$begingroup$
I'm confusing myself. Let's represent some state in the eigenbasis for Hydrogen:
$$|psirangle = sum_{n,l,m}|n,l,mranglelangle n,l,m|psirangle.$$
Now denote the initial state by $psi(t=0)equivpsi_o$, and hit this thing with time evolution:
$$U|psirangle = sum_{n,l,m}e^{-iE_nt/hbar}|n,l,mranglelangle n,l,m|psi_orangle.$$
I'm wanting to know what the probability is that I measure some specific $(l^*,m^*)$ at some later time $t$. Looking at this, we have
$$P(t,l=l^*,m=m^*)=sum_n|langle n,l^*,m^*|U|psirangle|^2 \ = sum_n|langle n,l^*,m^*|psi_orangle|^2.$$
This has no time dependence, and I feel I'm missing something obvious. For example, say we prepare the state to initially be $|psirangle = a|1,0,0rangle+b|2,1,1rangle+c|3,1,1rangle$, where all constants are real. This would imply from the above, after normalization, that
$$P(l=1,m=1) = (b^2+c^2)/(a^2+b^2+c^2),$$
independent of time. What am I missing here? Obviously the probability density function has cross terms, so I do not see why this should physically be the case, thus sparking my question.
quantum-mechanics atomic-physics
asked 4 hours ago
dm__dm__
1077
$endgroup$
add a comment |
$begingroup$
I'm confusing myself. Let's represent some state in the eigenbasis for Hydrogen:
$$|psirangle = sum_{n,l,m}|n,l,mranglelangle n,l,m|psirangle.$$
Now denote the initial state by $psi(t=0)equivpsi_o$, and hit this thing with time evolution:
$$U|psirangle = sum_{n,l,m}e^{-iE_nt/hbar}|n,l,mranglelangle n,l,m|psi_orangle.$$
I'm wanting to know what the probability is that I measure some specific $(l^*,m^*)$ at some later time $t$. Looking at this, we have
$$P(t,l=l^*,m=m^*)=sum_n|langle n,l^*,m^*|U|psirangle|^2 \ = sum_n|langle n,l^*,m^*|psi_orangle|^2.$$
This has no time dependence, and I feel I'm missing something obvious. For example, say we prepare the state to initially be $|psirangle = a|1,0,0rangle+b|2,1,1rangle+c|3,1,1rangle$, where all constants are real. This would imply from the above, after normalization, that
$$P(l=1,m=1) = (b^2+c^2)/(a^2+b^2+c^2),$$
independent of time. What am I missing here? Obviously the probability density function has cross terms, so I do not see why this should physically be the case, thus sparking my question.
quantum-mechanics atomic-physics
asked 4 hours ago
dm__dm__
1077
$endgroup$
add a comment |
$begingroup$
I'm confusing myself. Let's represent some state in the eigenbasis for Hydrogen:
$$|psirangle = sum_{n,l,m}|n,l,mranglelangle n,l,m|psirangle.$$
Now denote the initial state by $psi(t=0)equivpsi_o$, and hit this thing with time evolution:
$$U|psirangle = sum_{n,l,m}e^{-iE_nt/hbar}|n,l,mranglelangle n,l,m|psi_orangle.$$
I'm wanting to know what the probability is that I measure some specific $(l^*,m^*)$ at some later time $t$. Looking at this, we have
$$P(t,l=l^*,m=m^*)=sum_n|langle n,l^*,m^*|U|psirangle|^2 \ = sum_n|langle n,l^*,m^*|psi_orangle|^2.$$
This has no time dependence, and I feel I'm missing something obvious. For example, say we prepare the state to initially be $|psirangle = a|1,0,0rangle+b|2,1,1rangle+c|3,1,1rangle$, where all constants are real. This would imply from the above, after normalization, that
$$P(l=1,m=1) = (b^2+c^2)/(a^2+b^2+c^2),$$
independent of time. What am I missing here? Obviously the probability density function has cross terms, so I do not see why this should physically be the case, thus sparking my question.
