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An imaginary coefficient in heat equation ensures no special direction in time?
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In Pauli(1973): Wave Mechanics (Pauli lectures on physics Volume 5) he derives on page 4 the Schrödinger equation (similar to Tong: Lectures on QFT, p. 42)
$$
nabla^2 psi' + i frac2mhfracpartial psi'partial t - frac1c^2fracpartial^2 psi' partial t^2 = 0
$$
(the last term vanishes in the limit $c rightarrow infty$) and makes the following peculiar remark:
Aside from the imaginary coefficient, it corresponds to the equation
of heat conduction. The imaginary coefficient assures that there is no
special direction in time; [2.11] is invariant under the transformation $t rightarrow -t, psi' rightarrow psi'^*$, whereby $psi^*psi$ remains unchanged.
What does he mean by "the imaginary coefficient assures there is no special direction in time"? Does the real heat equation without imaginary coefficient have a special direction in time? What does direction mean? Forward or backward? Or a direction/angle relative to space?
I pretty confused. But knowing Pauli his remarks usually hold some deep insights I would like to understand.
quantum-mechanics thermodynamics quantum-field-theory time
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asmaier
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up vote
9
down vote
favorite
In Pauli(1973): Wave Mechanics (Pauli lectures on physics Volume 5) he derives on page 4 the Schrödinger equation (similar to Tong: Lectures on QFT, p. 42)
$$
nabla^2 psi' + i frac2mhfracpartial psi'partial t - frac1c^2fracpartial^2 psi' partial t^2 = 0
$$
(the last term vanishes in the limit $c rightarrow infty$) and makes the following peculiar remark:
Aside from the imaginary coefficient, it corresponds to the equation
of heat conduction. The imaginary coefficient assures that there is no
special direction in time; [2.11] is invariant under the transformation $t rightarrow -t, psi' rightarrow psi'^*$, whereby $psi^*psi$ remains unchanged.
What does he mean by "the imaginary coefficient assures there is no special direction in time"? Does the real heat equation without imaginary coefficient have a special direction in time? What does direction mean? Forward or backward? Or a direction/angle relative to space?
I pretty confused. But knowing Pauli his remarks usually hold some deep insights I would like to understand.
quantum-mechanics thermodynamics quantum-field-theory time
asked yesterday
asmaier
3,8573072
add a comment |Â
up vote
9
down vote
favorite
up vote
9
down vote
favorite
In Pauli(1973): Wave Mechanics (Pauli lectures on physics Volume 5) he derives on page 4 the Schrödinger equation (similar to Tong: Lectures on QFT, p. 42)
$$
nabla^2 psi' + i frac2mhfracpartial psi'partial t - frac1c^2fracpartial^2 psi' partial t^2 = 0
$$
(the last term vanishes in the limit $c rightarrow infty$) and makes the following peculiar remark:
Aside from the imaginary coefficient, it corresponds to the equation
of heat conduction. The imaginary coefficient assures that there is no
special direction in time; [2.11] is invariant under the transformation $t rightarrow -t, psi' rightarrow psi'^*$, whereby $psi^*psi$ remains unchanged.
What does he mean by "the imaginary coefficient assures there is no special direction in time"? Does the real heat equation without imaginary coefficient have a special direction in time? What does direction mean? Forward or backward? Or a direction/angle relative to space?
I pretty confused. But knowing Pauli his remarks usually hold some deep insights I would like to understand.
quantum-mechanics thermodynamics quantum-field-theory time
asked yesterday
asmaier
3,8573072
In Pauli(1973): Wave Mechanics (Pauli lectures on physics Volume 5) he derives on page 4 the Schrödinger equation (similar to Tong: Lectures on QFT, p. 42)
$$
nabla^2 psi' + i frac2mhfracpartial psi'partial t - frac1c^2fracpartial^2 psi' partial t^2 = 0
$$
(the last term vanishes in the limit $c rightarrow infty$) and makes the following peculiar remark:
Aside from the imaginary coefficient, it corresponds to the equation
of heat conduction. The imaginary coefficient assures that there is no
special direction in time; [2.11] is invariant under the transformation $t rightarrow -t, psi' rightarrow psi'^*$, whereby $psi^*psi$ remains unchanged.
