QM: Why is the geometric phase not zero?Unitary transformation between complete + orthonormal basesIs Ballentine's description of the Berry phase (in his book _Quantum Mechanics_) flawed?Unitary Transfomation from One Basis to AnotherPractical Calculation of Geometric PhaseProof: Elimination of the geometrical phase factors in the adiabatic approximationUnitary time conditionImportance of the Pancharatnam–Berry PhaseTransitionless quantum driving for specific eigenstatesperturbed wavefunctionsSolving the Schrodinger equation with a time-dependent Hamiltonian
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QM: Why is the geometric phase not zero?
Unitary transformation between complete + orthonormal basesIs Ballentine's description of the Berry phase (in his book _Quantum Mechanics_) flawed?Unitary Transfomation from One Basis to AnotherPractical Calculation of Geometric PhaseProof: Elimination of the geometrical phase factors in the adiabatic approximationUnitary time conditionImportance of the Pancharatnam–Berry PhaseTransitionless quantum driving for specific eigenstatesperturbed wavefunctionsSolving the Schrodinger equation with a time-dependent Hamiltonian
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I'm studying the adiabatic theorem and there is a derivation of a geometric phase factor which incorporates terms of the form $langle dot psi_n (t) | psi_n (t)rangle$ where the $psi_n(t)$ are orthonormal.
It seems clear to me that this should be zero. Differentiating the normalization condition $langle psi_n (t) | psi_n (t)rangle = 1$ (preserved with unitary evolution) causes the RHS to vanish. So zero for $langle dot psi_n (t) | psi_n (t)rangle$ and I'm at a loss as to how to understand the geometric phase which is heavily involved with these terms.
Does anyone know where I'm going wrong and why the geometric phase is not zero simply from differentiating the normalization condition?
quantum-mechanics
asked Jun 18 at 10:44
JamToastJamToast
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JamToast is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
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I'm studying the adiabatic theorem and there is a derivation of a geometric phase factor which incorporates terms of the form $langle dot psi_n (t) | psi_n (t)rangle$ where the $psi_n(t)$ are orthonormal.
It seems clear to me that this should be zero. Differentiating the normalization condition $langle psi_n (t) | psi_n (t)rangle = 1$ (preserved with unitary evolution) causes the RHS to vanish. So zero for $langle dot psi_n (t) | psi_n (t)rangle$ and I'm at a loss as to how to understand the geometric phase which is heavily involved with these terms.
Does anyone know where I'm going wrong and why the geometric phase is not zero simply from differentiating the normalization condition?
quantum-mechanics
asked Jun 18 at 10:44
JamToastJamToast
132 bronze badges
New contributor
JamToast is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
The product rule is $d/dt langle psi | psi rangle = langle dotpsi | psi rangle + langle psi | dotpsi rangle = 2Re(langle dotpsi|psirangle)$.
$endgroup$
– jacob1729
Jun 18 at 10:53
add a comment |
$begingroup$
I'm studying the adiabatic theorem and there is a derivation of a geometric phase factor which incorporates terms of the form $langle dot psi_n (t) | psi_n (t)rangle$ where the $psi_n(t)$ are orthonormal.
It seems clear to me that this should be zero. Differentiating the normalization condition $langle psi_n (t) | psi_n (t)rangle = 1$ (preserved with unitary evolution) causes the RHS to vanish. So zero for $langle dot psi_n (t) | psi_n (t)rangle$ and I'm at a loss as to how to understand the geometric phase which is heavily involved with these terms.
Does anyone know where I'm going wrong and why the geometric phase is not zero simply from differentiating the normalization condition?
quantum-mechanics
asked Jun 18 at 10:44
JamToastJamToast
132 bronze badges
New contributor
JamToast is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm studying the adiabatic theorem and there is a derivation of a geometric phase factor which incorporates terms of the form $langle dot psi_n (t) | psi_n (t)rangle$ where the $psi_n(t)$ are orthonormal.
