What happens to a wavefunction upon measurement when there's degeneracy?
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I'll use a hydrogen atom as an example. A hydrogen atom has multiple energy eigenstates for all but one of its energy levels. Suppose I measure a hydrogen atom to have an energy $E_n$ where $n > 1$. What's the state of the hydrogen atom right after my measurement?
quantum-mechanics quantum-states
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add a comment |
$begingroup$
I'll use a hydrogen atom as an example. A hydrogen atom has multiple energy eigenstates for all but one of its energy levels. Suppose I measure a hydrogen atom to have an energy $E_n$ where $n > 1$. What's the state of the hydrogen atom right after my measurement?
quantum-mechanics quantum-states
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add a comment |
$begingroup$
I'll use a hydrogen atom as an example. A hydrogen atom has multiple energy eigenstates for all but one of its energy levels. Suppose I measure a hydrogen atom to have an energy $E_n$ where $n > 1$. What's the state of the hydrogen atom right after my measurement?
quantum-mechanics quantum-states
$endgroup$
I'll use a hydrogen atom as an example. A hydrogen atom has multiple energy eigenstates for all but one of its energy levels. Suppose I measure a hydrogen atom to have an energy $E_n$ where $n > 1$. What's the state of the hydrogen atom right after my measurement?
quantum-mechanics quantum-states
quantum-mechanics quantum-states
asked 6 hours ago
PiKindOfGuyPiKindOfGuy
488515
488515
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2 Answers
2
active
oldest
votes
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This occurs in all measurements, because you don't measure the quantum number of every particle in the universe.
If you believe in the Copenhagen interpretation, then in your example the state after measurement is the projection of the original state onto the subspace with energy $E_n$. For example, let's use the quantum numbers $n$, $l$, where $l$ is the angular momentum. If the original state was $|1,0rangle+i|2,0rangle-|2,1rangle$, and you measure $n=2$, then after measurement the state becomes $i|2,0rangle-|2,1rangle$.
If you believe instead in standard quantum mechanics, then nothing happens to the wavefunction you're observing, but you become entangled with the wavefunction in a way that leads to the same facts about the results of your future measurements.
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I'm a beginner. Should I concern myself with what you call "standard quantum mechanics"? Entanglement is still pretty foreign to me.
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– PiKindOfGuy
6 hours ago
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Normalize your states so I can accept your answer.
$endgroup$
– PiKindOfGuy
5 hours ago
add a comment |
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The measurement is a projection so the appropriate projection operator would be
$$
hatPi= sum_{r=1}^k vert E_n;alpha_rranglelangle E_n;alpha_rvert
$$
where $vert E_n;alpha_rrangle$ is an eigenstate of $hat H$ with energy $E_n$ and $alpha_r$ denotes all the other eigenvalues required to fully label your states, given a complete set of commuting observables.
Thus, you’d get
$$
vertPhirangle=hat Pivert psirangle = sum_r vert E_n;alpha_rrangle c_{r}, ,tag{1}
$$
where $c_r=langle E_n;alpha rvertpsirangle$. Note that projection does not preserve the norm so (1) must be normalized by dividing “manually” by
$sqrt{sum_r vert c_rvert^2}$.
$endgroup$
$begingroup$
So Ben's answer is correct, except that the states ought to be normalized (for proper use)?
$endgroup$
– PiKindOfGuy
6 hours ago
1
$begingroup$
Yes his is correct albeit a little less general, and the normalization is missing.
$endgroup$
– ZeroTheHero
6 hours ago
$begingroup$
Should I pick the answer that I think is most complete or the one that helped me the most?
$endgroup$
– PiKindOfGuy
6 hours ago
$begingroup$
Frankly it doesn’t matter much. You can just let it happen and decide in 24hrs.
$endgroup$
– ZeroTheHero
6 hours ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
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$begingroup$
This occurs in all measurements, because you don't measure the quantum number of every particle in the universe.