What does he mean by "the imaginary coefficient assures there is no special direction in time"? Does the real heat equation without imaginary coefficient have a special direction in time? What does direction mean? Forward or backward? Or a direction/angle relative to space?
I pretty confused. But knowing Pauli his remarks usually hold some deep insights I would like to understand.
quantum-mechanics thermodynamics quantum-field-theory time
quantum-mechanics thermodynamics quantum-field-theory time
asked yesterday
asmaier
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asked yesterday
asmaier
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asked yesterday
asmaier
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asked yesterday
asmaier
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2 Answers
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Does the real heat equation have a special direction in time?
Yes, it does, because heat spreads out forward in time, and concentrates itself backwards in time. Mathematically, under time reversal $t to -t$ a first time derivative picks up a minus sign, which is why the heat equation is not time reversal invariant (TRI). It's the same principle as friction $F = - bv$ picking out a time direction.
What does he mean by "the imaginary coefficient assures there is no special direction in time"?
The wave equation only has second time derivatives, and hence is TRI. But it's hard to see how an equation that is first order in time could possibly be TRI. This is a problem, as the Schrodinger equation has to be first order in time, and must be TRI to reduce to classical mechanics.
The trick in quantum mechanics is that the gradient of the phase is tied to momentum. That is, classically we must flip $p to -p$ to reverse time, but since a quantum particle with momentum $p$ has wavefunction $e^ipx$, this must be realized by
$$psi to psi^*$$
under time reversal, or equivalently $i to -i$. The sign picked up by the first time derivative is thus canceled by the sign picked up by $i$, making the Schrodinger equation TRI as it must be.
answered yesterday
knzhou
33.8k897170
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up vote
8
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I will just paraphrase Pauli.
The Schrodinger equation has a time reversal symmetry $t mapsto -t$ combined with complex conjugation. It looks like
$$i partial_t psi = partial_xx psi.$$
The heat equation on the other hand does not have a time reversal symmetry. It looks like
$$partial_t u = partial_xx u.$$
There is no way to reconcile this with the time reversed equation
$$- partial_t u = partial_xx u.$$
Indeed, heat always flows from hot to cold in the forward time direction and vice versa in the backwards time direction. You can measure this smoothing in the Fourier domain. The Green's function for the heat equation is (schematically)
$$frac1sqrtt e^-x^2/t,$$
so one sees it doesn't even exist for $t<0$. On the other hand the Green's function for the Schrodinger equation looks like
$$frac1sqrtit e^-x^2/it.$$
(In general, the quantum mechanical and thermodynamic pictures are related by the replacement $t mapsto it$.)
answered yesterday
Ryan Thorngren
3,096929
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
15
down vote
Does the real heat equation have a special direction in time?
Yes, it does, because heat spreads out forward in time, and concentrates itself backwards in time. Mathematically, under time reversal $t to -t$ a first time derivative picks up a minus sign, which is why the heat equation is not time reversal invariant (TRI). It's the same principle as friction $F = - bv$ picking out a time direction.
What does he mean by "the imaginary coefficient assures there is no special direction in time"?
The wave equation only has second time derivatives, and hence is TRI. But it's hard to see how an equation that is first order in time could possibly be TRI. This is a problem, as the Schrodinger equation has to be first order in time, and must be TRI to reduce to classical mechanics.
The trick in quantum mechanics is that the gradient of the phase is tied to momentum. That is, classically we must flip $p to -p$ to reverse time, but since a quantum particle with momentum $p$ has wavefunction $e^ipx$, this must be realized by
$$psi to psi^*$$
under time reversal, or equivalently $i to -i$. The sign picked up by the first time derivative is thus canceled by the sign picked up by $i$, making the Schrodinger equation TRI as it must be.
answered yesterday
knzhou
33.8k897170
add a comment |Â
up vote
15
down vote
Does the real heat equation have a special direction in time?