It seems clear to me that this should be zero. Differentiating the normalization condition $langle psi_n (t) | psi_n (t)rangle = 1$ (preserved with unitary evolution) causes the RHS to vanish. So zero for $langle dot psi_n (t) | psi_n (t)rangle$ and I'm at a loss as to how to understand the geometric phase which is heavily involved with these terms.
Does anyone know where I'm going wrong and why the geometric phase is not zero simply from differentiating the normalization condition?
quantum-mechanics
quantum-mechanics
asked Jun 18 at 10:44
JamToastJamToast
132 bronze badges
New contributor
JamToast is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jun 18 at 10:44
JamToastJamToast
132 bronze badges
New contributor
JamToast is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jun 18 at 10:44
JamToastJamToast
132 bronze badges
New contributor
JamToast is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jun 18 at 10:44
JamToastJamToast
132 bronze badges
asked Jun 18 at 10:44
JamToastJamToast
132 bronze badges
132 bronze badges
New contributor
JamToast is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
JamToast is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
The product rule is $d/dt langle psi | psi rangle = langle dotpsi | psi rangle + langle psi | dotpsi rangle = 2Re(langle dotpsi|psirangle)$.
$endgroup$
– jacob1729
Jun 18 at 10:53
add a comment |
$begingroup$
The product rule is $d/dt langle psi | psi rangle = langle dotpsi | psi rangle + langle psi | dotpsi rangle = 2Re(langle dotpsi|psirangle)$.
$endgroup$
– jacob1729
Jun 18 at 10:53
$begingroup$
The product rule is $d/dt langle psi | psi rangle = langle dotpsi | psi rangle + langle psi | dotpsi rangle = 2Re(langle dotpsi|psirangle)$.
$endgroup$
– jacob1729
Jun 18 at 10:53
$begingroup$
The product rule is $d/dt langle psi | psi rangle = langle dotpsi | psi rangle + langle psi | dotpsi rangle = 2Re(langle dotpsi|psirangle)$.
$endgroup$
– jacob1729
Jun 18 at 10:53
add a comment |
1 Answer
1
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You are forgetting that both the bra and the ket have a $t$ dependence. We can differentiate $langle psi(t)|psi(t)rangle$ using the product rule
beginalign
fracddtlangle psi(t)|psi(t)rangle &= langledotpsi(t)|psi(t)rangle + langle psi(t)| dotpsi(t)rangle\
&= 0
endalign
therefore we can say
beginalign
langledotpsi(t)|psi(t)rangle &= -langle psi(t)| dotpsi(t)rangle\
&= -langledotpsi(t)|psi(t)rangle^*;.
endalign
This implies that the real part of $langledotpsi(t)|psi(t)rangle$ must be $0$, but the imaginary part can be non-zero.
answered Jun 18 at 10:54
By SymmetryBy Symmetry
4,8942 gold badges14 silver badges28 bronze badges
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I knew it would be something basic! Thank you. :)
$endgroup$
– JamToast
Jun 18 at 11:00
add a comment |
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$begingroup$
You are forgetting that both the bra and the ket have a $t$ dependence. We can differentiate $langle psi(t)|psi(t)rangle$ using the product rule
beginalign
fracddtlangle psi(t)|psi(t)rangle &= langledotpsi(t)|psi(t)rangle + langle psi(t)| dotpsi(t)rangle\
&= 0
endalign
therefore we can say
beginalign
langledotpsi(t)|psi(t)rangle &= -langle psi(t)| dotpsi(t)rangle\
&= -langledotpsi(t)|psi(t)rangle^*;.
endalign
This implies that the real part of $langledotpsi(t)|psi(t)rangle$ must be $0$, but the imaginary part can be non-zero.
