Probabilities in non-stationary statesIs there oscillating charge in a hydrogen atom?It appears that stationary states aren't so stationaryWhy is time evolution of wavefunctions non-trivial?Transition Probabilities for the Perturbed Harmonic OscillatorQuantum Harmonic Oscillator and the Classical LimitDoes angular momentum of hydrogen atom imply motion of electron around the nucleus?Motivation for transition probabilities in quantum mechanicsGeneral solution of states of time dependent HamiltonianGround state energy in terms of partition functionWhy do the matrix elements of an operator correspond to the Fourier components of the observable in Heisenberg's Matrix Mechanics?Find the function $G(x,y,t,t_0)$ such that $psi(x,t)=int_-infty^inftydy G(x,y,t,t_0) psi(y,t_0)$ is the time evolution equation
Do native speakers use "ultima" and "proxima" frequently in spoken English?
Error in master's thesis, I do not know what to do
Have any astronauts/cosmonauts died in space?
How can a new country break out from a developed country without war?
Homology of the fiber
Friend wants my recommendation but I don't want to
Can other pieces capture a threatening piece and prevent a checkmate?
Print a physical multiplication table
Imaginary part of expression too difficult to calculate
Why is participating in the European Parliamentary elections used as a threat?
"Marked down as someone wanting to sell shares." What does that mean?
Why is there so much iron?
10 year ban after applying for a UK student visa
Norwegian Refugee travel document
Why are there no stars visible in cislunar space?
Would mining huge amounts of resources on the Moon change its orbit?
Determine voltage drop over 10G resistors with cheap multimeter
What kind of footwear is suitable for walking in micro gravity environment?
Asserting that Atheism and Theism are both faith based positions
Are hand made posters acceptable in Academia?
Help with identifying unique aircraft over NE Pennsylvania
PTIJ: At the Passover Seder, is one allowed to speak more than once during Maggid?
Is this Pascal's Matrix?
Is there any common country to visit for uk and schengen visa?
Probabilities in non-stationary states
Is there oscillating charge in a hydrogen atom?It appears that stationary states aren't so stationaryWhy is time evolution of wavefunctions non-trivial?Transition Probabilities for the Perturbed Harmonic OscillatorQuantum Harmonic Oscillator and the Classical LimitDoes angular momentum of hydrogen atom imply motion of electron around the nucleus?Motivation for transition probabilities in quantum mechanicsGeneral solution of states of time dependent HamiltonianGround state energy in terms of partition functionWhy do the matrix elements of an operator correspond to the Fourier components of the observable in Heisenberg's Matrix Mechanics?Find the function $G(x,y,t,t_0)$ such that $psi(x,t)=int_-infty^inftydy G(x,y,t,t_0) psi(y,t_0)$ is the time evolution equation
$begingroup$
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:
$$U|psirangle = sum_n,l,me^-iE_nt/hbar|n,l,mranglelangle n,l,m|psi_orangle.$$
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.
====================================================================
Closure:
As pointed out by user 'The Vee', my confusion stemmed from this observable being an integral of the eigenbasis representation. I had internally generalized the time dependence of observable expectations, when this is not the case if that observable is also being used as a quantum number in the state representation. The general time evolution of some observable $Omega$ in this basis would be
$$langleOmega (t)rangle = langle psi|U^dagger Omega U|psirangle \ = sum_n',l',m'sum_n,l,me^i(E_n'-E_n)t/hbarlangle n',l',m'|Omega|n,l,mranglelangle n',l',m'|psi_orangle^*langle n,l,m|psi_orangle.$$
If $Omega = L^2$ or $L_z$, then orthogonality reduces this to
$$langle L^2rangle = sum_n,l,mhbar^2 l(l+1)|langle n,l,m|psi_orangle|^2 \
langle L_zrangle = sum_n,l,mhbar m|langle n,l,m|psi_orangle|^2$$
No time dependence of the expectations, hence no time dependence of observation probability; all is well. If $[H,Omega]neq 0$, then all of those cross terms do not drop out, and we see the oscillation in the exponential depending on the energy difference of states. I've kept it in this basis to provide consistency with the above question, but we can see how this generalizes to whatever CSCO we use, as user 'gented' does in his answer by using a collective notation $|arangle$.
quantum-mechanics atomic-physics probability time-evolution orbitals
$endgroup$
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:
$$U|psirangle = sum_n,l,me^-iE_nt/hbar|n,l,mranglelangle n,l,m|psi_orangle.$$
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.