If you believe in the Copenhagen interpretation, then in your example the state after measurement is the projection of the original state onto the subspace with energy $E_n$. For example, let's use the quantum numbers $n$, $l$, where $l$ is the angular momentum. If the original state was $|1,0rangle+i|2,0rangle-|2,1rangle$, and you measure $n=2$, then after measurement the state becomes $i|2,0rangle-|2,1rangle$.
If you believe instead in standard quantum mechanics, then nothing happens to the wavefunction you're observing, but you become entangled with the wavefunction in a way that leads to the same facts about the results of your future measurements.
$endgroup$
$begingroup$
I'm a beginner. Should I concern myself with what you call "standard quantum mechanics"? Entanglement is still pretty foreign to me.
$endgroup$
– PiKindOfGuy
6 hours ago
$begingroup$
Normalize your states so I can accept your answer.
$endgroup$
– PiKindOfGuy
5 hours ago
add a comment |
$begingroup$
This occurs in all measurements, because you don't measure the quantum number of every particle in the universe.
If you believe in the Copenhagen interpretation, then in your example the state after measurement is the projection of the original state onto the subspace with energy $E_n$. For example, let's use the quantum numbers $n$, $l$, where $l$ is the angular momentum. If the original state was $|1,0rangle+i|2,0rangle-|2,1rangle$, and you measure $n=2$, then after measurement the state becomes $i|2,0rangle-|2,1rangle$.
If you believe instead in standard quantum mechanics, then nothing happens to the wavefunction you're observing, but you become entangled with the wavefunction in a way that leads to the same facts about the results of your future measurements.
$endgroup$
$begingroup$
I'm a beginner. Should I concern myself with what you call "standard quantum mechanics"? Entanglement is still pretty foreign to me.
$endgroup$
– PiKindOfGuy
6 hours ago
$begingroup$
Normalize your states so I can accept your answer.
$endgroup$
– PiKindOfGuy
5 hours ago
add a comment |
$begingroup$
This occurs in all measurements, because you don't measure the quantum number of every particle in the universe.
If you believe in the Copenhagen interpretation, then in your example the state after measurement is the projection of the original state onto the subspace with energy $E_n$. For example, let's use the quantum numbers $n$, $l$, where $l$ is the angular momentum. If the original state was $|1,0rangle+i|2,0rangle-|2,1rangle$, and you measure $n=2$, then after measurement the state becomes $i|2,0rangle-|2,1rangle$.
If you believe instead in standard quantum mechanics, then nothing happens to the wavefunction you're observing, but you become entangled with the wavefunction in a way that leads to the same facts about the results of your future measurements.
$endgroup$
This occurs in all measurements, because you don't measure the quantum number of every particle in the universe.
If you believe in the Copenhagen interpretation, then in your example the state after measurement is the projection of the original state onto the subspace with energy $E_n$. For example, let's use the quantum numbers $n$, $l$, where $l$ is the angular momentum. If the original state was $|1,0rangle+i|2,0rangle-|2,1rangle$, and you measure $n=2$, then after measurement the state becomes $i|2,0rangle-|2,1rangle$.
If you believe instead in standard quantum mechanics, then nothing happens to the wavefunction you're observing, but you become entangled with the wavefunction in a way that leads to the same facts about the results of your future measurements.
answered 6 hours ago
Ben CrowellBen Crowell
50.2k5155296
50.2k5155296
$begingroup$
I'm a beginner. Should I concern myself with what you call "standard quantum mechanics"? Entanglement is still pretty foreign to me.
$endgroup$
– PiKindOfGuy
6 hours ago
$begingroup$
Normalize your states so I can accept your answer.
$endgroup$
– PiKindOfGuy
5 hours ago
add a comment |
$begingroup$
I'm a beginner. Should I concern myself with what you call "standard quantum mechanics"? Entanglement is still pretty foreign to me.
