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Is a bound state a stationary state?
The 2019 Stack Overflow Developer Survey Results Are InIt appears that stationary states aren't so stationaryBound states, scattering states and infinite potentialsOperator in Hilbert space of a spinHelp needed to understand “On the reality of the quantum state”Trace of density matrix for mixed stateUsing the Heisenberg Uncertainty Relation to Estimate Ground State EnergiesTime Derivative of Expectation Value - Stationary StateParticle in a Box, Expansion of Energy StateStates in QM and in the algebraic approachInfinite Series vs Integral Representation of State Vectors in QM?
$begingroup$
In Shankar's discussion on the 1D infinite square well in Principles of Quantum Mechanics (2nd edition), he made the following statement:
Now $langle P rangle = 0$ in any bound state for the following reason. Since a bound state is a stationary state, $langle P rangle$ is time independent. If this $langle Prangle ne 0$, the particle must (in the average sense) drift either to the right or to the left and eventually escape to infinity, which cannot happen in a bound state.
The final sentence makes sense to me, but his reasoning in the second sentence does not. Aren't bound states and stationary states entirely different things? Does the one in fact imply the other?
quantum-mechanics hilbert-space terminology definition quantum-states
edited Mar 24 at 2:06
Qmechanic♦
108k122001243
asked Mar 24 at 1:20
J-JJ-J
636
$endgroup$
add a comment |
$begingroup$
In Shankar's discussion on the 1D infinite square well in Principles of Quantum Mechanics (2nd edition), he made the following statement:
Now $langle P rangle = 0$ in any bound state for the following reason. Since a bound state is a stationary state, $langle P rangle$ is time independent. If this $langle Prangle ne 0$, the particle must (in the average sense) drift either to the right or to the left and eventually escape to infinity, which cannot happen in a bound state.
The final sentence makes sense to me, but his reasoning in the second sentence does not. Aren't bound states and stationary states entirely different things? Does the one in fact imply the other?
quantum-mechanics hilbert-space terminology definition quantum-states
edited Mar 24 at 2:06
Qmechanic♦
108k122001243
asked Mar 24 at 1:20
J-JJ-J
636
$endgroup$
2
$begingroup$
I find that puzzling too because I would think that a state moving around in a potential is still a bound state. I guess Shankar is just using the words in a particular way.
$endgroup$
– DanielSank
Mar 24 at 1:25
add a comment |
$begingroup$
In Shankar's discussion on the 1D infinite square well in Principles of Quantum Mechanics (2nd edition), he made the following statement:
Now $langle P rangle = 0$ in any bound state for the following reason. Since a bound state is a stationary state, $langle P rangle$ is time independent. If this $langle Prangle ne 0$, the particle must (in the average sense) drift either to the right or to the left and eventually escape to infinity, which cannot happen in a bound state.
The final sentence makes sense to me, but his reasoning in the second sentence does not. Aren't bound states and stationary states entirely different things? Does the one in fact imply the other?
quantum-mechanics hilbert-space terminology definition quantum-states
edited Mar 24 at 2:06
Qmechanic♦
108k122001243
asked Mar 24 at 1:20
J-JJ-J
636
$endgroup$
In Shankar's discussion on the 1D infinite square well in Principles of Quantum Mechanics (2nd edition), he made the following statement:
Now $langle P rangle = 0$ in any bound state for the following reason. Since a bound state is a stationary state, $langle P rangle$ is time independent. If this $langle Prangle ne 0$, the particle must (in the average sense) drift either to the right or to the left and eventually escape to infinity, which cannot happen in a bound state.
The final sentence makes sense to me, but his reasoning in the second sentence does not. Aren't bound states and stationary states entirely different things? Does the one in fact imply the other?
quantum-mechanics hilbert-space terminology definition quantum-states
quantum-mechanics hilbert-space terminology definition quantum-states
edited Mar 24 at 2:06
Qmechanic♦
108k122001243
asked Mar 24 at 1:20
J-JJ-J
636
edited Mar 24 at 2:06
Qmechanic♦
108k122001243
asked Mar 24 at 1:20
J-JJ-J
636
edited Mar 24 at 2:06
Qmechanic♦
108k122001243
edited Mar 24 at 2:06
Qmechanic♦
108k122001243
edited Mar 24 at 2:06
Qmechanic♦
108k122001243
108k122001243
asked Mar 24 at 1:20
J-JJ-J
636
asked Mar 24 at 1:20
J-JJ-J
636
asked Mar 24 at 1:20
J-JJ-J
636
636
2
$begingroup$
I find that puzzling too because I would think that a state moving around in a potential is still a bound state. I guess Shankar is just using the words in a particular way.
