Quantum Mechanics: Question on Angular Momentum

In summary, the system initially is in the state \psi\left(\theta,\phi\right)=\frac{1}{\sqrt{5}}Y_{1,-1}\left(\theta,\phi\right) + \frac{\sqrt{3}}{\sqrt{5}}Y_{1,0}\left(\theta,\phi\right)+\frac{1}{\sqrt{5}}Y_{1,1}\left(\theta,\phi\right). For Part 1, the value of <\psi|L_{+}|\psi> is found to be \sqrt{\frac{2}{5}}h+ \frac{\sqrt{6}}{\sqrt{5}}h. For Part 2, the
  • #1
Collisionman
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Homework Statement



Consider a system that is initially in the state:

[itex]\psi\left(\theta,\phi\right)=\frac{1}{\sqrt{5}}Y_{1,-1}\left(\theta,\phi\right) + \frac{\sqrt{3}}{5}Y_{1,0}\left(\theta,\phi\right)+\frac{1}{\sqrt{5}}Y_{1,1}\left(\theta,\phi\right)[/itex]

Part 1: Find [itex]<\psi|L_{+}|\psi>[/itex]
Part 2: If [itex]L_{z}[/itex] is measured, what values would one obtain and with what probabilities?

Homework Equations



  • [itex]L_{z}|lm>=mh|lm>[/itex]
  • [itex]L_{+}|lm>=[l(l+1)-m(m+1)]^{\frac{1}{2}}h|lm + 1>[/itex]
  • [itex]Probability = \frac{|<\varphi|\psi>|^{2}}{<\psi|\psi>}[/itex]

The Attempt at a Solution



For Part 1:

So I started off by putting the expression for [itex]\psi\left(\theta,\phi\right)[/itex] in Bra-Ket notation:

[itex]|\psi> = \frac{1}{\sqrt{5}}|1,-1> + \frac{\sqrt{3}}{5}|1,0> + \frac{1}{\sqrt{5}}|1,1>[/itex]

Then I applied [itex]L_{+}[/itex] to each individual component:
  • [itex]L_{+}|1,-1> = \frac{1}{\sqrt{5}}[1(1+1)-(-1)(-1+1)]^{\frac{1}{2}}h|1,0> = \sqrt{\frac{2}{5}}h|1,0>[/itex]
  • [itex]L_{+}|1,0> = \frac{\sqrt{3}}{5}[1(1+1)-0(0+1)]^{\frac{1}{2}}h|1,1> = \frac{\sqrt{6}}{2}h|1,1>[/itex]
  • [itex]L_{+}|1,1> = \frac{1}{\sqrt{5}}[1(1+1)-(-1)(-1+1)]^{\frac{1}{2}}h|1,2> = 0[/itex]

So, [itex]L_{+}|\psi> = \sqrt{\frac{2}{5}}h|1,0>[/itex][itex]+ \frac{\sqrt{6}}{2}h|1,1>[/itex]

And then,

[itex]<\psi|L_{+}|\psi> = <1,0|\sqrt{\frac{2}{5}}h|1,0>[/itex][itex]+ <1,1|\frac{\sqrt{6}}{2}h|1,1>[/itex]
As [itex]<1,1|1,1> = 1[/itex] and [itex]<1,0|1,0>=1[/itex]
[itex]<\psi|L_{+}|\psi> = \sqrt{\frac{2}{5}}h[/itex][itex]+ \frac{\sqrt{6}}{2}h[/itex]

I think I'm going wrong here somewhere. I think I'm using the wrong complex conjugate. Can someone verify if I am or not?

For Part 2:

I took [itex]\psi\left(\theta,\phi\right)[/itex] in Bra-Ket notation as before, i.e.,

[itex]|\psi> = \frac{1}{\sqrt{5}}|1,-1> + \frac{\sqrt{3}}{5}|1,0> + \frac{1}{\sqrt{5}}|1,1>[/itex]

And used [itex]L_{z}|lm>=mh|lm>[/itex] to try and obtain a value for [itex]L_{z}[/itex]. I used this on individual components as follows;

  • [itex]L_{z}|1,-1> = \frac{-h}{\sqrt{5}}|1,-1>[/itex]
  • [itex]L_{z}|1,0> = \frac{\sqrt{3}}{5}(0)h|1, 0> = 0[/itex]
  • [itex]L_{z}|1,1> = \frac{h}{\sqrt{5}}|1,1>[/itex]

Then I multiplied by the complex conjugate, i.e.,

  • [itex]<1,-1|L_{z}|1,-1> = <1,-1|\frac{-h}{\sqrt{5}}|1,-1>[/itex]
  • [itex]<1,1|L_{z}|1,1> = <1,1|\frac{h}{\sqrt{5}}|1,1>[/itex]

So, [itex]L_{z}=[/itex][itex]\frac{-h}{\sqrt{5}}[/itex][itex] + \frac{h}{\sqrt{5}} = 0[/itex]

Again, I'm not too sure if I'm right or wrong here. If someone could verify if I am or not, I'd really appreciate it. If I know where I'm going with [itex]L_{z}[/itex] I can continue on with finding the probabilities, which I understand how to do.