$endgroup$
– PiKindOfGuy
6 hours ago
$begingroup$
Normalize your states so I can accept your answer.
$endgroup$
– PiKindOfGuy
5 hours ago
$begingroup$
I'm a beginner. Should I concern myself with what you call "standard quantum mechanics"? Entanglement is still pretty foreign to me.
$endgroup$
– PiKindOfGuy
6 hours ago
$begingroup$
I'm a beginner. Should I concern myself with what you call "standard quantum mechanics"? Entanglement is still pretty foreign to me.
$endgroup$
– PiKindOfGuy
6 hours ago
$begingroup$
Normalize your states so I can accept your answer.
$endgroup$
– PiKindOfGuy
5 hours ago
$begingroup$
Normalize your states so I can accept your answer.
$endgroup$
– PiKindOfGuy
5 hours ago
add a comment |
$begingroup$
The measurement is a projection so the appropriate projection operator would be
$$
hatPi= sum_{r=1}^k vert E_n;alpha_rranglelangle E_n;alpha_rvert
$$
where $vert E_n;alpha_rrangle$ is an eigenstate of $hat H$ with energy $E_n$ and $alpha_r$ denotes all the other eigenvalues required to fully label your states, given a complete set of commuting observables.
Thus, you’d get
$$
vertPhirangle=hat Pivert psirangle = sum_r vert E_n;alpha_rrangle c_{r}, ,tag{1}
$$
where $c_r=langle E_n;alpha rvertpsirangle$. Note that projection does not preserve the norm so (1) must be normalized by dividing “manually” by
$sqrt{sum_r vert c_rvert^2}$.
$endgroup$
$begingroup$
So Ben's answer is correct, except that the states ought to be normalized (for proper use)?
$endgroup$
– PiKindOfGuy
6 hours ago
1
$begingroup$
Yes his is correct albeit a little less general, and the normalization is missing.
$endgroup$
– ZeroTheHero
6 hours ago
$begingroup$
Should I pick the answer that I think is most complete or the one that helped me the most?
$endgroup$
– PiKindOfGuy
6 hours ago
$begingroup$
Frankly it doesn’t matter much. You can just let it happen and decide in 24hrs.
$endgroup$
– ZeroTheHero
6 hours ago
add a comment |
$begingroup$
The measurement is a projection so the appropriate projection operator would be
$$
hatPi= sum_{r=1}^k vert E_n;alpha_rranglelangle E_n;alpha_rvert
$$
where $vert E_n;alpha_rrangle$ is an eigenstate of $hat H$ with energy $E_n$ and $alpha_r$ denotes all the other eigenvalues required to fully label your states, given a complete set of commuting observables.
Thus, you’d get
$$
vertPhirangle=hat Pivert psirangle = sum_r vert E_n;alpha_rrangle c_{r}, ,tag{1}
$$
where $c_r=langle E_n;alpha rvertpsirangle$. Note that projection does not preserve the norm so (1) must be normalized by dividing “manually” by
$sqrt{sum_r vert c_rvert^2}$.
$endgroup$
$begingroup$
So Ben's answer is correct, except that the states ought to be normalized (for proper use)?
$endgroup$
– PiKindOfGuy
6 hours ago
1
$begingroup$
Yes his is correct albeit a little less general, and the normalization is missing.
$endgroup$
– ZeroTheHero
6 hours ago
$begingroup$
Should I pick the answer that I think is most complete or the one that helped me the most?
$endgroup$
– PiKindOfGuy
6 hours ago
$begingroup$
Frankly it doesn’t matter much. You can just let it happen and decide in 24hrs.
$endgroup$
– ZeroTheHero
6 hours ago
add a comment |
$begingroup$
The measurement is a projection so the appropriate projection operator would be
$$
hatPi= sum_{r=1}^k vert E_n;alpha_rranglelangle E_n;alpha_rvert
$$
where $vert E_n;alpha_rrangle$ is an eigenstate of $hat H$ with energy $E_n$ and $alpha_r$ denotes all the other eigenvalues required to fully label your states, given a complete set of commuting observables.
