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
asked Feb 2 at 0:06
PiKindOfGuyPiKindOfGuy
400518
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add a comment |
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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
asked Feb 2 at 0:06
PiKindOfGuyPiKindOfGuy
400518
$endgroup$
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
asked Feb 2 at 0:06
PiKindOfGuyPiKindOfGuy
400518
$endgroup$
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 Feb 2 at 0:06
PiKindOfGuyPiKindOfGuy
400518
asked Feb 2 at 0:06
PiKindOfGuyPiKindOfGuy
400518
asked Feb 2 at 0:06
PiKindOfGuyPiKindOfGuy
400518
asked Feb 2 at 0:06
PiKindOfGuyPiKindOfGuy
400518
asked Feb 2 at 0:06
PiKindOfGuyPiKindOfGuy
400518
400518
add a comment |
add a comment |
2 Answers
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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.
answered Feb 2 at 0:10
Ben CrowellBen Crowell
51.3k6156302
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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
$$
vertPhirangle=hat Pivert psirangle = sum_r vert E_n;alpha_rrangle c_r, ,tag1
$$
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
$sqrtsum_r vert c_rvert^2$.
answered Feb 2 at 0:24
ZeroTheHeroZeroTheHero
20.3k53160
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So Ben's answer is correct, except that the states ought to be normalized (for proper use)?
$endgroup$
– PiKindOfGuy
Feb 2 at 0:26
1
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Yes his is correct albeit a little less general, and the normalization is missing.
$endgroup$
– ZeroTheHero
Feb 2 at 0:27
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Should I pick the answer that I think is most complete or the one that helped me the most?
$endgroup$
– PiKindOfGuy
Feb 2 at 0:31
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Frankly it doesn’t matter much. You can just let it happen and decide in 24hrs.
$endgroup$
– ZeroTheHero
Feb 2 at 0:32
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Feb 2 at 0:10
Ben CrowellBen Crowell
51.3k6156302
$endgroup$
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.
answered Feb 2 at 0:10
Ben CrowellBen Crowell
51.3k6156302
$endgroup$
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.
answered Feb 2 at 0:10
Ben CrowellBen Crowell
51.3k6156302
$endgroup$
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 Feb 2 at 0:10
Ben CrowellBen Crowell
51.3k6156302
answered Feb 2 at 0:10
Ben CrowellBen Crowell
51.3k6156302
answered Feb 2 at 0:10
Ben CrowellBen Crowell
51.3k6156302
answered Feb 2 at 0:10
Ben CrowellBen Crowell
51.3k6156302
51.3k6156302
add a comment |
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
$$
vertPhirangle=hat Pivert psirangle = sum_r vert E_n;alpha_rrangle c_r, ,tag1
$$
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
$sqrtsum_r vert c_rvert^2$.
answered Feb 2 at 0:24
ZeroTheHeroZeroTheHero
20.3k53160
$endgroup$
$begingroup$
So Ben's answer is correct, except that the states ought to be normalized (for proper use)?
$endgroup$
– PiKindOfGuy
Feb 2 at 0:26
1
$begingroup$
Yes his is correct albeit a little less general, and the normalization is missing.
$endgroup$
– ZeroTheHero
Feb 2 at 0:27
$begingroup$
Should I pick the answer that I think is most complete or the one that helped me the most?
$endgroup$
– PiKindOfGuy
Feb 2 at 0:31
$begingroup$
Frankly it doesn’t matter much. You can just let it happen and decide in 24hrs.
$endgroup$
– ZeroTheHero
Feb 2 at 0:32
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
$$
vertPhirangle=hat Pivert psirangle = sum_r vert E_n;alpha_rrangle c_r, ,tag1
$$
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
$sqrtsum_r vert c_rvert^2$.
answered Feb 2 at 0:24
ZeroTheHeroZeroTheHero
20.3k53160
$endgroup$
$begingroup$
So Ben's answer is correct, except that the states ought to be normalized (for proper use)?
$endgroup$
– PiKindOfGuy
Feb 2 at 0:26
1
$begingroup$
Yes his is correct albeit a little less general, and the normalization is missing.
