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Why does the position uncertainty of a harmonic oscillator not have the expectation value squared term?
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My question is about the derivation for the uncertainty of a quantum harmonic oscillator in the non-zero ground state energy. In my textbook A modern Approach to Quantum Mechanics by John S. Townsend there is a discussion about the position uncertainty $(Delta x)^2= big<x^2 big>+big<xbig>^2$ for a harmonic oscillator in the ground state energy.
We have established for a harmonic oscillator
$hat x=sqrt hbarover 2momega(hat a+hat a^dagger)$ so $big<xbig>=big<n big|sqrt hbarover 2momega(hat a+hat a^dagger) big|nbig>$ which gives us $big<x^2big>=big<nbig|hbarover 2momega(hat a+hat a^dagger)^2big|nbig>$.
And in a similar fashion $big<xbig>^2=big<n big|sqrt hbarover 2momega(hat a+hat a^dagger) big|nbig>^2$.
To me this means that $(Delta x)^2=big<nbig|hbarover 2momega(hat a+hat a^dagger)^2big|nbig> + big<n big|sqrt hbarover 2momega(hat a+hat a^dagger) big|nbig>^2$. However, the book seems to drop (with no explanation) $big<xbig>^2$ and comes up with $(Delta x)^2=big<0big|hbarover 2momega(hat a+hat a^dagger)^2big|0big>$. I know we are talking about ground state so I'm assuming that's why the $n$ was replaced with $0$, however as far as I can tell we just have $big<x^2big>$. Why is the other term dropped? Is this something to do with the oscillator being in the ground state?
quantum-mechanics harmonic-oscillator
asked 2 hours ago
matryoshka
284316
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up vote
1
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My question is about the derivation for the uncertainty of a quantum harmonic oscillator in the non-zero ground state energy. In my textbook A modern Approach to Quantum Mechanics by John S. Townsend there is a discussion about the position uncertainty $(Delta x)^2= big<x^2 big>+big<xbig>^2$ for a harmonic oscillator in the ground state energy.
We have established for a harmonic oscillator
$hat x=sqrt hbarover 2momega(hat a+hat a^dagger)$ so $big<xbig>=big<n big|sqrt hbarover 2momega(hat a+hat a^dagger) big|nbig>$ which gives us $big<x^2big>=big<nbig|hbarover 2momega(hat a+hat a^dagger)^2big|nbig>$.
And in a similar fashion $big<xbig>^2=big<n big|sqrt hbarover 2momega(hat a+hat a^dagger) big|nbig>^2$.
To me this means that $(Delta x)^2=big<nbig|hbarover 2momega(hat a+hat a^dagger)^2big|nbig> + big<n big|sqrt hbarover 2momega(hat a+hat a^dagger) big|nbig>^2$. However, the book seems to drop (with no explanation) $big<xbig>^2$ and comes up with $(Delta x)^2=big<0big|hbarover 2momega(hat a+hat a^dagger)^2big|0big>$. I know we are talking about ground state so I'm assuming that's why the $n$ was replaced with $0$, however as far as I can tell we just have $big<x^2big>$. Why is the other term dropped? Is this something to do with the oscillator being in the ground state?
quantum-mechanics harmonic-oscillator
asked 2 hours ago
matryoshka
284316
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
My question is about the derivation for the uncertainty of a quantum harmonic oscillator in the non-zero ground state energy. In my textbook A modern Approach to Quantum Mechanics by John S. Townsend there is a discussion about the position uncertainty $(Delta x)^2= big<x^2 big>+big<xbig>^2$ for a harmonic oscillator in the ground state energy.
We have established for a harmonic oscillator
$hat x=sqrt hbarover 2momega(hat a+hat a^dagger)$ so $big<xbig>=big<n big|sqrt hbarover 2momega(hat a+hat a^dagger) big|nbig>$ which gives us $big<x^2big>=big<nbig|hbarover 2momega(hat a+hat a^dagger)^2big|nbig>$.
And in a similar fashion $big<xbig>^2=big<n big|sqrt hbarover 2momega(hat a+hat a^dagger) big|nbig>^2$.
