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Is a bound state a stationary state?

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2

$begingroup$

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

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edited 2 hours ago

Qmechanic♦

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asked 3 hours ago

J-JJ-J

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$endgroup$

• 
  
  
  

  
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  $begingroup$
  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
  3 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

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

share|cite|improve this question

edited 2 hours ago

Qmechanic♦

106k121961226

asked 3 hours ago

J-JJ-J

586

$endgroup$

• 
  
  
  

  
  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
  3 hours ago
  

  
  
  
  

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

share|cite|improve this question

edited 2 hours ago

Qmechanic♦

106k121961226

asked 3 hours ago

J-JJ-J

586

$endgroup$

share|cite|improve this question

edited 2 hours ago

Qmechanic♦

106k121961226

asked 3 hours ago

J-JJ-J

586

share|cite|improve this question

edited 2 hours ago

Qmechanic♦

106k121961226

asked 3 hours ago

J-JJ-J

586

edited 2 hours ago

Qmechanic♦

106k121961226

edited 2 hours ago

Qmechanic♦

106k121961226

asked 3 hours ago

J-JJ-J

586

asked 3 hours ago

J-JJ-J

586

1 Answer
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2

$begingroup$

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.

share|cite|improve this answer

answered 3 hours ago

Chiral AnomalyChiral Anomaly

12.4k21541

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add a comment | 

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2

$begingroup$

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.

share|cite|improve this answer

answered 3 hours ago

Chiral AnomalyChiral Anomaly

12.4k21541

$endgroup$

add a comment | 

2

$begingroup$

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.

share|cite|improve this answer

answered 3 hours ago

Chiral AnomalyChiral Anomaly

12.4k21541

$endgroup$

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.

share|cite|improve this answer

answered 3 hours ago

Chiral AnomalyChiral Anomaly

12.4k21541

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Chiral AnomalyChiral Anomaly

12.4k21541

answered 3 hours ago

Chiral AnomalyChiral Anomaly

12.4k21541

answered 3 hours ago

Chiral AnomalyChiral Anomaly

12.4k21541