Mixing
Two different definitions of ladder operator for Harmonic Oscillators
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As it happened, I accidentally referred to two different editions of Introduction to QM by Griffiths. In the second chapter, while defining the ladder operator for harmonic oscillators, he used different terms. Now different definitions mean that their commutator change. Which of the two operators should I use? Also, why is there ambiguity in the definition of the same?
$$a_pm equiv frac 1 sqrt2mbigg( frac hbari frac ddx pm im omega xbigg)$$
$$a_pm equiv frac 1 sqrt2hbar m omegabigg(mp ip + m omega xbigg)$$
quantum-mechanics operators harmonic-oscillator
edited 1 hour ago
Hugo V
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asked 2 hours ago
Ashutosh Singh
236
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up vote
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As it happened, I accidentally referred to two different editions of Introduction to QM by Griffiths. In the second chapter, while defining the ladder operator for harmonic oscillators, he used different terms. Now different definitions mean that their commutator change. Which of the two operators should I use? Also, why is there ambiguity in the definition of the same?
$$a_pm equiv frac 1 sqrt2mbigg( frac hbari frac ddx pm im omega xbigg)$$
$$a_pm equiv frac 1 sqrt2hbar m omegabigg(mp ip + m omega xbigg)$$
quantum-mechanics operators harmonic-oscillator
edited 1 hour ago
Hugo V
1,053214
asked 2 hours ago
Ashutosh Singh
236
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
As it happened, I accidentally referred to two different editions of Introduction to QM by Griffiths. In the second chapter, while defining the ladder operator for harmonic oscillators, he used different terms. Now different definitions mean that their commutator change. Which of the two operators should I use? Also, why is there ambiguity in the definition of the same?
$$a_pm equiv frac 1 sqrt2mbigg( frac hbari frac ddx pm im omega xbigg)$$
$$a_pm equiv frac 1 sqrt2hbar m omegabigg(mp ip + m omega xbigg)$$
quantum-mechanics operators harmonic-oscillator
edited 1 hour ago
Hugo V
1,053214
asked 2 hours ago
Ashutosh Singh
236
As it happened, I accidentally referred to two different editions of Introduction to QM by Griffiths. In the second chapter, while defining the ladder operator for harmonic oscillators, he used different terms. Now different definitions mean that their commutator change. Which of the two operators should I use? Also, why is there ambiguity in the definition of the same?
$$a_pm equiv frac 1 sqrt2mbigg( frac hbari frac ddx pm im omega xbigg)$$
$$a_pm equiv frac 1 sqrt2hbar m omegabigg(mp ip + m omega xbigg)$$
quantum-mechanics operators harmonic-oscillator
quantum-mechanics operators harmonic-oscillator
edited 1 hour ago
Hugo V
1,053214
asked 2 hours ago
Ashutosh Singh
236
edited 1 hour ago
Hugo V
1,053214
asked 2 hours ago
Ashutosh Singh
236
edited 1 hour ago
Hugo V
1,053214
edited 1 hour ago
Hugo V
1,053214
edited 1 hour ago
Hugo V
1,053214
1,053214
asked 2 hours ago
Ashutosh Singh
236
asked 2 hours ago
Ashutosh Singh
236
asked 2 hours ago
Ashutosh Singh
236
236
add a comment |Â
add a comment |Â
2 Answers
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up vote
4
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Your second equation is the standard definition for $a$ ($a_-$) and $a^dagger$ ($a_+$) as found e.g. on Wikipedia. Note that the ladder operators here are dimensionless.
Griffiths is doing it slightly differently. I'm not sure why, but if you continue consistently with this alternative definition, you will get to the same correct results of course. For example, Griffiths (2.45) is
$$ [a_-, a_+] = hbaromega , $$
compare with the standard $[a, a^dagger] = 1$.
answered 1 hour ago
Noiralef
3,495926
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up vote
1
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In the first definition, $a_+$ and $a_-$ have been defined such that $[a_-,a_+]=hbaromega$. Here the position and momentum have been scaled as
$X=sqrtmx$ & $P=sqrtfrac1mp$ and $a_pm= frac1sqrt2(X pmiP)$.
Whereas, in the second definition, $a_+$ and $a_-$ have been defined such that $[a_-,a_+]=1$. Here the position and momentum have been scaled as
$X=sqrtfracmomegahbarx$ & $P=sqrtfrac1momegahbarp$ and $a_pm=frac1sqrt2(XpmiP)$.
answered 1 hour ago
Bhavesh Valecha
112
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Bhavesh Valecha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Your second equation is the standard definition for $a$ ($a_-$) and $a^dagger$ ($a_+$) as found e.g. on Wikipedia. Note that the ladder operators here are dimensionless.