quantum-mechanics atomic-physics
asked 4 hours ago
dm__dm__
1077
$endgroup$
I'm confusing myself. Let's represent some state in the eigenbasis for Hydrogen:
$$|psirangle = sum_{n,l,m}|n,l,mranglelangle n,l,m|psirangle.$$
Now denote the initial state by $psi(t=0)equivpsi_o$, and hit this thing with time evolution:
$$U|psirangle = sum_{n,l,m}e^{-iE_nt/hbar}|n,l,mranglelangle n,l,m|psi_orangle.$$
I'm wanting to know what the probability is that I measure some specific $(l^*,m^*)$ at some later time $t$. Looking at this, we have
$$P(t,l=l^*,m=m^*)=sum_n|langle n,l^*,m^*|U|psirangle|^2 \ = sum_n|langle n,l^*,m^*|psi_orangle|^2.$$
This has no time dependence, and I feel I'm missing something obvious. For example, say we prepare the state to initially be $|psirangle = a|1,0,0rangle+b|2,1,1rangle+c|3,1,1rangle$, where all constants are real. This would imply from the above, after normalization, that
$$P(l=1,m=1) = (b^2+c^2)/(a^2+b^2+c^2),$$
independent of time. What am I missing here? Obviously the probability density function has cross terms, so I do not see why this should physically be the case, thus sparking my question.
quantum-mechanics atomic-physics
quantum-mechanics atomic-physics
asked 4 hours ago
dm__dm__
1077
asked 4 hours ago
dm__dm__
1077
asked 4 hours ago
dm__dm__
1077
asked 4 hours ago
dm__dm__
1077
asked 4 hours ago
dm__dm__
1077
1077
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
This is in general true whenever you calculate the projection onto an eigenstate (and not a combination thereof). Let ${|arangle}$ be a set of
eigentstates for the Hamiltionian $hat{H}$, a state at time $t$ can be written as
$$
|psi(t)rangle = sum_{a}hat{U}(t)|aranglelangle|psi_0rangle.
$$
Its projection onto an eigenstate $|a'rangle$ is
$$
langle a'| psi(t)rangle = langle a'| Big(sum_{a}hat{U}(t)|aranglelangle|psi_0rangleBig)=hat{U}(t)_{a' a'}langle a'|psi_0rangle
$$
whose norm does not depend on time. This is because once a state collapses into an eigenstate, it remains there indefinitely.
answered 3 hours ago
gentedgented
4,508916
$endgroup$
add a comment |
$begingroup$
Your calculations are correct.
The spectrum is degenerate with respect to angular momentum. All states with equal radial quantum number pick up the same phase during time evolution. And if you only care about angular momentum quantum numbers (as you do by marginalizing over $n$!), any time dependence is naturally lost.
Do the same calculation again for a spectrum that does depend on e.g. $l$ as well and you'll see how the lifted degeneracy causes superpositions of different $l$s to evolve with an explicit time dependence.
answered 3 hours ago
NephenteNephente
2,16811020
$endgroup$
$begingroup$
How would that change if the energy depends on $n,l$? As long as the time evolution is diagonal on the energy states they always only pick up a phase factor, don't they?
$endgroup$
– gented
2 mins ago
add a comment |
$begingroup$
I think you're mixing up two different things. We define a stationary state as one in which all observables are time independent, and you are quite correct that if we write:
$$ Psi = sum_i a_{nlm} ,psi_{nlm} $$
then this is not a stationary state. There is a nice example of this in Emilio Pisanty's answer to Is there oscillating charge in a hydrogen atom?
But what you appear to be expecting is that the coefficients $a_{nlm}$ will change with time, and they will not. All that happens is that the relative phases of the $psi_{nlm}$ functions changes. In practice of course the state will emit photons and collapse to the ground state, but you'd have to add a coupling to a photon for that to happen to your function.
answered 2 hours ago
John RennieJohn Rennie
278k44553798
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is in general true whenever you calculate the projection onto an eigenstate (and not a combination thereof). Let ${|arangle}$ be a set of
eigentstates for the Hamiltionian $hat{H}$, a state at time $t$ can be written as
$$
|psi(t)rangle = sum_{a}hat{U}(t)|aranglelangle|psi_0rangle.
$$
Its projection onto an eigenstate $|a'rangle$ is
$$
langle a'| psi(t)rangle = langle a'| Big(sum_{a}hat{U}(t)|aranglelangle|psi_0rangleBig)=hat{U}(t)_{a' a'}langle a'|psi_0rangle
$$
whose norm does not depend on time. This is because once a state collapses into an eigenstate, it remains there indefinitely.
answered 3 hours ago
gentedgented
4,508916
$endgroup$
add a comment |
$begingroup$
This is in general true whenever you calculate the projection onto an eigenstate (and not a combination thereof). Let ${|arangle}$ be a set of
eigentstates for the Hamiltionian $hat{H}$, a state at time $t$ can be written as
$$
|psi(t)rangle = sum_{a}hat{U}(t)|aranglelangle|psi_0rangle.