$endgroup$
– DanielSank
Mar 24 at 1:25
add a comment |
2
$begingroup$
I find that puzzling too because I would think that a state moving around in a potential is still a bound state. I guess Shankar is just using the words in a particular way.
$endgroup$
– DanielSank
Mar 24 at 1:25
2
2
$begingroup$
I find that puzzling too because I would think that a state moving around in a potential is still a bound state. I guess Shankar is just using the words in a particular way.
$endgroup$
– DanielSank
Mar 24 at 1:25
$begingroup$
I find that puzzling too because I would think that a state moving around in a potential is still a bound state. I guess Shankar is just using the words in a particular way.
$endgroup$
– DanielSank
Mar 24 at 1:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think most of us would agree that superposition of bound states — say, of an electron in an atom — still deserves to be called a bound state, even though most such superpositions are time-dependent. The electron is still bound to the atom.
Based on the context from which the excerpt shown in the OP was extracted, it looks like Shankar is specifically talking about the ground state. The paragraph begins with
Let us now ... discuss the fact that the lowest energy is not zero...
(emphasis added by me), and the following paragraph ends with
The uncertainty principle is often used in this fashion to provide a quick order-of-magnitude estimate for the ground-state energy.
So although Shankar doesn't say it directly, the whole derivation seems to be focused on a particular stationary state, not a generic bound state. This inference is consistent with the fact that, just a few paragraphs earlier, Shankar writes
Bound states are thus characterized by $psi(x)to 0$ [as $|x|toinfty$] ... The energy levels of bound states are always quantized.
Shankar doesn't say that bound states always have sharply-defined energies, so none of this contradicts the usual convention that a superposition of bound states is still called a bound state, whether or not it happens to be stationary.
answered Mar 24 at 1:55
Chiral AnomalyChiral Anomaly
13.5k21845
$endgroup$
$begingroup$
What a thorough answer! Thank you so much!
$endgroup$
– user3518839
Mar 24 at 16:01
$begingroup$
Thanks, this makes much more sense
$endgroup$
– J-J
Mar 25 at 23:14
add a comment |
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$begingroup$
I think most of us would agree that superposition of bound states — say, of an electron in an atom — still deserves to be called a bound state, even though most such superpositions are time-dependent. The electron is still bound to the atom.
Based on the context from which the excerpt shown in the OP was extracted, it looks like Shankar is specifically talking about the ground state. The paragraph begins with
Let us now ... discuss the fact that the lowest energy is not zero...
(emphasis added by me), and the following paragraph ends with
The uncertainty principle is often used in this fashion to provide a quick order-of-magnitude estimate for the ground-state energy.
So although Shankar doesn't say it directly, the whole derivation seems to be focused on a particular stationary state, not a generic bound state. This inference is consistent with the fact that, just a few paragraphs earlier, Shankar writes
Bound states are thus characterized by $psi(x)to 0$ [as $|x|toinfty$] ... The energy levels of bound states are always quantized.
Shankar doesn't say that bound states always have sharply-defined energies, so none of this contradicts the usual convention that a superposition of bound states is still called a bound state, whether or not it happens to be stationary.
answered Mar 24 at 1:55
Chiral AnomalyChiral Anomaly
13.5k21845
$endgroup$
$begingroup$
What a thorough answer! Thank you so much!
$endgroup$
– user3518839
Mar 24 at 16:01
$begingroup$
Thanks, this makes much more sense
$endgroup$
– J-J
Mar 25 at 23:14
add a comment |
$begingroup$
I think most of us would agree that superposition of bound states — say, of an electron in an atom — still deserves to be called a bound state, even though most such superpositions are time-dependent. The electron is still bound to the atom.
Based on the context from which the excerpt shown in the OP was extracted, it looks like Shankar is specifically talking about the ground state. The paragraph begins with
Let us now ... discuss the fact that the lowest energy is not zero...
(emphasis added by me), and the following paragraph ends with
The uncertainty principle is often used in this fashion to provide a quick order-of-magnitude estimate for the ground-state energy.