====================================================================
Closure:
As pointed out by user 'The Vee', my confusion stemmed from this observable being an integral of the eigenbasis representation. I had internally generalized the time dependence of observable expectations, when this is not the case if that observable is also being used as a quantum number in the state representation. The general time evolution of some observable $Omega$ in this basis would be
$$langleOmega (t)rangle = langle psi|U^dagger Omega U|psirangle \ = sum_n',l',m'sum_n,l,me^i(E_n'-E_n)t/hbarlangle n',l',m'|Omega|n,l,mranglelangle n',l',m'|psi_orangle^*langle n,l,m|psi_orangle.$$
If $Omega = L^2$ or $L_z$, then orthogonality reduces this to
$$langle L^2rangle = sum_n,l,mhbar^2 l(l+1)|langle n,l,m|psi_orangle|^2 \
langle L_zrangle = sum_n,l,mhbar m|langle n,l,m|psi_orangle|^2$$
No time dependence of the expectations, hence no time dependence of observation probability; all is well. If $[H,Omega]neq 0$, then all of those cross terms do not drop out, and we see the oscillation in the exponential depending on the energy difference of states. I've kept it in this basis to provide consistency with the above question, but we can see how this generalizes to whatever CSCO we use, as user 'gented' does in his answer by using a collective notation $|arangle$.
quantum-mechanics atomic-physics probability time-evolution orbitals
$endgroup$
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:
$$U|psirangle = sum_n,l,me^-iE_nt/hbar|n,l,mranglelangle n,l,m|psi_orangle.$$
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.
====================================================================
Closure:
As pointed out by user 'The Vee', my confusion stemmed from this observable being an integral of the eigenbasis representation. I had internally generalized the time dependence of observable expectations, when this is not the case if that observable is also being used as a quantum number in the state representation. The general time evolution of some observable $Omega$ in this basis would be
$$langleOmega (t)rangle = langle psi|U^dagger Omega U|psirangle \ = sum_n',l',m'sum_n,l,me^i(E_n'-E_n)t/hbarlangle n',l',m'|Omega|n,l,mranglelangle n',l',m'|psi_orangle^*langle n,l,m|psi_orangle.$$
If $Omega = L^2$ or $L_z$, then orthogonality reduces this to
$$langle L^2rangle = sum_n,l,mhbar^2 l(l+1)|langle n,l,m|psi_orangle|^2 \
langle L_zrangle = sum_n,l,mhbar m|langle n,l,m|psi_orangle|^2$$
No time dependence of the expectations, hence no time dependence of observation probability; all is well. If $[H,Omega]neq 0$, then all of those cross terms do not drop out, and we see the oscillation in the exponential depending on the energy difference of states. I've kept it in this basis to provide consistency with the above question, but we can see how this generalizes to whatever CSCO we use, as user 'gented' does in his answer by using a collective notation $|arangle$.
quantum-mechanics atomic-physics probability time-evolution orbitals
$endgroup$
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:
$$U|psirangle = sum_n,l,me^-iE_nt/hbar|n,l,mranglelangle n,l,m|psi_orangle.$$
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.