$$
Its projection onto an eigenstate $|a'rangle$ is
$$
langle a'| psi(t)rangle = langle a'| Big(sum_{a}hat{U}(t)|aranglelangle|psi_0rangleBig)=hat{U}(t)_{a' a'}langle a'|psi_0rangle
$$
whose norm does not depend on time. This is because once a state collapses into an eigenstate, it remains there indefinitely.
answered 3 hours ago
gentedgented
4,508916
$endgroup$
add a comment |
$begingroup$
This is in general true whenever you calculate the projection onto an eigenstate (and not a combination thereof). Let ${|arangle}$ be a set of
eigentstates for the Hamiltionian $hat{H}$, a state at time $t$ can be written as
$$
|psi(t)rangle = sum_{a}hat{U}(t)|aranglelangle|psi_0rangle.
$$
Its projection onto an eigenstate $|a'rangle$ is
$$
langle a'| psi(t)rangle = langle a'| Big(sum_{a}hat{U}(t)|aranglelangle|psi_0rangleBig)=hat{U}(t)_{a' a'}langle a'|psi_0rangle
$$
whose norm does not depend on time. This is because once a state collapses into an eigenstate, it remains there indefinitely.
answered 3 hours ago
gentedgented
4,508916
$endgroup$
This is in general true whenever you calculate the projection onto an eigenstate (and not a combination thereof). Let ${|arangle}$ be a set of
eigentstates for the Hamiltionian $hat{H}$, a state at time $t$ can be written as
$$
|psi(t)rangle = sum_{a}hat{U}(t)|aranglelangle|psi_0rangle.
$$
Its projection onto an eigenstate $|a'rangle$ is
$$
langle a'| psi(t)rangle = langle a'| Big(sum_{a}hat{U}(t)|aranglelangle|psi_0rangleBig)=hat{U}(t)_{a' a'}langle a'|psi_0rangle
$$
whose norm does not depend on time. This is because once a state collapses into an eigenstate, it remains there indefinitely.
answered 3 hours ago
gentedgented
4,508916
answered 3 hours ago
gentedgented
4,508916
answered 3 hours ago
gentedgented
4,508916
answered 3 hours ago
gentedgented
4,508916
4,508916
add a comment |
add a comment |
$begingroup$
Your calculations are correct.
The spectrum is degenerate with respect to angular momentum. All states with equal radial quantum number pick up the same phase during time evolution. And if you only care about angular momentum quantum numbers (as you do by marginalizing over $n$!), any time dependence is naturally lost.
Do the same calculation again for a spectrum that does depend on e.g. $l$ as well and you'll see how the lifted degeneracy causes superpositions of different $l$s to evolve with an explicit time dependence.
answered 3 hours ago
NephenteNephente
2,16811020
$endgroup$
$begingroup$
How would that change if the energy depends on $n,l$? As long as the time evolution is diagonal on the energy states they always only pick up a phase factor, don't they?
$endgroup$
– gented
2 mins ago
add a comment |
$begingroup$
Your calculations are correct.
The spectrum is degenerate with respect to angular momentum. All states with equal radial quantum number pick up the same phase during time evolution. And if you only care about angular momentum quantum numbers (as you do by marginalizing over $n$!), any time dependence is naturally lost.
Do the same calculation again for a spectrum that does depend on e.g. $l$ as well and you'll see how the lifted degeneracy causes superpositions of different $l$s to evolve with an explicit time dependence.
answered 3 hours ago
NephenteNephente
2,16811020
$endgroup$
$begingroup$
How would that change if the energy depends on $n,l$? As long as the time evolution is diagonal on the energy states they always only pick up a phase factor, don't they?
$endgroup$
– gented
2 mins ago
add a comment |
$begingroup$
Your calculations are correct.
The spectrum is degenerate with respect to angular momentum. All states with equal radial quantum number pick up the same phase during time evolution. And if you only care about angular momentum quantum numbers (as you do by marginalizing over $n$!), any time dependence is naturally lost.
Do the same calculation again for a spectrum that does depend on e.g. $l$ as well and you'll see how the lifted degeneracy causes superpositions of different $l$s to evolve with an explicit time dependence.
answered 3 hours ago
NephenteNephente
2,16811020
$endgroup$
Your calculations are correct.