To me this means that $(Delta x)^2=big<nbig|hbarover 2momega(hat a+hat a^dagger)^2big|nbig> + big<n big|sqrt hbarover 2momega(hat a+hat a^dagger) big|nbig>^2$. However, the book seems to drop (with no explanation) $big<xbig>^2$ and comes up with $(Delta x)^2=big<0big|hbarover 2momega(hat a+hat a^dagger)^2big|0big>$. I know we are talking about ground state so I'm assuming that's why the $n$ was replaced with $0$, however as far as I can tell we just have $big<x^2big>$. Why is the other term dropped? Is this something to do with the oscillator being in the ground state?
quantum-mechanics harmonic-oscillator
asked 2 hours ago
matryoshka
284316
My question is about the derivation for the uncertainty of a quantum harmonic oscillator in the non-zero ground state energy. In my textbook A modern Approach to Quantum Mechanics by John S. Townsend there is a discussion about the position uncertainty $(Delta x)^2= big<x^2 big>+big<xbig>^2$ for a harmonic oscillator in the ground state energy.
We have established for a harmonic oscillator
$hat x=sqrt hbarover 2momega(hat a+hat a^dagger)$ so $big<xbig>=big<n big|sqrt hbarover 2momega(hat a+hat a^dagger) big|nbig>$ which gives us $big<x^2big>=big<nbig|hbarover 2momega(hat a+hat a^dagger)^2big|nbig>$.
And in a similar fashion $big<xbig>^2=big<n big|sqrt hbarover 2momega(hat a+hat a^dagger) big|nbig>^2$.
To me this means that $(Delta x)^2=big<nbig|hbarover 2momega(hat a+hat a^dagger)^2big|nbig> + big<n big|sqrt hbarover 2momega(hat a+hat a^dagger) big|nbig>^2$. However, the book seems to drop (with no explanation) $big<xbig>^2$ and comes up with $(Delta x)^2=big<0big|hbarover 2momega(hat a+hat a^dagger)^2big|0big>$. I know we are talking about ground state so I'm assuming that's why the $n$ was replaced with $0$, however as far as I can tell we just have $big<x^2big>$. Why is the other term dropped? Is this something to do with the oscillator being in the ground state?
quantum-mechanics harmonic-oscillator
quantum-mechanics harmonic-oscillator
asked 2 hours ago
matryoshka
284316
asked 2 hours ago
matryoshka
284316
asked 2 hours ago
matryoshka
284316
asked 2 hours ago
matryoshka
284316
asked 2 hours ago
matryoshka
284316
284316
add a comment |Â
add a comment |Â
1 Answer
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::chuckles::
I remember being baffled by how this works out mathematically myself, and it's one of those "I can't believe it's that simple!" ones.
Three facts:
$hata$ and $hata^dagger$ are the lowering and raising ladder operators: they take a numbered state to one numbered either one less or one more that the starting state.- The numbered states are a set of eigenstates, so they are orthogonal to one another.
- Expand $left( hata + hata^dagger right)^2$, and see why it has a very different character than $left( hata + hata^dagger right)$.
answered 1 hour ago
dmckee♦
72.9k6128260
I just expanded them and I see exactly what you mean, I need to be more careful about how I use operators. Looking at this I can't see any way $<x>^2$ would ever contribute to the uncertainty, no matter what energy state it is in because of the lowering and raising ladder operations. Is that the case?
– matryoshka
1 hour ago
1
Yup. Or at least, "Yup for the harmonic oscillator". The situation where $hatx$ is composed of a sum of simple raising and lowering operators is special to the SHO.
– dmckee♦
1 hour ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
::chuckles::
I remember being baffled by how this works out mathematically myself, and it's one of those "I can't believe it's that simple!" ones.
Three facts:
$hata$ and $hata^dagger$ are the lowering and raising ladder operators: they take a numbered state to one numbered either one less or one more that the starting state.- The numbered states are a set of eigenstates, so they are orthogonal to one another.
- Expand $left( hata + hata^dagger right)^2$, and see why it has a very different character than $left( hata + hata^dagger right)$.
answered 1 hour ago
dmckee♦
72.9k6128260
I just expanded them and I see exactly what you mean, I need to be more careful about how I use operators. Looking at this I can't see any way $<x>^2$ would ever contribute to the uncertainty, no matter what energy state it is in because of the lowering and raising ladder operations. Is that the case?
– matryoshka
1 hour ago
1
Yup. Or at least, "Yup for the harmonic oscillator". The situation where $hatx$ is composed of a sum of simple raising and lowering operators is special to the SHO.