Yes, it does, because heat spreads out forward in time, and concentrates itself backwards in time. Mathematically, under time reversal $t to -t$ a first time derivative picks up a minus sign, which is why the heat equation is not time reversal invariant (TRI). It's the same principle as friction $F = - bv$ picking out a time direction.
What does he mean by "the imaginary coefficient assures there is no special direction in time"?
The wave equation only has second time derivatives, and hence is TRI. But it's hard to see how an equation that is first order in time could possibly be TRI. This is a problem, as the Schrodinger equation has to be first order in time, and must be TRI to reduce to classical mechanics.
The trick in quantum mechanics is that the gradient of the phase is tied to momentum. That is, classically we must flip $p to -p$ to reverse time, but since a quantum particle with momentum $p$ has wavefunction $e^ipx$, this must be realized by
$$psi to psi^*$$
under time reversal, or equivalently $i to -i$. The sign picked up by the first time derivative is thus canceled by the sign picked up by $i$, making the Schrodinger equation TRI as it must be.
answered yesterday
knzhou
33.8k897170
add a comment |Â
up vote
15
down vote
up vote
15
down vote
Does the real heat equation have a special direction in time?
Yes, it does, because heat spreads out forward in time, and concentrates itself backwards in time. Mathematically, under time reversal $t to -t$ a first time derivative picks up a minus sign, which is why the heat equation is not time reversal invariant (TRI). It's the same principle as friction $F = - bv$ picking out a time direction.
What does he mean by "the imaginary coefficient assures there is no special direction in time"?
The wave equation only has second time derivatives, and hence is TRI. But it's hard to see how an equation that is first order in time could possibly be TRI. This is a problem, as the Schrodinger equation has to be first order in time, and must be TRI to reduce to classical mechanics.
The trick in quantum mechanics is that the gradient of the phase is tied to momentum. That is, classically we must flip $p to -p$ to reverse time, but since a quantum particle with momentum $p$ has wavefunction $e^ipx$, this must be realized by
$$psi to psi^*$$
under time reversal, or equivalently $i to -i$. The sign picked up by the first time derivative is thus canceled by the sign picked up by $i$, making the Schrodinger equation TRI as it must be.
answered yesterday
knzhou
33.8k897170
Does the real heat equation have a special direction in time?
Yes, it does, because heat spreads out forward in time, and concentrates itself backwards in time. Mathematically, under time reversal $t to -t$ a first time derivative picks up a minus sign, which is why the heat equation is not time reversal invariant (TRI). It's the same principle as friction $F = - bv$ picking out a time direction.
What does he mean by "the imaginary coefficient assures there is no special direction in time"?
The wave equation only has second time derivatives, and hence is TRI. But it's hard to see how an equation that is first order in time could possibly be TRI. This is a problem, as the Schrodinger equation has to be first order in time, and must be TRI to reduce to classical mechanics.
The trick in quantum mechanics is that the gradient of the phase is tied to momentum. That is, classically we must flip $p to -p$ to reverse time, but since a quantum particle with momentum $p$ has wavefunction $e^ipx$, this must be realized by
$$psi to psi^*$$
under time reversal, or equivalently $i to -i$. The sign picked up by the first time derivative is thus canceled by the sign picked up by $i$, making the Schrodinger equation TRI as it must be.
answered yesterday
knzhou
33.8k897170
answered yesterday
knzhou
33.8k897170
answered yesterday
knzhou
33.8k897170
answered yesterday
knzhou
33.8k897170
33.8k897170
add a comment |Â
add a comment |Â
up vote
8
down vote
I will just paraphrase Pauli.