Thus, you’d get
$$
vertPhirangle=hat Pivert psirangle = sum_r vert E_n;alpha_rrangle c_{r}, ,tag{1}
$$
where $c_r=langle E_n;alpha rvertpsirangle$. Note that projection does not preserve the norm so (1) must be normalized by dividing “manually” by
$sqrt{sum_r vert c_rvert^2}$.
$endgroup$
The measurement is a projection so the appropriate projection operator would be
$$
hatPi= sum_{r=1}^k vert E_n;alpha_rranglelangle E_n;alpha_rvert
$$
where $vert E_n;alpha_rrangle$ is an eigenstate of $hat H$ with energy $E_n$ and $alpha_r$ denotes all the other eigenvalues required to fully label your states, given a complete set of commuting observables.
Thus, you’d get
$$
vertPhirangle=hat Pivert psirangle = sum_r vert E_n;alpha_rrangle c_{r}, ,tag{1}
$$
where $c_r=langle E_n;alpha rvertpsirangle$. Note that projection does not preserve the norm so (1) must be normalized by dividing “manually” by
$sqrt{sum_r vert c_rvert^2}$.
answered 6 hours ago
ZeroTheHeroZeroTheHero
19.6k53159
19.6k53159
$begingroup$
So Ben's answer is correct, except that the states ought to be normalized (for proper use)?
$endgroup$
– PiKindOfGuy
6 hours ago
1
$begingroup$
Yes his is correct albeit a little less general, and the normalization is missing.
$endgroup$
– ZeroTheHero
6 hours ago
$begingroup$
Should I pick the answer that I think is most complete or the one that helped me the most?
$endgroup$
– PiKindOfGuy
6 hours ago
$begingroup$
Frankly it doesn’t matter much. You can just let it happen and decide in 24hrs.
$endgroup$
– ZeroTheHero
6 hours ago
add a comment |
$begingroup$
So Ben's answer is correct, except that the states ought to be normalized (for proper use)?
$endgroup$
– PiKindOfGuy
6 hours ago
1
$begingroup$
Yes his is correct albeit a little less general, and the normalization is missing.
$endgroup$
– ZeroTheHero
6 hours ago
$begingroup$
Should I pick the answer that I think is most complete or the one that helped me the most?
$endgroup$
– PiKindOfGuy
6 hours ago
$begingroup$
Frankly it doesn’t matter much. You can just let it happen and decide in 24hrs.
$endgroup$
– ZeroTheHero
6 hours ago
$begingroup$
So Ben's answer is correct, except that the states ought to be normalized (for proper use)?
$endgroup$
– PiKindOfGuy
6 hours ago
$begingroup$
So Ben's answer is correct, except that the states ought to be normalized (for proper use)?
$endgroup$
– PiKindOfGuy
6 hours ago
1
1
$begingroup$
Yes his is correct albeit a little less general, and the normalization is missing.
$endgroup$
– ZeroTheHero
6 hours ago
$begingroup$
Yes his is correct albeit a little less general, and the normalization is missing.
$endgroup$
– ZeroTheHero
6 hours ago
$begingroup$
Should I pick the answer that I think is most complete or the one that helped me the most?
$endgroup$
– PiKindOfGuy
6 hours ago
$begingroup$
Should I pick the answer that I think is most complete or the one that helped me the most?
$endgroup$
– PiKindOfGuy
6 hours ago
$begingroup$
Frankly it doesn’t matter much. You can just let it happen and decide in 24hrs.
$endgroup$
– ZeroTheHero
6 hours ago
$begingroup$
Frankly it doesn’t matter much. You can just let it happen and decide in 24hrs.
$endgroup$
– ZeroTheHero
6 hours ago
add a comment |
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