====================================================================
Closure:
As pointed out by user 'The Vee', my confusion stemmed from this observable being an integral of the eigenbasis representation. I had internally generalized the time dependence of observable expectations, when this is not the case if that observable is also being used as a quantum number in the state representation. The general time evolution of some observable $Omega$ in this basis would be
$$langleOmega (t)rangle = langle psi|U^dagger Omega U|psirangle \ = sum_n',l',m'sum_n,l,me^i(E_n'-E_n)t/hbarlangle n',l',m'|Omega|n,l,mranglelangle n',l',m'|psi_orangle^*langle n,l,m|psi_orangle.$$
If $Omega = L^2$ or $L_z$, then orthogonality reduces this to
$$langle L^2rangle = sum_n,l,mhbar^2 l(l+1)|langle n,l,m|psi_orangle|^2 \
langle L_zrangle = sum_n,l,mhbar m|langle n,l,m|psi_orangle|^2$$
No time dependence of the expectations, hence no time dependence of observation probability; all is well. If $[H,Omega]neq 0$, then all of those cross terms do not drop out, and we see the oscillation in the exponential depending on the energy difference of states. I've kept it in this basis to provide consistency with the above question, but we can see how this generalizes to whatever CSCO we use, as user 'gented' does in his answer by using a collective notation $|arangle$.
quantum-mechanics atomic-physics probability time-evolution orbitals
quantum-mechanics atomic-physics probability time-evolution orbitals
edited 5 hours ago
dm__
asked 18 hours ago
dm__dm__
1197
1197
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 $left_ain A$ be a set of
eigentstates for the Hamiltionian $hatH$, a state at time $t$ can be written as
$$
|psi(t)rangle = sum_ahatU(t)|aranglelangle a |psi_0rangle.
$$
Its projection onto an eigenstate $|a'rangle$ is
$$
langle a'| psi(t)rangle = langle a'| Big(sum_ahatU(t)|aranglelangle a |psi_0rangleBig)=hatU(t)_a' a'langle a'|psi_0rangle
$$
whose norm does not depend on time as long as $hatU(t)$ only picks up a phase factor when acting onto eigenstates. This is because once a state collapses into an eigenstate, it remains there indefinitely.
$endgroup$
$begingroup$
Accepted because it nicely shows the generality of this. It should be noted that you are explicitly assuming $|arangle$ to be collective notation for an eigenstate of the hamiltonian. Thanks for the answer!
$endgroup$
– dm__
6 hours ago
$begingroup$
Also, you dropped the $a$ from your bra in the summation. I tried to edit, but it won't let me. Perhaps you can modify it. It is clear from the top equation, but it could be confusing.
$endgroup$
– dm__
6 hours ago
$begingroup$
Yes, I noticed and fixed it :)
$endgroup$
– gented
5 hours 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.
$endgroup$
1
$begingroup$
I'm not sure what distinction you're trying to draw at the beginning of the second paragraph. Can't the change in phase of a $psi_nlm$ function be equally well viewed as a change in the coefficient $a_nlm$?
$endgroup$
– Michael Seifert
13 hours ago
add a comment |
$begingroup$
You just happened to consider an observable (or rather, a pair of observables) that is, in fact, an integral (integrals) of motion of the system. In other words, the probability of measuring any value of $(l,m)$ is in fact not expected to change during time evolution.
This is not true for other observables in general, but it does hold for any time-independent $A$ which commutes with the Hamiltonian. Since both $L^2$ and $L_z$ have this property, both $l$ and $m$ are integrals of motion and your result follows. (They also commute with each other which enables you to use both the measured values simultaneously.)
For a counterexample, you may consider the probability of measuring something that is not an eigenstate of the Hamiltonian, like $|varphirangle := (|1,0,0rangle + |2,0,0rangle)/sqrt2$. I won't try to come up with an observable of which this is an eigenvector – that would only obscure the idea and at the end of the day you only need the eigenvector anyway. If you want, examples of common observables that don't commute with the hydrogen Hamiltonian are any component of position or of momentum, but there the direct calculation is complicated by the fact that these do not have eigenvalues.