– dmckee♦
1 hour ago
add a comment |Â
up vote
3
down vote
accepted
::chuckles::
I remember being baffled by how this works out mathematically myself, and it's one of those "I can't believe it's that simple!" ones.
Three facts:
$hata$ and $hata^dagger$ are the lowering and raising ladder operators: they take a numbered state to one numbered either one less or one more that the starting state.- The numbered states are a set of eigenstates, so they are orthogonal to one another.
- Expand $left( hata + hata^dagger right)^2$, and see why it has a very different character than $left( hata + hata^dagger right)$.
answered 1 hour ago
dmckee♦
72.9k6128260
I just expanded them and I see exactly what you mean, I need to be more careful about how I use operators. Looking at this I can't see any way $<x>^2$ would ever contribute to the uncertainty, no matter what energy state it is in because of the lowering and raising ladder operations. Is that the case?
– matryoshka
1 hour ago
1
Yup. Or at least, "Yup for the harmonic oscillator". The situation where $hatx$ is composed of a sum of simple raising and lowering operators is special to the SHO.
– dmckee♦
1 hour ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
::chuckles::
I remember being baffled by how this works out mathematically myself, and it's one of those "I can't believe it's that simple!" ones.
Three facts:
$hata$ and $hata^dagger$ are the lowering and raising ladder operators: they take a numbered state to one numbered either one less or one more that the starting state.- The numbered states are a set of eigenstates, so they are orthogonal to one another.
- Expand $left( hata + hata^dagger right)^2$, and see why it has a very different character than $left( hata + hata^dagger right)$.
answered 1 hour ago
dmckee♦
72.9k6128260
::chuckles::
I remember being baffled by how this works out mathematically myself, and it's one of those "I can't believe it's that simple!" ones.
Three facts:
$hata$ and $hata^dagger$ are the lowering and raising ladder operators: they take a numbered state to one numbered either one less or one more that the starting state.- The numbered states are a set of eigenstates, so they are orthogonal to one another.
- Expand $left( hata + hata^dagger right)^2$, and see why it has a very different character than $left( hata + hata^dagger right)$.
answered 1 hour ago
dmckee♦
72.9k6128260
edited 1 hour ago
answered 1 hour ago
dmckee♦
72.9k6128260
answered 1 hour ago
dmckee♦
72.9k6128260
answered 1 hour ago
dmckee♦
72.9k6128260
72.9k6128260
I just expanded them and I see exactly what you mean, I need to be more careful about how I use operators. Looking at this I can't see any way $<x>^2$ would ever contribute to the uncertainty, no matter what energy state it is in because of the lowering and raising ladder operations. Is that the case?
– matryoshka
1 hour ago
1
Yup. Or at least, "Yup for the harmonic oscillator". The situation where $hatx$ is composed of a sum of simple raising and lowering operators is special to the SHO.
– dmckee♦
1 hour ago
add a comment |Â
I just expanded them and I see exactly what you mean, I need to be more careful about how I use operators. Looking at this I can't see any way $<x>^2$ would ever contribute to the uncertainty, no matter what energy state it is in because of the lowering and raising ladder operations. Is that the case?
– matryoshka
1 hour ago
1
Yup. Or at least, "Yup for the harmonic oscillator". The situation where $hatx$ is composed of a sum of simple raising and lowering operators is special to the SHO.
– dmckee♦
1 hour ago
I just expanded them and I see exactly what you mean, I need to be more careful about how I use operators. Looking at this I can't see any way $<x>^2$ would ever contribute to the uncertainty, no matter what energy state it is in because of the lowering and raising ladder operations. Is that the case?
– matryoshka
1 hour ago
I just expanded them and I see exactly what you mean, I need to be more careful about how I use operators. Looking at this I can't see any way $<x>^2$ would ever contribute to the uncertainty, no matter what energy state it is in because of the lowering and raising ladder operations. Is that the case?
– matryoshka
1 hour ago
1
1
Yup. Or at least, "Yup for the harmonic oscillator". The situation where $hatx$ is composed of a sum of simple raising and lowering operators is special to the SHO.
– dmckee♦
1 hour ago
Yup. Or at least, "Yup for the harmonic oscillator". The situation where $hatx$ is composed of a sum of simple raising and lowering operators is special to the SHO.
– dmckee♦
1 hour ago
add a comment |Â
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