Thanks again in advance. Any help appreciated!
 
Last edited:
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  • #2
Anyone ... please?
 
  • #3
Generally, your work on part 1 looks ok, but I think there are some minor errors. You wrote the wavefunction as
[itex]\psi\left(\theta,\phi\right)=\frac{1}{\sqrt{5}}Y_{1,-1}\left(\theta,\phi\right) + \frac{\sqrt{3}}{5}Y_{1,0}\left(\theta,\phi\right)+\frac{1}{\sqrt{5}}Y_{1,1}\left(\theta,\phi\right)[/itex]

First, I suspect that the numerical coefficient of the Y1,0 term should have a denominator of √5 rather than 5.

Also, I think you need to check the numerical factors in the following:
[itex]L_{+}|1,0> = \frac{\sqrt{3}}{5}[1(1+1)-0(0+1)]^{\frac{1}{2}}h|1,1> = \frac{\sqrt{6}}{2}h|1,1>[/itex]

Finally, when constructing ##<\psi|## you left out the numerical coefficients contained in ##|\psi>##

For part 2, you have found the "expectation value" of ##L_z##. But that won't give you much information about what values are possible for individual measurements of ##L_z##. The only possible value that you can get for a measurement of an operator in QM is one of the eigenvalues of that operator. Note that your wavefunction is a superposition of three eigenstates of ##L_z##. Each eigenstate corresponds to a specific eigenvalue of ##L_z##.

So, what are the possible values of a measurement of ##L_z## for your wavefunction?

The numerical coefficients of each of the terms in the wavefunction have something to do with the probability of measuring a particular eigenvalue of ##L_z##.
 
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  • #4
Thanks TSny, that helped a lot!
 
  • #5


Your solution for Part 1 is correct. You have correctly applied the ladder operators L_{+} and L_{-} to the individual components of \psi and then used the definition of the inner product to find <\psi|L_{+}|\psi>.

For Part 2, you have the right idea, but your calculations are not quite correct. Remember that the operator L_{z} is Hermitian, meaning that it is equal to its own Hermitian conjugate. So, when you take the inner product <\psi|L_{z}|\psi>, you need to use the complex conjugate of |\psi>, which is <\psi|. Also, when you use the definition of the inner product, you need to take the complex conjugate of the first term, not the second. So, your final answer for L_{z} should be:

<\psi|L_{z}|\psi> = <\psi|\frac{-h}{\sqrt{5}}|1,-1> + <\psi|\frac{\sqrt{3}}{5}(0)h|1, 0> + <\psi|\frac{h}{\sqrt{5}}|1,1> = \frac{-h}{\sqrt{5}} + \frac{h}{\sqrt{5}} = 0

This means that the only possible measurement for L_{z} is 0, with a probability of 1. This makes sense because the state \psi is a superposition of only m=1, m=0, and m=-1 states, all with equal coefficients. This means that there is no preferred direction for the angular momentum, and it is equally likely to be measured in any direction, giving a result of 0.

Hope this helps clarify your understanding of angular momentum in quantum mechanics!
 

1. What is angular momentum in quantum mechanics?

Angular momentum is a physical quantity that describes the rotation or spinning of a particle. In quantum mechanics, it is a fundamental property that is quantized, meaning it can only have certain discrete values.

2. How is angular momentum measured in quantum mechanics?

In quantum mechanics, angular momentum is measured using the operators Lx, Ly, and Lz, which represent the components of angular momentum along the x, y, and z axes, respectively. The eigenvalues of these operators correspond to the allowed values of angular momentum for a given system.

3. What is the relationship between angular momentum and spin in quantum mechanics?

In quantum mechanics, spin is a type of intrinsic angular momentum possessed by elementary particles such as electrons. It is not related to the physical spinning of a particle, but rather is a quantum property that has discrete values, such as 1/2, 1, or 3/2. Spin is often represented as the third component of angular momentum (Sz) and is an important concept in understanding the behavior of subatomic particles.

4. How does angular momentum affect the energy levels of a quantum system?

In quantum mechanics, angular momentum is a conserved quantity, meaning it cannot be created or destroyed. This has significant implications for the energy levels of a quantum system, as the allowed values of angular momentum determine the possible energy levels that a particle can have. Additionally, angular momentum can affect the degeneracy or spacing of energy levels in a system.

5. What is the significance of angular momentum in quantum mechanics?

Angular momentum is a crucial concept in understanding the behavior of particles at the quantum level. It is related to many other important quantities, such as energy and spin, and is used to describe the properties of atoms, molecules, and other subatomic particles. Understanding angular momentum is essential for understanding the physical laws that govern the behavior of matter at a very small scale.

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