Griffiths is doing it slightly differently. I'm not sure why, but if you continue consistently with this alternative definition, you will get to the same correct results of course. For example, Griffiths (2.45) is
$$ [a_-, a_+] = hbaromega , $$
compare with the standard $[a, a^dagger] = 1$.
answered 1 hour ago
Noiralef
3,495926
add a comment |Â
up vote
4
down vote
Your second equation is the standard definition for $a$ ($a_-$) and $a^dagger$ ($a_+$) as found e.g. on Wikipedia. Note that the ladder operators here are dimensionless.
Griffiths is doing it slightly differently. I'm not sure why, but if you continue consistently with this alternative definition, you will get to the same correct results of course. For example, Griffiths (2.45) is
$$ [a_-, a_+] = hbaromega , $$
compare with the standard $[a, a^dagger] = 1$.
answered 1 hour ago
Noiralef
3,495926
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Your second equation is the standard definition for $a$ ($a_-$) and $a^dagger$ ($a_+$) as found e.g. on Wikipedia. Note that the ladder operators here are dimensionless.
Griffiths is doing it slightly differently. I'm not sure why, but if you continue consistently with this alternative definition, you will get to the same correct results of course. For example, Griffiths (2.45) is
$$ [a_-, a_+] = hbaromega , $$
compare with the standard $[a, a^dagger] = 1$.
answered 1 hour ago
Noiralef
3,495926
Your second equation is the standard definition for $a$ ($a_-$) and $a^dagger$ ($a_+$) as found e.g. on Wikipedia. Note that the ladder operators here are dimensionless.
Griffiths is doing it slightly differently. I'm not sure why, but if you continue consistently with this alternative definition, you will get to the same correct results of course. For example, Griffiths (2.45) is
$$ [a_-, a_+] = hbaromega , $$
compare with the standard $[a, a^dagger] = 1$.
answered 1 hour ago
Noiralef
3,495926
answered 1 hour ago
Noiralef
3,495926
answered 1 hour ago
Noiralef
3,495926
answered 1 hour ago
Noiralef
3,495926
3,495926
add a comment |Â
add a comment |Â
up vote
1
down vote
In the first definition, $a_+$ and $a_-$ have been defined such that $[a_-,a_+]=hbaromega$. Here the position and momentum have been scaled as
$X=sqrtmx$ & $P=sqrtfrac1mp$ and $a_pm= frac1sqrt2(X pmiP)$.
Whereas, in the second definition, $a_+$ and $a_-$ have been defined such that $[a_-,a_+]=1$. Here the position and momentum have been scaled as
$X=sqrtfracmomegahbarx$ & $P=sqrtfrac1momegahbarp$ and $a_pm=frac1sqrt2(XpmiP)$.
answered 1 hour ago
Bhavesh Valecha
112
New contributor
Bhavesh Valecha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
In the first definition, $a_+$ and $a_-$ have been defined such that $[a_-,a_+]=hbaromega$. Here the position and momentum have been scaled as
$X=sqrtmx$ & $P=sqrtfrac1mp$ and $a_pm= frac1sqrt2(X pmiP)$.
Whereas, in the second definition, $a_+$ and $a_-$ have been defined such that $[a_-,a_+]=1$. Here the position and momentum have been scaled as
$X=sqrtfracmomegahbarx$ & $P=sqrtfrac1momegahbarp$ and $a_pm=frac1sqrt2(XpmiP)$.
answered 1 hour ago
Bhavesh Valecha
112
New contributor
Bhavesh Valecha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
In the first definition, $a_+$ and $a_-$ have been defined such that $[a_-,a_+]=hbaromega$. Here the position and momentum have been scaled as
$X=sqrtmx$ & $P=sqrtfrac1mp$ and $a_pm= frac1sqrt2(X pmiP)$.
Whereas, in the second definition, $a_+$ and $a_-$ have been defined such that $[a_-,a_+]=1$. Here the position and momentum have been scaled as
$X=sqrtfracmomegahbarx$ & $P=sqrtfrac1momegahbarp$ and $a_pm=frac1sqrt2(XpmiP)$.
answered 1 hour ago
Bhavesh Valecha
112
New contributor
Bhavesh Valecha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In the first definition, $a_+$ and $a_-$ have been defined such that $[a_-,a_+]=hbaromega$. Here the position and momentum have been scaled as
$X=sqrtmx$ & $P=sqrtfrac1mp$ and $a_pm= frac1sqrt2(X pmiP)$.
Whereas, in the second definition, $a_+$ and $a_-$ have been defined such that $[a_-,a_+]=1$. Here the position and momentum have been scaled as
$X=sqrtfracmomegahbarx$ & $P=sqrtfrac1momegahbarp$ and $a_pm=frac1sqrt2(XpmiP)$.
answered 1 hour ago
Bhavesh Valecha
112
New contributor
Bhavesh Valecha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 14 mins ago
answered 1 hour ago
Bhavesh Valecha
112
New contributor
Bhavesh Valecha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
Bhavesh Valecha
112
answered 1 hour ago
Bhavesh Valecha
112
112
New contributor
Bhavesh Valecha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Bhavesh Valecha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Bhavesh Valecha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
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