
 
  • #19
No, your expression should have units of length. The factors of ##\alpha## cancel out, so your expression is unitless. Also, normalization should get rid of many of the constants you still have there.
 
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  • #20
milkism said:
Oh! Wait I can rewrite x as $$\frac{\chi}{\alpha}$$ then my integral will look something like
View attachment 337863
Oh, nevermind I misread this formula I thought there was an extra ##\chi## there
 
  • #21
Do I literally have to integrate the four terms that there will be? I am lost.
 
  • #22
1703877759304.png
i guess this is the formula, am unsure how to find the coefficients c_i tho.
 
  • #23
Is it
$$2 \gamma \sqrt{1-\gamma ^2} \sqrt{\frac{\hbar}{2m \omega}} cos(\omega t)$$?
@vela
 
  • #24
Please show your work.
 
  • #25
milkism said:
Do I literally have to integrate the four terms that there will be? I am lost.
You can find the needed integrals without doing any integration if you use the raising and lowering opertors as suggested by @vela.

First write down what you get when they operate on the eigenstates ##\psi_n##
##a_{+}\psi_n=~?## ##~~~a_{-}\psi_n=~?##

Second, noting that ##~~\hat x=\sqrt{\dfrac{\hbar}{2m\omega}}(a_{+}+a_{-})##, find ##~~\hat x \left(\gamma~ \psi_n+\sqrt{1-\gamma^2}~\psi_{n+1}\right)=~?## You should get a linear combination of four eigenstates.

Third, find $$\int\left(\gamma~ \psi^*_n+\sqrt{1-\gamma^2}~\psi^*_{n+1}\right)\left[\hat x \left(\gamma~ \psi_n+\sqrt{1-\gamma^2}~\psi_{n+1}\right)\right]dx=~?$$ Do NOT integrate explicitly. Use the orthonormality condition ##\int \psi^*_i\psi_j dx=\delta_{ij}## to find the answer.

Show your work!
 
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Related to Finding <x> using raising and lowering operators and orthonormality

What are raising and lowering operators?

Raising and lowering operators, often denoted as \(a^\dagger\) and \(a\) respectively, are tools used in quantum mechanics to move between different quantum states. The raising operator \(a^\dagger\) increases the quantum number of a state, while the lowering operator \(a\) decreases it. They are particularly useful in the context of the quantum harmonic oscillator and angular momentum problems.

How do raising and lowering operators relate to the orthonormality of states?

Orthonormality means that the quantum states are orthogonal and normalized. When using raising and lowering operators, the orthonormality ensures that the states generated by these operators are still orthogonal and normalized. This property is crucial for maintaining the physical validity of the quantum states and for simplifying calculations.

What is the significance of finding using these operators?

Finding the expectation value of a position operator \(x\) using raising and lowering operators is significant because it provides a way to calculate physical observables in quantum mechanics. This method leverages the algebraic properties of these operators to simplify the computation of expectation values and matrix elements, which are essential for understanding the behavior of quantum systems.

Can you provide a step-by-step example of finding using raising and lowering operators?

Sure! For a quantum harmonic oscillator, the position operator \(x\) can be expressed in terms of raising and lowering operators as \(x = \sqrt{\frac{\hbar}{2m\omega}} (a + a^\dagger)\). To find the expectation value \(\langle n | x | n \rangle\) in the \(n\)-th state, we use the orthonormality of the states and the properties of the operators:1. Express \(x\) in terms of \(a\) and \(a^\dagger\).2. Apply \(x\) to the state \(|n\rangle\).3. Use the commutation relations and orthonormality to simplify the expression.The result will show that \(\langle n | x | n \rangle = 0\) for all \(n\), due to the symmetry of the harmonic oscillator potential.

What are the commutation relations for raising and lowering operators?

The commutation relations for raising and lowering operators are fundamental to their use in quantum mechanics. For the harmonic oscillator, they are given by \([a, a^\dagger] = 1\) and \([a, a] = [a^\dagger, a^\dagger] = 0\). These relations help in deriving the algebraic properties of the operators and in simplifying calculations involving them.

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