$endgroup$
$begingroup$
Hi there, thanks for the answer, I appreciate it. That was exactly my confusion. I added a 'closure' to the bottom mathematically showing this.
$endgroup$
– dm__
6 hours ago
$begingroup$
This is actually a good description of why at the end of the day integral of motion must commute with the Hamiltonian (because otherwise not being diagonal on the eigenstates, the probabilities may not be independent of time).
$endgroup$
– gented
5 hours ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "151"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f467136%2fprobabilities-in-non-stationary-states%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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 $left_ain A$ be a set of
eigentstates for the Hamiltionian $hatH$, a state at time $t$ can be written as
$$
|psi(t)rangle = sum_ahatU(t)|aranglelangle a |psi_0rangle.
$$
Its projection onto an eigenstate $|a'rangle$ is
$$
langle a'| psi(t)rangle = langle a'| Big(sum_ahatU(t)|aranglelangle a |psi_0rangleBig)=hatU(t)_a' a'langle a'|psi_0rangle
$$
whose norm does not depend on time as long as $hatU(t)$ only picks up a phase factor when acting onto eigenstates. This is because once a state collapses into an eigenstate, it remains there indefinitely.
$endgroup$
$begingroup$
Accepted because it nicely shows the generality of this. It should be noted that you are explicitly assuming $|arangle$ to be collective notation for an eigenstate of the hamiltonian. Thanks for the answer!
$endgroup$
– dm__
6 hours ago
$begingroup$
Also, you dropped the $a$ from your bra in the summation. I tried to edit, but it won't let me. Perhaps you can modify it. It is clear from the top equation, but it could be confusing.
$endgroup$
– dm__
6 hours ago
$begingroup$
Yes, I noticed and fixed it :)
$endgroup$
– gented
5 hours ago
add a comment |
$begingroup$
This is in general true whenever you calculate the projection onto an eigenstate (and not a combination thereof). Let $left_ain A$ be a set of
eigentstates for the Hamiltionian $hatH$, a state at time $t$ can be written as
$$
|psi(t)rangle = sum_ahatU(t)|aranglelangle a |psi_0rangle.
$$
Its projection onto an eigenstate $|a'rangle$ is
$$
langle a'| psi(t)rangle = langle a'| Big(sum_ahatU(t)|aranglelangle a |psi_0rangleBig)=hatU(t)_a' a'langle a'|psi_0rangle
$$
whose norm does not depend on time as long as $hatU(t)$ only picks up a phase factor when acting onto eigenstates. This is because once a state collapses into an eigenstate, it remains there indefinitely.
$endgroup$
$begingroup$
Accepted because it nicely shows the generality of this. It should be noted that you are explicitly assuming $|arangle$ to be collective notation for an eigenstate of the hamiltonian. Thanks for the answer!
$endgroup$
– dm__
6 hours ago
$begingroup$
Also, you dropped the $a$ from your bra in the summation. I tried to edit, but it won't let me. Perhaps you can modify it. It is clear from the top equation, but it could be confusing.
$endgroup$
– dm__
6 hours ago
$begingroup$
Yes, I noticed and fixed it :)
$endgroup$
– gented
5 hours ago
add a comment |
$begingroup$
This is in general true whenever you calculate the projection onto an eigenstate (and not a combination thereof). Let $left_ain A$ be a set of
eigentstates for the Hamiltionian $hatH$, a state at time $t$ can be written as
$$
|psi(t)rangle = sum_ahatU(t)|aranglelangle a |psi_0rangle.
$$
Its projection onto an eigenstate $|a'rangle$ is
$$
langle a'| psi(t)rangle = langle a'| Big(sum_ahatU(t)|aranglelangle a |psi_0rangleBig)=hatU(t)_a' a'langle a'|psi_0rangle
$$
whose norm does not depend on time as long as $hatU(t)$ only picks up a phase factor when acting onto eigenstates. This is because once a state collapses into an eigenstate, it remains there indefinitely.