The Schrodinger equation has a time reversal symmetry $t mapsto -t$ combined with complex conjugation. It looks like
$$i partial_t psi = partial_xx psi.$$
The heat equation on the other hand does not have a time reversal symmetry. It looks like
$$partial_t u = partial_xx u.$$
There is no way to reconcile this with the time reversed equation
$$- partial_t u = partial_xx u.$$
Indeed, heat always flows from hot to cold in the forward time direction and vice versa in the backwards time direction. You can measure this smoothing in the Fourier domain. The Green's function for the heat equation is (schematically)
$$frac1sqrtt e^-x^2/t,$$
so one sees it doesn't even exist for $t<0$. On the other hand the Green's function for the Schrodinger equation looks like
$$frac1sqrtit e^-x^2/it.$$
(In general, the quantum mechanical and thermodynamic pictures are related by the replacement $t mapsto it$.)
answered yesterday
Ryan Thorngren
3,096929
add a comment |Â
up vote
8
down vote
I will just paraphrase Pauli.
The Schrodinger equation has a time reversal symmetry $t mapsto -t$ combined with complex conjugation. It looks like
$$i partial_t psi = partial_xx psi.$$
The heat equation on the other hand does not have a time reversal symmetry. It looks like
$$partial_t u = partial_xx u.$$
There is no way to reconcile this with the time reversed equation
$$- partial_t u = partial_xx u.$$
Indeed, heat always flows from hot to cold in the forward time direction and vice versa in the backwards time direction. You can measure this smoothing in the Fourier domain. The Green's function for the heat equation is (schematically)
$$frac1sqrtt e^-x^2/t,$$
so one sees it doesn't even exist for $t<0$. On the other hand the Green's function for the Schrodinger equation looks like
$$frac1sqrtit e^-x^2/it.$$
(In general, the quantum mechanical and thermodynamic pictures are related by the replacement $t mapsto it$.)
answered yesterday
Ryan Thorngren
3,096929
add a comment |Â
up vote
8
down vote
up vote
8
down vote
I will just paraphrase Pauli.
The Schrodinger equation has a time reversal symmetry $t mapsto -t$ combined with complex conjugation. It looks like
$$i partial_t psi = partial_xx psi.$$
The heat equation on the other hand does not have a time reversal symmetry. It looks like
$$partial_t u = partial_xx u.$$
There is no way to reconcile this with the time reversed equation
$$- partial_t u = partial_xx u.$$
Indeed, heat always flows from hot to cold in the forward time direction and vice versa in the backwards time direction. You can measure this smoothing in the Fourier domain. The Green's function for the heat equation is (schematically)
$$frac1sqrtt e^-x^2/t,$$
so one sees it doesn't even exist for $t<0$. On the other hand the Green's function for the Schrodinger equation looks like
$$frac1sqrtit e^-x^2/it.$$
(In general, the quantum mechanical and thermodynamic pictures are related by the replacement $t mapsto it$.)
answered yesterday
Ryan Thorngren
3,096929
I will just paraphrase Pauli.
The Schrodinger equation has a time reversal symmetry $t mapsto -t$ combined with complex conjugation. It looks like
$$i partial_t psi = partial_xx psi.$$
The heat equation on the other hand does not have a time reversal symmetry. It looks like
$$partial_t u = partial_xx u.$$
There is no way to reconcile this with the time reversed equation
$$- partial_t u = partial_xx u.$$
Indeed, heat always flows from hot to cold in the forward time direction and vice versa in the backwards time direction. You can measure this smoothing in the Fourier domain. The Green's function for the heat equation is (schematically)
$$frac1sqrtt e^-x^2/t,$$
so one sees it doesn't even exist for $t<0$. On the other hand the Green's function for the Schrodinger equation looks like
$$frac1sqrtit e^-x^2/it.$$
(In general, the quantum mechanical and thermodynamic pictures are related by the replacement $t mapsto it$.)
answered yesterday
Ryan Thorngren
3,096929
answered yesterday
Ryan Thorngren
3,096929
answered yesterday
Ryan Thorngren
3,096929
answered yesterday
Ryan Thorngren
3,096929
3,096929
add a comment |Â
add a comment |Â
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