Sfdwhf

How to implement a feedback to keep the DC gain at zero for this conceptual passive filter?

How can Trident be so inexpensive? Will it orbit Triton or just do a (slow) flyby?

Why do compilers behave differently when static_cast(ing) a function to void*?

A social experiment. What is the worst that can happen?

Should I stop contributing to retirement accounts?

Does a 'pending' US visa application constitute a denial?

Why should universal income be universal?

Multiplicative persistence

What was the exact wording from Ivanhoe of this advice on how to free yourself from slavery?

Travelling outside the UK without a passport

What is Cash Advance APR?

why `nmap 192.168.1.97` returns less services than `nmap 127.0.0.1`?

Drawing ramified coverings with tikz

Where does the bonus feat in the cleric starting package come from?

How should I respond when I lied about my education and the company finds out through background check?

GraphicsGrid with a Label for each Column and Row

Is it possible to have a strip of cold climate in the middle of a planet?

Store Credit Card Information in Password Manager?

"Spoil" vs "Ruin"

What is this called? Old film camera viewer?

Lowest total scrabble score

Open a doc from terminal, but not by its name

Removing files under particular conditions (number of files, file age)

The screen of my macbook suddenly broken down how can I do to recover

Is a bound state a stationary state?

It 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?

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 | 

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
1

active

oldest

votes

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 2 hours ago

Chiral AnomalyChiral Anomaly

12.4k21541

$endgroup$

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
});


}
});

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f468307%2fis-a-bound-state-a-stationary-state%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

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 2 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 2 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 2 hours ago

Chiral AnomalyChiral Anomaly

12.4k21541

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Chiral AnomalyChiral Anomaly

12.4k21541

answered 2 hours ago

Chiral AnomalyChiral Anomaly

12.4k21541

answered 2 hours ago

Chiral AnomalyChiral Anomaly

12.4k21541