$endgroup$
This is in general true whenever you calculate the projection onto an eigenstate (and not a combination thereof). Let $left_ain A$ be a set of
eigentstates for the Hamiltionian $hatH$, a state at time $t$ can be written as
$$
|psi(t)rangle = sum_ahatU(t)|aranglelangle a |psi_0rangle.
$$
Its projection onto an eigenstate $|a'rangle$ is
$$
langle a'| psi(t)rangle = langle a'| Big(sum_ahatU(t)|aranglelangle a |psi_0rangleBig)=hatU(t)_a' a'langle a'|psi_0rangle
$$
whose norm does not depend on time as long as $hatU(t)$ only picks up a phase factor when acting onto eigenstates. This is because once a state collapses into an eigenstate, it remains there indefinitely.
edited 5 hours ago
answered 16 hours ago
gentedgented
4,562916
4,562916
$begingroup$
Accepted because it nicely shows the generality of this. It should be noted that you are explicitly assuming $|arangle$ to be collective notation for an eigenstate of the hamiltonian. Thanks for the answer!
$endgroup$
– dm__
6 hours ago
$begingroup$
Also, you dropped the $a$ from your bra in the summation. I tried to edit, but it won't let me. Perhaps you can modify it. It is clear from the top equation, but it could be confusing.
$endgroup$
– dm__
6 hours ago
$begingroup$
Yes, I noticed and fixed it :)
$endgroup$
– gented
5 hours ago
add a comment |
$begingroup$
Accepted because it nicely shows the generality of this. It should be noted that you are explicitly assuming $|arangle$ to be collective notation for an eigenstate of the hamiltonian. Thanks for the answer!
$endgroup$
– dm__
6 hours ago
$begingroup$
Also, you dropped the $a$ from your bra in the summation. I tried to edit, but it won't let me. Perhaps you can modify it. It is clear from the top equation, but it could be confusing.
$endgroup$
– dm__
6 hours ago
$begingroup$
Yes, I noticed and fixed it :)
$endgroup$
– gented
5 hours ago
$begingroup$
Accepted because it nicely shows the generality of this. It should be noted that you are explicitly assuming $|arangle$ to be collective notation for an eigenstate of the hamiltonian. Thanks for the answer!
$endgroup$
– dm__
6 hours ago
$begingroup$
Accepted because it nicely shows the generality of this. It should be noted that you are explicitly assuming $|arangle$ to be collective notation for an eigenstate of the hamiltonian. Thanks for the answer!
$endgroup$
– dm__
6 hours ago
$begingroup$
Also, you dropped the $a$ from your bra in the summation. I tried to edit, but it won't let me. Perhaps you can modify it. It is clear from the top equation, but it could be confusing.
$endgroup$
– dm__
6 hours ago
$begingroup$
Also, you dropped the $a$ from your bra in the summation. I tried to edit, but it won't let me. Perhaps you can modify it. It is clear from the top equation, but it could be confusing.
$endgroup$
– dm__
6 hours ago
$begingroup$
Yes, I noticed and fixed it :)
$endgroup$
– gented
5 hours ago
$begingroup$
Yes, I noticed and fixed it :)
$endgroup$
– gented
5 hours 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.
$endgroup$
1
$begingroup$
I'm not sure what distinction you're trying to draw at the beginning of the second paragraph. Can't the change in phase of a $psi_nlm$ function be equally well viewed as a change in the coefficient $a_nlm$?
$endgroup$
– Michael Seifert
13 hours 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.
$endgroup$
1
$begingroup$
I'm not sure what distinction you're trying to draw at the beginning of the second paragraph. Can't the change in phase of a $psi_nlm$ function be equally well viewed as a change in the coefficient $a_nlm$?
$endgroup$
– Michael Seifert
13 hours 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.
