Projecting onto an eigenstate when we make a measurement

In summary, the fundamental result of quantum mechanics is that measuring a certain observable will collapse the state onto an eigenstate of that observable's eigenvalue. In the case of measuring L_z and getting 0, the state will collapse to the eigenstate |1 0>, as there is only one possible eigenstate for this measurement due to the normalization and degeneracy of the system. Other observables, such as total angular momentum, can still be measured to distinguish between different eigenstates.
  • #1
AxiomOfChoice
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Suppose we prepare a system in some properly normalized superposition of the spherical harmonics: A|11> + B|10> + C|1-1>. One of the fundamental results of quantum mechanics is that, if we measure L_z, we will collapse the state of the system onto an eigenstate of the eigenvalue we measure. My question is this: Suppose we measure L_z and we get 0. Which eigenstate of L_z do we collapse to? There are infinitely many spherical harmonics for which m = 0!

Thanks.
 
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  • #2
If the original wave function is A|11> + B|10> + C|1-1> and it is normalizated, eigenfunction you can get when measurement of Lz is zero is l1 0>

I hope my reply help you
 
  • #3
The general rule is that when your measurement returns one of the eigenvalues, the state gets projected onto the corresponding eigenspace. That is, if there are n (orthonormal) eigenstates [itex]|\lambda_i>[/itex] corresponding to the eigenvalue [itex]\lambda[/itex], then if we define:

[tex] P_\lambda = |\lambda_1><\lambda_1| + |\lambda_2><\lambda_2| + ... + |\lambda_n><\lambda_n| [/tex]

Then if a measurement of [itex]|\psi>[/itex] returns [itex]\lambda[/itex], then the state collapses to [itex]P_\lambda |\psi>[/itex].

If you have this degeneracy, it means there are other commuting observables you can still measure to distinguish between the different [itex]|\lambda_i>[/itex]. In your case, one such observable would be the total angular momentum, and your state is already in an eigenstate of this, so there's only one [itex]|\lambda_i>[/itex] possible, |1 0>.
 

1. What is an eigenstate?

An eigenstate is a state in quantum mechanics that is associated with a particular observable quantity, such as position or momentum. When a measurement is made on a system in an eigenstate, the result of the measurement is always the corresponding eigenvalue.

2. How does projecting onto an eigenstate work?

When a measurement is made on a system, it collapses into one of its eigenstates. This means that the system now exists only in that particular eigenstate, and all other possible states disappear. The measurement result is then determined by the eigenvalue associated with the measured eigenstate.

3. What is the significance of projecting onto an eigenstate?

Projecting onto an eigenstate is a fundamental concept in quantum mechanics and is essential for understanding how measurements are made. It allows us to determine the properties of a system at a specific moment in time and is a key aspect of the probabilistic nature of quantum mechanics.

4. Can a system be in more than one eigenstate at a time?

No, a system can only exist in one eigenstate at a time. This is known as the principle of superposition, which states that a quantum system can exist in a combination of multiple eigenstates, but when a measurement is made, it collapses into a single eigenstate.

5. How does projecting onto an eigenstate affect the uncertainty principle?

The uncertainty principle states that the more precisely we know the position of a particle, the less we know about its momentum, and vice versa. When we project onto an eigenstate, we are measuring the properties of the system with a high degree of precision, therefore increasing our knowledge of the system and decreasing the uncertainty of its properties.

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