The spectrum is degenerate with respect to angular momentum. All states with equal radial quantum number pick up the same phase during time evolution. And if you only care about angular momentum quantum numbers (as you do by marginalizing over $n$!), any time dependence is naturally lost.
Do the same calculation again for a spectrum that does depend on e.g. $l$ as well and you'll see how the lifted degeneracy causes superpositions of different $l$s to evolve with an explicit time dependence.
answered 3 hours ago
NephenteNephente
2,16811020
edited 2 hours ago
answered 3 hours ago
NephenteNephente
2,16811020
answered 3 hours ago
NephenteNephente
2,16811020
answered 3 hours ago
NephenteNephente
2,16811020
2,16811020
$begingroup$
How would that change if the energy depends on $n,l$? As long as the time evolution is diagonal on the energy states they always only pick up a phase factor, don't they?
$endgroup$
– gented
2 mins ago
add a comment |
$begingroup$
How would that change if the energy depends on $n,l$? As long as the time evolution is diagonal on the energy states they always only pick up a phase factor, don't they?
$endgroup$
– gented
2 mins ago
$begingroup$
How would that change if the energy depends on $n,l$? As long as the time evolution is diagonal on the energy states they always only pick up a phase factor, don't they?
$endgroup$
– gented
2 mins ago
$begingroup$
How would that change if the energy depends on $n,l$? As long as the time evolution is diagonal on the energy states they always only pick up a phase factor, don't they?
$endgroup$
– gented
2 mins ago
add a comment |
$begingroup$
I think you're mixing up two different things. We define a stationary state as one in which all observables are time independent, and you are quite correct that if we write:
$$ Psi = sum_i a_{nlm} ,psi_{nlm} $$
then this is not a stationary state. There is a nice example of this in Emilio Pisanty's answer to Is there oscillating charge in a hydrogen atom?
But what you appear to be expecting is that the coefficients $a_{nlm}$ will change with time, and they will not. All that happens is that the relative phases of the $psi_{nlm}$ functions changes. In practice of course the state will emit photons and collapse to the ground state, but you'd have to add a coupling to a photon for that to happen to your function.
answered 2 hours ago
John RennieJohn Rennie
278k44553798
$endgroup$
add a comment |
$begingroup$
I think you're mixing up two different things. We define a stationary state as one in which all observables are time independent, and you are quite correct that if we write:
$$ Psi = sum_i a_{nlm} ,psi_{nlm} $$
then this is not a stationary state. There is a nice example of this in Emilio Pisanty's answer to Is there oscillating charge in a hydrogen atom?
But what you appear to be expecting is that the coefficients $a_{nlm}$ will change with time, and they will not. All that happens is that the relative phases of the $psi_{nlm}$ functions changes. In practice of course the state will emit photons and collapse to the ground state, but you'd have to add a coupling to a photon for that to happen to your function.
answered 2 hours ago
John RennieJohn Rennie
278k44553798
$endgroup$
add a comment |
$begingroup$
I think you're mixing up two different things. We define a stationary state as one in which all observables are time independent, and you are quite correct that if we write:
$$ Psi = sum_i a_{nlm} ,psi_{nlm} $$
then this is not a stationary state. There is a nice example of this in Emilio Pisanty's answer to Is there oscillating charge in a hydrogen atom?
But what you appear to be expecting is that the coefficients $a_{nlm}$ will change with time, and they will not. All that happens is that the relative phases of the $psi_{nlm}$ functions changes. In practice of course the state will emit photons and collapse to the ground state, but you'd have to add a coupling to a photon for that to happen to your function.
answered 2 hours ago
John RennieJohn Rennie
278k44553798
$endgroup$
I think you're mixing up two different things. We define a stationary state as one in which all observables are time independent, and you are quite correct that if we write:
$$ Psi = sum_i a_{nlm} ,psi_{nlm} $$
then this is not a stationary state. There is a nice example of this in Emilio Pisanty's answer to Is there oscillating charge in a hydrogen atom?
But what you appear to be expecting is that the coefficients $a_{nlm}$ will change with time, and they will not. All that happens is that the relative phases of the $psi_{nlm}$ functions changes. In practice of course the state will emit photons and collapse to the ground state, but you'd have to add a coupling to a photon for that to happen to your function.
answered 2 hours ago
John RennieJohn Rennie
278k44553798
answered 2 hours ago
John RennieJohn Rennie
278k44553798
answered 2 hours ago
John RennieJohn Rennie
278k44553798
answered 2 hours ago
John RennieJohn Rennie
278k44553798
278k44553798
add a comment |
add a comment |
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