$endgroup$
– ZeroTheHero
Feb 2 at 0:27
$begingroup$
Should I pick the answer that I think is most complete or the one that helped me the most?
$endgroup$
– PiKindOfGuy
Feb 2 at 0:31
$begingroup$
Frankly it doesn’t matter much. You can just let it happen and decide in 24hrs.
$endgroup$
– ZeroTheHero
Feb 2 at 0:32
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
$$
vertPhirangle=hat Pivert psirangle = sum_r vert E_n;alpha_rrangle c_r, ,tag1
$$
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
$sqrtsum_r vert c_rvert^2$.
answered Feb 2 at 0:24
ZeroTheHeroZeroTheHero
20.3k53160
$endgroup$
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
$$
vertPhirangle=hat Pivert psirangle = sum_r vert E_n;alpha_rrangle c_r, ,tag1
$$
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
$sqrtsum_r vert c_rvert^2$.
answered Feb 2 at 0:24
ZeroTheHeroZeroTheHero
20.3k53160
answered Feb 2 at 0:24
ZeroTheHeroZeroTheHero
20.3k53160
answered Feb 2 at 0:24
ZeroTheHeroZeroTheHero
20.3k53160
answered Feb 2 at 0:24
ZeroTheHeroZeroTheHero
20.3k53160
20.3k53160
$begingroup$
So Ben's answer is correct, except that the states ought to be normalized (for proper use)?
$endgroup$
– PiKindOfGuy
Feb 2 at 0:26
1
$begingroup$
Yes his is correct albeit a little less general, and the normalization is missing.
$endgroup$
– ZeroTheHero
Feb 2 at 0:27
$begingroup$
Should I pick the answer that I think is most complete or the one that helped me the most?
$endgroup$
– PiKindOfGuy
Feb 2 at 0:31
$begingroup$
Frankly it doesn’t matter much. You can just let it happen and decide in 24hrs.
$endgroup$
– ZeroTheHero
Feb 2 at 0:32
add a comment |
$begingroup$
So Ben's answer is correct, except that the states ought to be normalized (for proper use)?
$endgroup$
– PiKindOfGuy
Feb 2 at 0:26
1
$begingroup$
Yes his is correct albeit a little less general, and the normalization is missing.
$endgroup$
– ZeroTheHero
Feb 2 at 0:27
$begingroup$
Should I pick the answer that I think is most complete or the one that helped me the most?
$endgroup$
– PiKindOfGuy
Feb 2 at 0:31
$begingroup$
Frankly it doesn’t matter much. You can just let it happen and decide in 24hrs.
$endgroup$
– ZeroTheHero
Feb 2 at 0:32
$begingroup$
So Ben's answer is correct, except that the states ought to be normalized (for proper use)?
$endgroup$
– PiKindOfGuy
Feb 2 at 0:26
$begingroup$
So Ben's answer is correct, except that the states ought to be normalized (for proper use)?
$endgroup$
– PiKindOfGuy
Feb 2 at 0:26
1
1
$begingroup$
Yes his is correct albeit a little less general, and the normalization is missing.
$endgroup$
– ZeroTheHero
Feb 2 at 0:27
$begingroup$
Yes his is correct albeit a little less general, and the normalization is missing.
$endgroup$
– ZeroTheHero
Feb 2 at 0:27
$begingroup$
Should I pick the answer that I think is most complete or the one that helped me the most?
$endgroup$
– PiKindOfGuy
Feb 2 at 0:31
$begingroup$
Should I pick the answer that I think is most complete or the one that helped me the most?
$endgroup$
– PiKindOfGuy
Feb 2 at 0:31
$begingroup$
Frankly it doesn’t matter much. You can just let it happen and decide in 24hrs.
$endgroup$
– ZeroTheHero
Feb 2 at 0:32
$begingroup$
Frankly it doesn’t matter much. You can just let it happen and decide in 24hrs.
$endgroup$
– ZeroTheHero
Feb 2 at 0:32
add a comment |
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