$endgroup$
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 16 hours ago
John RennieJohn Rennie
278k44553799
278k44553799
1
$begingroup$
I'm not sure what distinction you're trying to draw at the beginning of the second paragraph. Can't the change in phase of a $psi_nlm$ function be equally well viewed as a change in the coefficient $a_nlm$?
$endgroup$
– Michael Seifert
13 hours ago
add a comment |
1
$begingroup$
I'm not sure what distinction you're trying to draw at the beginning of the second paragraph. Can't the change in phase of a $psi_nlm$ function be equally well viewed as a change in the coefficient $a_nlm$?
$endgroup$
– Michael Seifert
13 hours ago
1
1
$begingroup$
I'm not sure what distinction you're trying to draw at the beginning of the second paragraph. Can't the change in phase of a $psi_nlm$ function be equally well viewed as a change in the coefficient $a_nlm$?
$endgroup$
– Michael Seifert
13 hours ago
$begingroup$
I'm not sure what distinction you're trying to draw at the beginning of the second paragraph. Can't the change in phase of a $psi_nlm$ function be equally well viewed as a change in the coefficient $a_nlm$?
$endgroup$
– Michael Seifert
13 hours ago
add a comment |
$begingroup$
You just happened to consider an observable (or rather, a pair of observables) that is, in fact, an integral (integrals) of motion of the system. In other words, the probability of measuring any value of $(l,m)$ is in fact not expected to change during time evolution.
This is not true for other observables in general, but it does hold for any time-independent $A$ which commutes with the Hamiltonian. Since both $L^2$ and $L_z$ have this property, both $l$ and $m$ are integrals of motion and your result follows. (They also commute with each other which enables you to use both the measured values simultaneously.)
For a counterexample, you may consider the probability of measuring something that is not an eigenstate of the Hamiltonian, like $|varphirangle := (|1,0,0rangle + |2,0,0rangle)/sqrt2$. I won't try to come up with an observable of which this is an eigenvector – that would only obscure the idea and at the end of the day you only need the eigenvector anyway. If you want, examples of common observables that don't commute with the hydrogen Hamiltonian are any component of position or of momentum, but there the direct calculation is complicated by the fact that these do not have eigenvalues.
$endgroup$
$begingroup$
Hi there, thanks for the answer, I appreciate it. That was exactly my confusion. I added a 'closure' to the bottom mathematically showing this.
$endgroup$
– dm__
6 hours ago
$begingroup$
This is actually a good description of why at the end of the day integral of motion must commute with the Hamiltonian (because otherwise not being diagonal on the eigenstates, the probabilities may not be independent of time).
$endgroup$
– gented
5 hours ago
add a comment |
$begingroup$
You just happened to consider an observable (or rather, a pair of observables) that is, in fact, an integral (integrals) of motion of the system. In other words, the probability of measuring any value of $(l,m)$ is in fact not expected to change during time evolution.
This is not true for other observables in general, but it does hold for any time-independent $A$ which commutes with the Hamiltonian. Since both $L^2$ and $L_z$ have this property, both $l$ and $m$ are integrals of motion and your result follows. (They also commute with each other which enables you to use both the measured values simultaneously.)
For a counterexample, you may consider the probability of measuring something that is not an eigenstate of the Hamiltonian, like $|varphirangle := (|1,0,0rangle + |2,0,0rangle)/sqrt2$. I won't try to come up with an observable of which this is an eigenvector – that would only obscure the idea and at the end of the day you only need the eigenvector anyway. If you want, examples of common observables that don't commute with the hydrogen Hamiltonian are any component of position or of momentum, but there the direct calculation is complicated by the fact that these do not have eigenvalues.
$endgroup$
$begingroup$
Hi there, thanks for the answer, I appreciate it. That was exactly my confusion. I added a 'closure' to the bottom mathematically showing this.
$endgroup$
– dm__
6 hours ago
$begingroup$
This is actually a good description of why at the end of the day integral of motion must commute with the Hamiltonian (because otherwise not being diagonal on the eigenstates, the probabilities may not be independent of time).