answered Jun 18 at 10:54
By SymmetryBy Symmetry
4,8942 gold badges14 silver badges28 bronze badges
$endgroup$
$begingroup$
I knew it would be something basic! Thank you. :)
$endgroup$
– JamToast
Jun 18 at 11:00
add a comment |
$begingroup$
You are forgetting that both the bra and the ket have a $t$ dependence. We can differentiate $langle psi(t)|psi(t)rangle$ using the product rule
beginalign
fracddtlangle psi(t)|psi(t)rangle &= langledotpsi(t)|psi(t)rangle + langle psi(t)| dotpsi(t)rangle\
&= 0
endalign
therefore we can say
beginalign
langledotpsi(t)|psi(t)rangle &= -langle psi(t)| dotpsi(t)rangle\
&= -langledotpsi(t)|psi(t)rangle^*;.
endalign
This implies that the real part of $langledotpsi(t)|psi(t)rangle$ must be $0$, but the imaginary part can be non-zero.
answered Jun 18 at 10:54
By SymmetryBy Symmetry
4,8942 gold badges14 silver badges28 bronze badges
$endgroup$
$begingroup$
I knew it would be something basic! Thank you. :)
$endgroup$
– JamToast
Jun 18 at 11:00
add a comment |
$begingroup$
You are forgetting that both the bra and the ket have a $t$ dependence. We can differentiate $langle psi(t)|psi(t)rangle$ using the product rule
beginalign
fracddtlangle psi(t)|psi(t)rangle &= langledotpsi(t)|psi(t)rangle + langle psi(t)| dotpsi(t)rangle\
&= 0
endalign
therefore we can say
beginalign
langledotpsi(t)|psi(t)rangle &= -langle psi(t)| dotpsi(t)rangle\
&= -langledotpsi(t)|psi(t)rangle^*;.
endalign
This implies that the real part of $langledotpsi(t)|psi(t)rangle$ must be $0$, but the imaginary part can be non-zero.
answered Jun 18 at 10:54
By SymmetryBy Symmetry
4,8942 gold badges14 silver badges28 bronze badges
$endgroup$
You are forgetting that both the bra and the ket have a $t$ dependence. We can differentiate $langle psi(t)|psi(t)rangle$ using the product rule
beginalign
fracddtlangle psi(t)|psi(t)rangle &= langledotpsi(t)|psi(t)rangle + langle psi(t)| dotpsi(t)rangle\
&= 0
endalign
therefore we can say
beginalign
langledotpsi(t)|psi(t)rangle &= -langle psi(t)| dotpsi(t)rangle\
&= -langledotpsi(t)|psi(t)rangle^*;.
endalign
This implies that the real part of $langledotpsi(t)|psi(t)rangle$ must be $0$, but the imaginary part can be non-zero.
answered Jun 18 at 10:54
By SymmetryBy Symmetry
4,8942 gold badges14 silver badges28 bronze badges
answered Jun 18 at 10:54
By SymmetryBy Symmetry
4,8942 gold badges14 silver badges28 bronze badges
answered Jun 18 at 10:54
By SymmetryBy Symmetry
4,8942 gold badges14 silver badges28 bronze badges
answered Jun 18 at 10:54
By SymmetryBy Symmetry
4,8942 gold badges14 silver badges28 bronze badges
4,8942 gold badges14 silver badges28 bronze badges
$begingroup$
I knew it would be something basic! Thank you. :)
$endgroup$
– JamToast
Jun 18 at 11:00
add a comment |
$begingroup$
I knew it would be something basic! Thank you. :)
$endgroup$
– JamToast
Jun 18 at 11:00
$begingroup$
I knew it would be something basic! Thank you. :)
$endgroup$
– JamToast
Jun 18 at 11:00
$begingroup$
I knew it would be something basic! Thank you. :)
$endgroup$
– JamToast
Jun 18 at 11:00
add a comment |
JamToast is a new contributor. Be nice, and check out our Code of Conduct.
JamToast is a new contributor. Be nice, and check out our Code of Conduct.
JamToast is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
The product rule is $d/dt langle psi | psi rangle = langle dotpsi | psi rangle + langle psi | dotpsi rangle = 2Re(langle dotpsi|psirangle)$.
$endgroup$
– jacob1729
Jun 18 at 10:53