So although Shankar doesn't say it directly, the whole derivation seems to be focused on a particular stationary state, not a generic bound state. This inference is consistent with the fact that, just a few paragraphs earlier, Shankar writes
Bound states are thus characterized by $psi(x)to 0$ [as $|x|toinfty$] ... The energy levels of bound states are always quantized.
Shankar doesn't say that bound states always have sharply-defined energies, so none of this contradicts the usual convention that a superposition of bound states is still called a bound state, whether or not it happens to be stationary.
answered Mar 24 at 1:55
Chiral AnomalyChiral Anomaly
13.5k21845
$endgroup$
$begingroup$
What a thorough answer! Thank you so much!
$endgroup$
– user3518839
Mar 24 at 16:01
$begingroup$
Thanks, this makes much more sense
$endgroup$
– J-J
Mar 25 at 23:14
add a comment |
$begingroup$
I think most of us would agree that superposition of bound states — say, of an electron in an atom — still deserves to be called a bound state, even though most such superpositions are time-dependent. The electron is still bound to the atom.
Based on the context from which the excerpt shown in the OP was extracted, it looks like Shankar is specifically talking about the ground state. The paragraph begins with
Let us now ... discuss the fact that the lowest energy is not zero...
(emphasis added by me), and the following paragraph ends with
The uncertainty principle is often used in this fashion to provide a quick order-of-magnitude estimate for the ground-state energy.
So although Shankar doesn't say it directly, the whole derivation seems to be focused on a particular stationary state, not a generic bound state. This inference is consistent with the fact that, just a few paragraphs earlier, Shankar writes
Bound states are thus characterized by $psi(x)to 0$ [as $|x|toinfty$] ... The energy levels of bound states are always quantized.
Shankar doesn't say that bound states always have sharply-defined energies, so none of this contradicts the usual convention that a superposition of bound states is still called a bound state, whether or not it happens to be stationary.
answered Mar 24 at 1:55
Chiral AnomalyChiral Anomaly
13.5k21845
$endgroup$
I think most of us would agree that superposition of bound states — say, of an electron in an atom — still deserves to be called a bound state, even though most such superpositions are time-dependent. The electron is still bound to the atom.
Based on the context from which the excerpt shown in the OP was extracted, it looks like Shankar is specifically talking about the ground state. The paragraph begins with
Let us now ... discuss the fact that the lowest energy is not zero...
(emphasis added by me), and the following paragraph ends with
The uncertainty principle is often used in this fashion to provide a quick order-of-magnitude estimate for the ground-state energy.
So although Shankar doesn't say it directly, the whole derivation seems to be focused on a particular stationary state, not a generic bound state. This inference is consistent with the fact that, just a few paragraphs earlier, Shankar writes
Bound states are thus characterized by $psi(x)to 0$ [as $|x|toinfty$] ... The energy levels of bound states are always quantized.
Shankar doesn't say that bound states always have sharply-defined energies, so none of this contradicts the usual convention that a superposition of bound states is still called a bound state, whether or not it happens to be stationary.
answered Mar 24 at 1:55
Chiral AnomalyChiral Anomaly
13.5k21845
answered Mar 24 at 1:55
Chiral AnomalyChiral Anomaly
13.5k21845
answered Mar 24 at 1:55
Chiral AnomalyChiral Anomaly
13.5k21845
answered Mar 24 at 1:55
Chiral AnomalyChiral Anomaly
13.5k21845
13.5k21845
$begingroup$
What a thorough answer! Thank you so much!
$endgroup$
– user3518839
Mar 24 at 16:01
$begingroup$
Thanks, this makes much more sense
$endgroup$
– J-J
Mar 25 at 23:14
add a comment |
$begingroup$
What a thorough answer! Thank you so much!
$endgroup$
– user3518839
Mar 24 at 16:01
$begingroup$
Thanks, this makes much more sense
$endgroup$
– J-J
Mar 25 at 23:14
$begingroup$
What a thorough answer! Thank you so much!
$endgroup$
– user3518839
Mar 24 at 16:01
$begingroup$
What a thorough answer! Thank you so much!
$endgroup$
– user3518839
Mar 24 at 16:01
$begingroup$
Thanks, this makes much more sense
$endgroup$
– J-J
Mar 25 at 23:14
$begingroup$
Thanks, this makes much more sense
$endgroup$
– J-J
Mar 25 at 23:14
add a comment |
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I find that puzzling too because I would think that a state moving around in a potential is still a bound state. I guess Shankar is just using the words in a particular way.
$endgroup$
– DanielSank
Mar 24 at 1:25