$endgroup$
– gented
5 hours ago
add a comment |
$begingroup$
You just happened to consider an observable (or rather, a pair of observables) that is, in fact, an integral (integrals) of motion of the system. In other words, the probability of measuring any value of $(l,m)$ is in fact not expected to change during time evolution.
This is not true for other observables in general, but it does hold for any time-independent $A$ which commutes with the Hamiltonian. Since both $L^2$ and $L_z$ have this property, both $l$ and $m$ are integrals of motion and your result follows. (They also commute with each other which enables you to use both the measured values simultaneously.)
For a counterexample, you may consider the probability of measuring something that is not an eigenstate of the Hamiltonian, like $|varphirangle := (|1,0,0rangle + |2,0,0rangle)/sqrt2$. I won't try to come up with an observable of which this is an eigenvector – that would only obscure the idea and at the end of the day you only need the eigenvector anyway. If you want, examples of common observables that don't commute with the hydrogen Hamiltonian are any component of position or of momentum, but there the direct calculation is complicated by the fact that these do not have eigenvalues.
$endgroup$
You just happened to consider an observable (or rather, a pair of observables) that is, in fact, an integral (integrals) of motion of the system. In other words, the probability of measuring any value of $(l,m)$ is in fact not expected to change during time evolution.
This is not true for other observables in general, but it does hold for any time-independent $A$ which commutes with the Hamiltonian. Since both $L^2$ and $L_z$ have this property, both $l$ and $m$ are integrals of motion and your result follows. (They also commute with each other which enables you to use both the measured values simultaneously.)
For a counterexample, you may consider the probability of measuring something that is not an eigenstate of the Hamiltonian, like $|varphirangle := (|1,0,0rangle + |2,0,0rangle)/sqrt2$. I won't try to come up with an observable of which this is an eigenvector – that would only obscure the idea and at the end of the day you only need the eigenvector anyway. If you want, examples of common observables that don't commute with the hydrogen Hamiltonian are any component of position or of momentum, but there the direct calculation is complicated by the fact that these do not have eigenvalues.
answered 11 hours ago
The VeeThe Vee
833412
833412
$begingroup$
Hi there, thanks for the answer, I appreciate it. That was exactly my confusion. I added a 'closure' to the bottom mathematically showing this.
$endgroup$
– dm__
6 hours ago
$begingroup$
This is actually a good description of why at the end of the day integral of motion must commute with the Hamiltonian (because otherwise not being diagonal on the eigenstates, the probabilities may not be independent of time).
$endgroup$
– gented
5 hours ago
add a comment |
$begingroup$
Hi there, thanks for the answer, I appreciate it. That was exactly my confusion. I added a 'closure' to the bottom mathematically showing this.
$endgroup$
– dm__
6 hours ago
$begingroup$
This is actually a good description of why at the end of the day integral of motion must commute with the Hamiltonian (because otherwise not being diagonal on the eigenstates, the probabilities may not be independent of time).
$endgroup$
– gented
5 hours ago
$begingroup$
Hi there, thanks for the answer, I appreciate it. That was exactly my confusion. I added a 'closure' to the bottom mathematically showing this.
$endgroup$
– dm__
6 hours ago
$begingroup$
Hi there, thanks for the answer, I appreciate it. That was exactly my confusion. I added a 'closure' to the bottom mathematically showing this.
$endgroup$
– dm__
6 hours ago
$begingroup$
This is actually a good description of why at the end of the day integral of motion must commute with the Hamiltonian (because otherwise not being diagonal on the eigenstates, the probabilities may not be independent of time).
$endgroup$
– gented
5 hours ago
$begingroup$
This is actually a good description of why at the end of the day integral of motion must commute with the Hamiltonian (because otherwise not being diagonal on the eigenstates, the probabilities may not be independent of time).
$endgroup$
– gented
5 hours ago
add a comment |
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f467136%2fprobabilities-in-non-stationary-states%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown