Conserved Current for a PDE
Let $(x,t) in mathbb{R}^2$, $W(x)$ be a (smooth enough) real-valued function and consider the following partial differential equation for the real-valued function $U(x,t)$
begin{equation}
frac{partial^2 U}{partial t^2} = - frac{hbar^2}{4 m^2} frac{partial^4 U}{partial x^4}+ frac{W}{m} frac{partial^2 U}{partial x^2} +frac{W’}{m} frac{partial U}{partial x} + left( frac{W’’}{2m} - frac{W^2}{hbar^2} right) U qquad (I),
end{equation}
where $m$ and $hbar$ are positive constants.
In the following we shall be quite sloppy, and we shall assume that given (smooth enough) initial conditions $U(x,0)$ and $frac{partial U}{partial t}(x,0)$ (lying in some space) there exists a unique (smooth enough) solution $U$ (lying in some space) to (I). Let us call the set of solutions $mathcal{E}$.
Let $D_{x}^k F$ be the set of all partial derivatives of $F$ with respect to $x$ from order 1 to order $k$. I ask whether there exist (smooth enough) real-valued functions $p geq 0$ and $j$ such that, by setting
begin{equation}
P(x,t)=p left(U(x,t),(D_{x}^k U)(x,t), frac{partial U}{partial t}(x,t), left(D_{x}^{k} frac{partial U}{partial t}right)(x,t) right), \
J(x,t)=j left(U(x,t),(D_{x}^k U)(x,t), frac{partial U}{partial t}(x,t), left(D_{x}^{k} frac{partial U}{partial t}right)(x,t) right),
end{equation}
the following properties hold:
(i) if $U$ is the solution of (I) corresponding to a given function $W(x)$ and given initial conditions $U(x,0)$ and $frac{partial U}{partial t}(x,0)$, and $tilde{U}$ is the solution of (I) corresponding to the same initial conditions, but to $W(x)+c$, with $c in mathbb{R}$, then $P(x,t)$ is the same when computed for $U$ and $tilde{U}$;
(ii) for every $U in mathcal{E}$ the following conservation law holds
begin{equation}
frac{partial P}{partial t} + frac{partial J}{partial x}=0;
end{equation}
(iii) $p$ is not the constant function.
The answer I think should be negative, but I don't how to "prove" this: since we have not formulated the problem in a rigorous way, we do not expect to get a rigorous proof, but some heuristic, but convincing argument in this direction.
NOTE (1) This problem, as the notation shows, has a physical background, and the mathematical formulation of the problem that I give here is my personal interpretation of a physical exposition given by the great XXth century physicist David Bohm in his wonderful treatise $mathit{Quantum}$ $mathit{Theory}$ published in 1951. For all the physical details about this problem see my post Nonexistence of a Probability for Real Wave Functions.
NOTE (2) Bohm's physical discussion is not very clear, so that it can admit different mathematical interpretations. A simpler interpretation of Bohm's original statement is the following. Consider the following equation
begin{equation}
frac{partial^2 U}{partial t^2} = - frac{hbar^2}{4m^2} frac{partial^4 U}{partial x^4} qquad{(II)},
end{equation}
and let $mathcal{F}$ the set of all (smooth enough) solutions of this equation.
Do there exists (smooth enough) real-valued functions $p geq 0$ and $j$ such that, by setting
begin{equation}
P(x,t)=p left(U(x,t),(D_{x}^k U)(x,t), frac{partial U}{partial t}(x,t), left(D_{x}^{k} frac{partial U}{partial t}right)(x,t) right), \
J(x,t)=j left(U(x,t),(D_{x}^k U)(x,t), frac{partial U}{partial t}(x,t), left(D_{x}^{k} frac{partial U}{partial t}right)(x,t) right),
end{equation}
the following properties hold:
(i) for every $U in mathcal{F}$ we have
begin{equation}
frac{partial P}{partial t} + frac{partial J}{partial x}=0;
end{equation}
(ii) for the special solution $U(x,t)=cosleft(sqrt{frac{2m omega}{hbar}}x-omega t right)$, we have that $P(x,t)$ is independent of $omega > 0$;
(iii) $p$ is not a constant function?
Maybe this mathematical problem is more easily seen to have a negative answer than the one I formulated above.
real-analysis pde mathematical-physics
|
show 6 more comments
Let $(x,t) in mathbb{R}^2$, $W(x)$ be a (smooth enough) real-valued function and consider the following partial differential equation for the real-valued function $U(x,t)$
begin{equation}
frac{partial^2 U}{partial t^2} = - frac{hbar^2}{4 m^2} frac{partial^4 U}{partial x^4}+ frac{W}{m} frac{partial^2 U}{partial x^2} +frac{W’}{m} frac{partial U}{partial x} + left( frac{W’’}{2m} - frac{W^2}{hbar^2} right) U qquad (I),
end{equation}
where $m$ and $hbar$ are positive constants.
In the following we shall be quite sloppy, and we shall assume that given (smooth enough) initial conditions $U(x,0)$ and $frac{partial U}{partial t}(x,0)$ (lying in some space) there exists a unique (smooth enough) solution $U$ (lying in some space) to (I). Let us call the set of solutions $mathcal{E}$.
Let $D_{x}^k F$ be the set of all partial derivatives of $F$ with respect to $x$ from order 1 to order $k$. I ask whether there exist (smooth enough) real-valued functions $p geq 0$ and $j$ such that, by setting
begin{equation}
P(x,t)=p left(U(x,t),(D_{x}^k U)(x,t), frac{partial U}{partial t}(x,t), left(D_{x}^{k} frac{partial U}{partial t}right)(x,t) right), \
J(x,t)=j left(U(x,t),(D_{x}^k U)(x,t), frac{partial U}{partial t}(x,t), left(D_{x}^{k} frac{partial U}{partial t}right)(x,t) right),
end{equation}
the following properties hold:
(i) if $U$ is the solution of (I) corresponding to a given function $W(x)$ and given initial conditions $U(x,0)$ and $frac{partial U}{partial t}(x,0)$, and $tilde{U}$ is the solution of (I) corresponding to the same initial conditions, but to $W(x)+c$, with $c in mathbb{R}$, then $P(x,t)$ is the same when computed for $U$ and $tilde{U}$;
(ii) for every $U in mathcal{E}$ the following conservation law holds
begin{equation}
frac{partial P}{partial t} + frac{partial J}{partial x}=0;
end{equation}
(iii) $p$ is not the constant function.
The answer I think should be negative, but I don't how to "prove" this: since we have not formulated the problem in a rigorous way, we do not expect to get a rigorous proof, but some heuristic, but convincing argument in this direction.
NOTE (1) This problem, as the notation shows, has a physical background, and the mathematical formulation of the problem that I give here is my personal interpretation of a physical exposition given by the great XXth century physicist David Bohm in his wonderful treatise $mathit{Quantum}$ $mathit{Theory}$ published in 1951. For all the physical details about this problem see my post Nonexistence of a Probability for Real Wave Functions.
NOTE (2) Bohm's physical discussion is not very clear, so that it can admit different mathematical interpretations. A simpler interpretation of Bohm's original statement is the following. Consider the following equation
begin{equation}
frac{partial^2 U}{partial t^2} = - frac{hbar^2}{4m^2} frac{partial^4 U}{partial x^4} qquad{(II)},
end{equation}
and let $mathcal{F}$ the set of all (smooth enough) solutions of this equation.
Do there exists (smooth enough) real-valued functions $p geq 0$ and $j$ such that, by setting
begin{equation}
P(x,t)=p left(U(x,t),(D_{x}^k U)(x,t), frac{partial U}{partial t}(x,t), left(D_{x}^{k} frac{partial U}{partial t}right)(x,t) right), \
J(x,t)=j left(U(x,t),(D_{x}^k U)(x,t), frac{partial U}{partial t}(x,t), left(D_{x}^{k} frac{partial U}{partial t}right)(x,t) right),
end{equation}
the following properties hold:
(i) for every $U in mathcal{F}$ we have
begin{equation}
frac{partial P}{partial t} + frac{partial J}{partial x}=0;
end{equation}
(ii) for the special solution $U(x,t)=cosleft(sqrt{frac{2m omega}{hbar}}x-omega t right)$, we have that $P(x,t)$ is independent of $omega > 0$;
(iii) $p$ is not a constant function?
Maybe this mathematical problem is more easily seen to have a negative answer than the one I formulated above.
real-analysis pde mathematical-physics
If a constant $c$ is added to $W$, the wave function $psi$ is multiplied by $exp(-ict/hbar)$. The real part of the wave function $U$ changes in a nontrivial way: is becomes a linear combination of $U$ and $V$. Therefore it is not clear what you mean by same initial conditions in property (i).
– atarasenko
Nov 2 at 13:05
Also an interesting point: if $W(x)$ is real, then there always exists a solution $psi=iV(t,x)$, with $U=0$. In this case any function $p$ that depends only on $U$ and its derivatives is constant. But from the physical point of view solutions $psi=U(t,x)$ and $psi=iU(t,x)$ are equivalent, because the wave functions for them differ only by a constant phase.
– atarasenko
Nov 2 at 13:24
@atarasenko As for the first observation, I mean exactly the same $U(x,0)$ and $frac{partial U}{partial t}(x,0)$. As for the second observation it is simply wrong. Write down the Schrödinger equation and you will see that $i V(x,t)$ cannot be a solution.
– Maurizio Barbato
Nov 2 at 13:44
If $psi=U$ is a solution, then $ipsi$ is a solution as well. It is due to linearity of the Schrödinger equation.
– atarasenko
Nov 2 at 13:53
Regarding the first observation - if the initial conditions $U(0,x)$ and $D_tU(0,x)$ are the same but functions $W$ differ by a constant, then the wave functions $psi$ will be completely different (not just by a phase factor $exp(−ict/hbar)$). And there is no reason for $p$ to be the same for 2 physically different solutions.
– atarasenko
Nov 2 at 13:56
|
show 6 more comments
Let $(x,t) in mathbb{R}^2$, $W(x)$ be a (smooth enough) real-valued function and consider the following partial differential equation for the real-valued function $U(x,t)$
begin{equation}
frac{partial^2 U}{partial t^2} = - frac{hbar^2}{4 m^2} frac{partial^4 U}{partial x^4}+ frac{W}{m} frac{partial^2 U}{partial x^2} +frac{W’}{m} frac{partial U}{partial x} + left( frac{W’’}{2m} - frac{W^2}{hbar^2} right) U qquad (I),
end{equation}
where $m$ and $hbar$ are positive constants.
In the following we shall be quite sloppy, and we shall assume that given (smooth enough) initial conditions $U(x,0)$ and $frac{partial U}{partial t}(x,0)$ (lying in some space) there exists a unique (smooth enough) solution $U$ (lying in some space) to (I). Let us call the set of solutions $mathcal{E}$.
Let $D_{x}^k F$ be the set of all partial derivatives of $F$ with respect to $x$ from order 1 to order $k$. I ask whether there exist (smooth enough) real-valued functions $p geq 0$ and $j$ such that, by setting
begin{equation}
P(x,t)=p left(U(x,t),(D_{x}^k U)(x,t), frac{partial U}{partial t}(x,t), left(D_{x}^{k} frac{partial U}{partial t}right)(x,t) right), \
J(x,t)=j left(U(x,t),(D_{x}^k U)(x,t), frac{partial U}{partial t}(x,t), left(D_{x}^{k} frac{partial U}{partial t}right)(x,t) right),
end{equation}
the following properties hold:
(i) if $U$ is the solution of (I) corresponding to a given function $W(x)$ and given initial conditions $U(x,0)$ and $frac{partial U}{partial t}(x,0)$, and $tilde{U}$ is the solution of (I) corresponding to the same initial conditions, but to $W(x)+c$, with $c in mathbb{R}$, then $P(x,t)$ is the same when computed for $U$ and $tilde{U}$;
(ii) for every $U in mathcal{E}$ the following conservation law holds
begin{equation}
frac{partial P}{partial t} + frac{partial J}{partial x}=0;
end{equation}
(iii) $p$ is not the constant function.
The answer I think should be negative, but I don't how to "prove" this: since we have not formulated the problem in a rigorous way, we do not expect to get a rigorous proof, but some heuristic, but convincing argument in this direction.
NOTE (1) This problem, as the notation shows, has a physical background, and the mathematical formulation of the problem that I give here is my personal interpretation of a physical exposition given by the great XXth century physicist David Bohm in his wonderful treatise $mathit{Quantum}$ $mathit{Theory}$ published in 1951. For all the physical details about this problem see my post Nonexistence of a Probability for Real Wave Functions.
NOTE (2) Bohm's physical discussion is not very clear, so that it can admit different mathematical interpretations. A simpler interpretation of Bohm's original statement is the following. Consider the following equation
begin{equation}
frac{partial^2 U}{partial t^2} = - frac{hbar^2}{4m^2} frac{partial^4 U}{partial x^4} qquad{(II)},
end{equation}
and let $mathcal{F}$ the set of all (smooth enough) solutions of this equation.
Do there exists (smooth enough) real-valued functions $p geq 0$ and $j$ such that, by setting
begin{equation}
P(x,t)=p left(U(x,t),(D_{x}^k U)(x,t), frac{partial U}{partial t}(x,t), left(D_{x}^{k} frac{partial U}{partial t}right)(x,t) right), \
J(x,t)=j left(U(x,t),(D_{x}^k U)(x,t), frac{partial U}{partial t}(x,t), left(D_{x}^{k} frac{partial U}{partial t}right)(x,t) right),
end{equation}
the following properties hold:
(i) for every $U in mathcal{F}$ we have
begin{equation}
frac{partial P}{partial t} + frac{partial J}{partial x}=0;
end{equation}
(ii) for the special solution $U(x,t)=cosleft(sqrt{frac{2m omega}{hbar}}x-omega t right)$, we have that $P(x,t)$ is independent of $omega > 0$;
(iii) $p$ is not a constant function?
Maybe this mathematical problem is more easily seen to have a negative answer than the one I formulated above.
real-analysis pde mathematical-physics
Let $(x,t) in mathbb{R}^2$, $W(x)$ be a (smooth enough) real-valued function and consider the following partial differential equation for the real-valued function $U(x,t)$
begin{equation}
frac{partial^2 U}{partial t^2} = - frac{hbar^2}{4 m^2} frac{partial^4 U}{partial x^4}+ frac{W}{m} frac{partial^2 U}{partial x^2} +frac{W’}{m} frac{partial U}{partial x} + left( frac{W’’}{2m} - frac{W^2}{hbar^2} right) U qquad (I),
end{equation}
where $m$ and $hbar$ are positive constants.
In the following we shall be quite sloppy, and we shall assume that given (smooth enough) initial conditions $U(x,0)$ and $frac{partial U}{partial t}(x,0)$ (lying in some space) there exists a unique (smooth enough) solution $U$ (lying in some space) to (I). Let us call the set of solutions $mathcal{E}$.
Let $D_{x}^k F$ be the set of all partial derivatives of $F$ with respect to $x$ from order 1 to order $k$. I ask whether there exist (smooth enough) real-valued functions $p geq 0$ and $j$ such that, by setting
begin{equation}
P(x,t)=p left(U(x,t),(D_{x}^k U)(x,t), frac{partial U}{partial t}(x,t), left(D_{x}^{k} frac{partial U}{partial t}right)(x,t) right), \
J(x,t)=j left(U(x,t),(D_{x}^k U)(x,t), frac{partial U}{partial t}(x,t), left(D_{x}^{k} frac{partial U}{partial t}right)(x,t) right),
end{equation}
the following properties hold:
(i) if $U$ is the solution of (I) corresponding to a given function $W(x)$ and given initial conditions $U(x,0)$ and $frac{partial U}{partial t}(x,0)$, and $tilde{U}$ is the solution of (I) corresponding to the same initial conditions, but to $W(x)+c$, with $c in mathbb{R}$, then $P(x,t)$ is the same when computed for $U$ and $tilde{U}$;
(ii) for every $U in mathcal{E}$ the following conservation law holds
begin{equation}
frac{partial P}{partial t} + frac{partial J}{partial x}=0;
end{equation}
(iii) $p$ is not the constant function.
The answer I think should be negative, but I don't how to "prove" this: since we have not formulated the problem in a rigorous way, we do not expect to get a rigorous proof, but some heuristic, but convincing argument in this direction.
NOTE (1) This problem, as the notation shows, has a physical background, and the mathematical formulation of the problem that I give here is my personal interpretation of a physical exposition given by the great XXth century physicist David Bohm in his wonderful treatise $mathit{Quantum}$ $mathit{Theory}$ published in 1951. For all the physical details about this problem see my post Nonexistence of a Probability for Real Wave Functions.
NOTE (2) Bohm's physical discussion is not very clear, so that it can admit different mathematical interpretations. A simpler interpretation of Bohm's original statement is the following. Consider the following equation
begin{equation}
frac{partial^2 U}{partial t^2} = - frac{hbar^2}{4m^2} frac{partial^4 U}{partial x^4} qquad{(II)},
end{equation}
and let $mathcal{F}$ the set of all (smooth enough) solutions of this equation.
Do there exists (smooth enough) real-valued functions $p geq 0$ and $j$ such that, by setting
begin{equation}
P(x,t)=p left(U(x,t),(D_{x}^k U)(x,t), frac{partial U}{partial t}(x,t), left(D_{x}^{k} frac{partial U}{partial t}right)(x,t) right), \
J(x,t)=j left(U(x,t),(D_{x}^k U)(x,t), frac{partial U}{partial t}(x,t), left(D_{x}^{k} frac{partial U}{partial t}right)(x,t) right),
end{equation}
the following properties hold:
(i) for every $U in mathcal{F}$ we have
begin{equation}
frac{partial P}{partial t} + frac{partial J}{partial x}=0;
end{equation}
(ii) for the special solution $U(x,t)=cosleft(sqrt{frac{2m omega}{hbar}}x-omega t right)$, we have that $P(x,t)$ is independent of $omega > 0$;
(iii) $p$ is not a constant function?
Maybe this mathematical problem is more easily seen to have a negative answer than the one I formulated above.
real-analysis pde mathematical-physics
real-analysis pde mathematical-physics
edited Nov 5 at 10:39
asked Oct 28 at 10:18
Maurizio Barbato
1,239620
1,239620
If a constant $c$ is added to $W$, the wave function $psi$ is multiplied by $exp(-ict/hbar)$. The real part of the wave function $U$ changes in a nontrivial way: is becomes a linear combination of $U$ and $V$. Therefore it is not clear what you mean by same initial conditions in property (i).
– atarasenko
Nov 2 at 13:05
Also an interesting point: if $W(x)$ is real, then there always exists a solution $psi=iV(t,x)$, with $U=0$. In this case any function $p$ that depends only on $U$ and its derivatives is constant. But from the physical point of view solutions $psi=U(t,x)$ and $psi=iU(t,x)$ are equivalent, because the wave functions for them differ only by a constant phase.
– atarasenko
Nov 2 at 13:24
@atarasenko As for the first observation, I mean exactly the same $U(x,0)$ and $frac{partial U}{partial t}(x,0)$. As for the second observation it is simply wrong. Write down the Schrödinger equation and you will see that $i V(x,t)$ cannot be a solution.
– Maurizio Barbato
Nov 2 at 13:44
If $psi=U$ is a solution, then $ipsi$ is a solution as well. It is due to linearity of the Schrödinger equation.
– atarasenko
Nov 2 at 13:53
Regarding the first observation - if the initial conditions $U(0,x)$ and $D_tU(0,x)$ are the same but functions $W$ differ by a constant, then the wave functions $psi$ will be completely different (not just by a phase factor $exp(−ict/hbar)$). And there is no reason for $p$ to be the same for 2 physically different solutions.
– atarasenko
Nov 2 at 13:56
|
show 6 more comments
If a constant $c$ is added to $W$, the wave function $psi$ is multiplied by $exp(-ict/hbar)$. The real part of the wave function $U$ changes in a nontrivial way: is becomes a linear combination of $U$ and $V$. Therefore it is not clear what you mean by same initial conditions in property (i).
– atarasenko
Nov 2 at 13:05
Also an interesting point: if $W(x)$ is real, then there always exists a solution $psi=iV(t,x)$, with $U=0$. In this case any function $p$ that depends only on $U$ and its derivatives is constant. But from the physical point of view solutions $psi=U(t,x)$ and $psi=iU(t,x)$ are equivalent, because the wave functions for them differ only by a constant phase.
– atarasenko
Nov 2 at 13:24
@atarasenko As for the first observation, I mean exactly the same $U(x,0)$ and $frac{partial U}{partial t}(x,0)$. As for the second observation it is simply wrong. Write down the Schrödinger equation and you will see that $i V(x,t)$ cannot be a solution.
– Maurizio Barbato
Nov 2 at 13:44
If $psi=U$ is a solution, then $ipsi$ is a solution as well. It is due to linearity of the Schrödinger equation.
– atarasenko
Nov 2 at 13:53
Regarding the first observation - if the initial conditions $U(0,x)$ and $D_tU(0,x)$ are the same but functions $W$ differ by a constant, then the wave functions $psi$ will be completely different (not just by a phase factor $exp(−ict/hbar)$). And there is no reason for $p$ to be the same for 2 physically different solutions.
– atarasenko
Nov 2 at 13:56
If a constant $c$ is added to $W$, the wave function $psi$ is multiplied by $exp(-ict/hbar)$. The real part of the wave function $U$ changes in a nontrivial way: is becomes a linear combination of $U$ and $V$. Therefore it is not clear what you mean by same initial conditions in property (i).
– atarasenko
Nov 2 at 13:05
If a constant $c$ is added to $W$, the wave function $psi$ is multiplied by $exp(-ict/hbar)$. The real part of the wave function $U$ changes in a nontrivial way: is becomes a linear combination of $U$ and $V$. Therefore it is not clear what you mean by same initial conditions in property (i).
– atarasenko
Nov 2 at 13:05
Also an interesting point: if $W(x)$ is real, then there always exists a solution $psi=iV(t,x)$, with $U=0$. In this case any function $p$ that depends only on $U$ and its derivatives is constant. But from the physical point of view solutions $psi=U(t,x)$ and $psi=iU(t,x)$ are equivalent, because the wave functions for them differ only by a constant phase.
– atarasenko
Nov 2 at 13:24
Also an interesting point: if $W(x)$ is real, then there always exists a solution $psi=iV(t,x)$, with $U=0$. In this case any function $p$ that depends only on $U$ and its derivatives is constant. But from the physical point of view solutions $psi=U(t,x)$ and $psi=iU(t,x)$ are equivalent, because the wave functions for them differ only by a constant phase.
– atarasenko
Nov 2 at 13:24
@atarasenko As for the first observation, I mean exactly the same $U(x,0)$ and $frac{partial U}{partial t}(x,0)$. As for the second observation it is simply wrong. Write down the Schrödinger equation and you will see that $i V(x,t)$ cannot be a solution.
– Maurizio Barbato
Nov 2 at 13:44
@atarasenko As for the first observation, I mean exactly the same $U(x,0)$ and $frac{partial U}{partial t}(x,0)$. As for the second observation it is simply wrong. Write down the Schrödinger equation and you will see that $i V(x,t)$ cannot be a solution.
– Maurizio Barbato
Nov 2 at 13:44
If $psi=U$ is a solution, then $ipsi$ is a solution as well. It is due to linearity of the Schrödinger equation.
– atarasenko
Nov 2 at 13:53
If $psi=U$ is a solution, then $ipsi$ is a solution as well. It is due to linearity of the Schrödinger equation.
– atarasenko
Nov 2 at 13:53
Regarding the first observation - if the initial conditions $U(0,x)$ and $D_tU(0,x)$ are the same but functions $W$ differ by a constant, then the wave functions $psi$ will be completely different (not just by a phase factor $exp(−ict/hbar)$). And there is no reason for $p$ to be the same for 2 physically different solutions.
– atarasenko
Nov 2 at 13:56
Regarding the first observation - if the initial conditions $U(0,x)$ and $D_tU(0,x)$ are the same but functions $W$ differ by a constant, then the wave functions $psi$ will be completely different (not just by a phase factor $exp(−ict/hbar)$). And there is no reason for $p$ to be the same for 2 physically different solutions.
– atarasenko
Nov 2 at 13:56
|
show 6 more comments
2 Answers
2
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votes
Proof:
Consider equation (I) with $W(x)=const$ and the following initial conditions:
$$U(x,0)=0,$$
$$D_tU(x,0)=A,$$
and periodic boundary conditions:
$$U(x,t)=U(x+a,t),$$
where $a$ and $A$ are constants.
Solution of equation (I) depends only in $t$ in this case: $U=U(t)$, and $p=p(t)$, $j=0$. It has the following form:
$$U=frac{Ahbar}{W}sin{frac{Wt}{hbar}}$$
The first derivative is:
$$D_tU=Acos{frac{Wt}{hbar}}$$
Function $p=p(U,D_tU)$ has 2 properties:
1 It must not depend on $t$ due to the conservation law.
2 It must not depend on $W$ due to property (i) - $p$ does not change after transformation $Wrightarrow W+c$.
However, such function $p$ is constant: one can invert expressions for $U$ and $D_t U$ to find functions $W(U,D_tU)$ and $t(U,D_tU)$ and use
$$frac{partial p}{partial U}=frac{partial p}{partial W}frac{partial W}{partial U}+frac{partial p}{partial t}frac{partial t}{partial U}=0,$$
$$frac{partial p}{partial (D_tU)}=frac{partial p}{partial W}frac{partial W}{partial (D_tU)}+frac{partial p}{partial t}frac{partial t}{partial (D_tU)}=0,$$
because $frac{partial p}{partial W}=0$ and $frac{partial p}{partial t}=0$.
UPDATE: answering the updated question:
Assume that $U=f(x-vt)$, where $v=(hbaromega/2m)^{1/2}$ (it follows from (ii) that $f(x)=cos{((2momega/hbar)^{1/2}x)}$). Assume that there is a function $p$ that depends only on $U$ and its derivatives, and does not depend on $omega$. In this case, $P(x,t)$ is also a function of $x-vt$: $P(x,t)=F(x-vt)$.
As a consequence:
$$frac{D_t P}{D_x P}=-v=-sqrt{frac{hbaromega}{2m}},$$
which contradicts the assumption that $P(x,t)$ does not depend on $omega$.
Thank you very much for having answered my question. Unfortunately, this argument does not work. To see why note that $p$ is a function also of the spatial derivatives of $U$ and of $frac{partial U}{partial t}$. Such spatial derivatives are zero for the special solution you chose. So you cannot conclude that $p$ is a constant function.
– Maurizio Barbato
Nov 2 at 16:24
Dear @atarasenko, your argument about the second formulation is interesting, but only proves the following fact. Consider $P(x,t)$ for the special solution $U(x,t)=cosleft(sqrt{frac{2m omega}{hbar}}x-omega t right)$. We have $P(x,t)=F(x-vt)$, with $v=[hbar omega/(2m)]^{1/2}$. Under the $mathbf{hypothesis}$ that $F$ is not constant, then $P(x,t)$ must depend on $omega$. We can say it in another equivalent way: if $P(x,t)$ turns out to be independent of $omega$, then it is a constant function.
– Maurizio Barbato
Nov 3 at 10:06
Note that this is not the thesis I formulated in the post, since you can find some non-constant function $p$ such that $P(x,t)$ turns out to be constant for the special solution $U(x,t)=cosleft(sqrt{frac{2m omega}{hbar}}x-omega t right)$ (think of appropriate ratios of partial derivatives). The point to prove (which is difficult) is that such functions cannot satisfy a continuity equation.
– Maurizio Barbato
Nov 3 at 10:13
I have posted the question also on Math OverFlow: see Conserved Positive Charge for a PDE, but no one has given any answer up to now. So it is for sure a difficult problem, and frankly I do not imagine it can be given a simple and clear solution.
– Maurizio Barbato
Nov 3 at 10:22
add a comment |
After some careful thought I must conclude that the mathematical formulation of Bohm's statement I have given in my post above is wrong for the reason that atarasenko identified in his comment.
Indeed, if we start from the usual on-dimensional Schrödinger equation for a particle in the case of a potential $W(x)$:
begin{equation}
i hbar frac{partial psi}{partial t}(x,t) = -frac{hbar^2}{2m} frac{partial^2}{partial x^2}psi(x,t) + W(x) psi(x,t),
end{equation}
and we change $W(x)$ by adding a constant $W_0$, the new solution $psi$ corresponding to the initial data $x mapsto psi(x,0)$ is $tilde{psi}(x,t)=psi(x,t) exp(-iW_0 t/ hbar)$. If $U$ and $V$ are respectively the real and imaginary part of $psi$, and $tilde{U}$ and $tilde{V}$ are respectively the real and imaginary part of $tilde{psi}$, then from
begin{equation}
frac{partial tilde{psi}}{partial t} = exp(-iW_0t/ hbar) left[ frac{partial psi}{partial t} - frac{i W_0}{hbar} psi right],
end{equation}
we get
begin{equation}
frac{partial tilde{U}}{partial t}(x,0)= frac{partial U}{partial t}(x,0) + frac{W_0}{hbar} V(x,0),
end{equation}
so $x mapsto frac{partial tilde{U}}{partial t}(x,0)$ and $x mapsto frac{partial U}{partial t}(x,0)$ do not coincide. Since Bohm requires that the usual theory based on the complex-valued wave function $psi$ and the (eventual) theory based on the real-valued wave function $U$ should be equivalent, we conclude that the mathematical formulation I have given above is incorrect.
As for the second mathematical formulation of Bohm's statement I gave in my NOTE (2) above, I have discussed it at some length in my post on MathOverflow Conserved Positive Charge for a PDE, but no one could give some answer. This problem seems not trivial at all, and maybe it could have a positive answer, in the sense that there exist some functions $p geq 0$ and $j$ which satisfy properties (i),(ii) and (iii) I listed in NOTE (2) above. For this reason, I suspect that this mathematical formulation is flawed too, this time probably because Bohm had in his mind some kind of invariance argument based on physical considerations (hidden in Condition (3) required for the probability function in Section (4.2) of his book Quantum Theory) which we do not know and which would make the non-existence result transparent.
add a comment |
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Proof:
Consider equation (I) with $W(x)=const$ and the following initial conditions:
$$U(x,0)=0,$$
$$D_tU(x,0)=A,$$
and periodic boundary conditions:
$$U(x,t)=U(x+a,t),$$
where $a$ and $A$ are constants.
Solution of equation (I) depends only in $t$ in this case: $U=U(t)$, and $p=p(t)$, $j=0$. It has the following form:
$$U=frac{Ahbar}{W}sin{frac{Wt}{hbar}}$$
The first derivative is:
$$D_tU=Acos{frac{Wt}{hbar}}$$
Function $p=p(U,D_tU)$ has 2 properties:
1 It must not depend on $t$ due to the conservation law.
2 It must not depend on $W$ due to property (i) - $p$ does not change after transformation $Wrightarrow W+c$.
However, such function $p$ is constant: one can invert expressions for $U$ and $D_t U$ to find functions $W(U,D_tU)$ and $t(U,D_tU)$ and use
$$frac{partial p}{partial U}=frac{partial p}{partial W}frac{partial W}{partial U}+frac{partial p}{partial t}frac{partial t}{partial U}=0,$$
$$frac{partial p}{partial (D_tU)}=frac{partial p}{partial W}frac{partial W}{partial (D_tU)}+frac{partial p}{partial t}frac{partial t}{partial (D_tU)}=0,$$
because $frac{partial p}{partial W}=0$ and $frac{partial p}{partial t}=0$.
UPDATE: answering the updated question:
Assume that $U=f(x-vt)$, where $v=(hbaromega/2m)^{1/2}$ (it follows from (ii) that $f(x)=cos{((2momega/hbar)^{1/2}x)}$). Assume that there is a function $p$ that depends only on $U$ and its derivatives, and does not depend on $omega$. In this case, $P(x,t)$ is also a function of $x-vt$: $P(x,t)=F(x-vt)$.
As a consequence:
$$frac{D_t P}{D_x P}=-v=-sqrt{frac{hbaromega}{2m}},$$
which contradicts the assumption that $P(x,t)$ does not depend on $omega$.
Thank you very much for having answered my question. Unfortunately, this argument does not work. To see why note that $p$ is a function also of the spatial derivatives of $U$ and of $frac{partial U}{partial t}$. Such spatial derivatives are zero for the special solution you chose. So you cannot conclude that $p$ is a constant function.
– Maurizio Barbato
Nov 2 at 16:24
Dear @atarasenko, your argument about the second formulation is interesting, but only proves the following fact. Consider $P(x,t)$ for the special solution $U(x,t)=cosleft(sqrt{frac{2m omega}{hbar}}x-omega t right)$. We have $P(x,t)=F(x-vt)$, with $v=[hbar omega/(2m)]^{1/2}$. Under the $mathbf{hypothesis}$ that $F$ is not constant, then $P(x,t)$ must depend on $omega$. We can say it in another equivalent way: if $P(x,t)$ turns out to be independent of $omega$, then it is a constant function.
– Maurizio Barbato
Nov 3 at 10:06
Note that this is not the thesis I formulated in the post, since you can find some non-constant function $p$ such that $P(x,t)$ turns out to be constant for the special solution $U(x,t)=cosleft(sqrt{frac{2m omega}{hbar}}x-omega t right)$ (think of appropriate ratios of partial derivatives). The point to prove (which is difficult) is that such functions cannot satisfy a continuity equation.
– Maurizio Barbato
Nov 3 at 10:13
I have posted the question also on Math OverFlow: see Conserved Positive Charge for a PDE, but no one has given any answer up to now. So it is for sure a difficult problem, and frankly I do not imagine it can be given a simple and clear solution.
– Maurizio Barbato
Nov 3 at 10:22
add a comment |
Proof:
Consider equation (I) with $W(x)=const$ and the following initial conditions:
$$U(x,0)=0,$$
$$D_tU(x,0)=A,$$
and periodic boundary conditions:
$$U(x,t)=U(x+a,t),$$
where $a$ and $A$ are constants.
Solution of equation (I) depends only in $t$ in this case: $U=U(t)$, and $p=p(t)$, $j=0$. It has the following form:
$$U=frac{Ahbar}{W}sin{frac{Wt}{hbar}}$$
The first derivative is:
$$D_tU=Acos{frac{Wt}{hbar}}$$
Function $p=p(U,D_tU)$ has 2 properties:
1 It must not depend on $t$ due to the conservation law.
2 It must not depend on $W$ due to property (i) - $p$ does not change after transformation $Wrightarrow W+c$.
However, such function $p$ is constant: one can invert expressions for $U$ and $D_t U$ to find functions $W(U,D_tU)$ and $t(U,D_tU)$ and use
$$frac{partial p}{partial U}=frac{partial p}{partial W}frac{partial W}{partial U}+frac{partial p}{partial t}frac{partial t}{partial U}=0,$$
$$frac{partial p}{partial (D_tU)}=frac{partial p}{partial W}frac{partial W}{partial (D_tU)}+frac{partial p}{partial t}frac{partial t}{partial (D_tU)}=0,$$
because $frac{partial p}{partial W}=0$ and $frac{partial p}{partial t}=0$.
UPDATE: answering the updated question:
Assume that $U=f(x-vt)$, where $v=(hbaromega/2m)^{1/2}$ (it follows from (ii) that $f(x)=cos{((2momega/hbar)^{1/2}x)}$). Assume that there is a function $p$ that depends only on $U$ and its derivatives, and does not depend on $omega$. In this case, $P(x,t)$ is also a function of $x-vt$: $P(x,t)=F(x-vt)$.
As a consequence:
$$frac{D_t P}{D_x P}=-v=-sqrt{frac{hbaromega}{2m}},$$
which contradicts the assumption that $P(x,t)$ does not depend on $omega$.
Thank you very much for having answered my question. Unfortunately, this argument does not work. To see why note that $p$ is a function also of the spatial derivatives of $U$ and of $frac{partial U}{partial t}$. Such spatial derivatives are zero for the special solution you chose. So you cannot conclude that $p$ is a constant function.
– Maurizio Barbato
Nov 2 at 16:24
Dear @atarasenko, your argument about the second formulation is interesting, but only proves the following fact. Consider $P(x,t)$ for the special solution $U(x,t)=cosleft(sqrt{frac{2m omega}{hbar}}x-omega t right)$. We have $P(x,t)=F(x-vt)$, with $v=[hbar omega/(2m)]^{1/2}$. Under the $mathbf{hypothesis}$ that $F$ is not constant, then $P(x,t)$ must depend on $omega$. We can say it in another equivalent way: if $P(x,t)$ turns out to be independent of $omega$, then it is a constant function.
– Maurizio Barbato
Nov 3 at 10:06
Note that this is not the thesis I formulated in the post, since you can find some non-constant function $p$ such that $P(x,t)$ turns out to be constant for the special solution $U(x,t)=cosleft(sqrt{frac{2m omega}{hbar}}x-omega t right)$ (think of appropriate ratios of partial derivatives). The point to prove (which is difficult) is that such functions cannot satisfy a continuity equation.
– Maurizio Barbato
Nov 3 at 10:13
I have posted the question also on Math OverFlow: see Conserved Positive Charge for a PDE, but no one has given any answer up to now. So it is for sure a difficult problem, and frankly I do not imagine it can be given a simple and clear solution.
– Maurizio Barbato
Nov 3 at 10:22
add a comment |
Proof:
Consider equation (I) with $W(x)=const$ and the following initial conditions:
$$U(x,0)=0,$$
$$D_tU(x,0)=A,$$
and periodic boundary conditions:
$$U(x,t)=U(x+a,t),$$
where $a$ and $A$ are constants.
Solution of equation (I) depends only in $t$ in this case: $U=U(t)$, and $p=p(t)$, $j=0$. It has the following form:
$$U=frac{Ahbar}{W}sin{frac{Wt}{hbar}}$$
The first derivative is:
$$D_tU=Acos{frac{Wt}{hbar}}$$
Function $p=p(U,D_tU)$ has 2 properties:
1 It must not depend on $t$ due to the conservation law.
2 It must not depend on $W$ due to property (i) - $p$ does not change after transformation $Wrightarrow W+c$.
However, such function $p$ is constant: one can invert expressions for $U$ and $D_t U$ to find functions $W(U,D_tU)$ and $t(U,D_tU)$ and use
$$frac{partial p}{partial U}=frac{partial p}{partial W}frac{partial W}{partial U}+frac{partial p}{partial t}frac{partial t}{partial U}=0,$$
$$frac{partial p}{partial (D_tU)}=frac{partial p}{partial W}frac{partial W}{partial (D_tU)}+frac{partial p}{partial t}frac{partial t}{partial (D_tU)}=0,$$
because $frac{partial p}{partial W}=0$ and $frac{partial p}{partial t}=0$.
UPDATE: answering the updated question:
Assume that $U=f(x-vt)$, where $v=(hbaromega/2m)^{1/2}$ (it follows from (ii) that $f(x)=cos{((2momega/hbar)^{1/2}x)}$). Assume that there is a function $p$ that depends only on $U$ and its derivatives, and does not depend on $omega$. In this case, $P(x,t)$ is also a function of $x-vt$: $P(x,t)=F(x-vt)$.
As a consequence:
$$frac{D_t P}{D_x P}=-v=-sqrt{frac{hbaromega}{2m}},$$
which contradicts the assumption that $P(x,t)$ does not depend on $omega$.
Proof:
Consider equation (I) with $W(x)=const$ and the following initial conditions:
$$U(x,0)=0,$$
$$D_tU(x,0)=A,$$
and periodic boundary conditions:
$$U(x,t)=U(x+a,t),$$
where $a$ and $A$ are constants.
Solution of equation (I) depends only in $t$ in this case: $U=U(t)$, and $p=p(t)$, $j=0$. It has the following form:
$$U=frac{Ahbar}{W}sin{frac{Wt}{hbar}}$$
The first derivative is:
$$D_tU=Acos{frac{Wt}{hbar}}$$
Function $p=p(U,D_tU)$ has 2 properties:
1 It must not depend on $t$ due to the conservation law.
2 It must not depend on $W$ due to property (i) - $p$ does not change after transformation $Wrightarrow W+c$.
However, such function $p$ is constant: one can invert expressions for $U$ and $D_t U$ to find functions $W(U,D_tU)$ and $t(U,D_tU)$ and use
$$frac{partial p}{partial U}=frac{partial p}{partial W}frac{partial W}{partial U}+frac{partial p}{partial t}frac{partial t}{partial U}=0,$$
$$frac{partial p}{partial (D_tU)}=frac{partial p}{partial W}frac{partial W}{partial (D_tU)}+frac{partial p}{partial t}frac{partial t}{partial (D_tU)}=0,$$
because $frac{partial p}{partial W}=0$ and $frac{partial p}{partial t}=0$.
UPDATE: answering the updated question:
Assume that $U=f(x-vt)$, where $v=(hbaromega/2m)^{1/2}$ (it follows from (ii) that $f(x)=cos{((2momega/hbar)^{1/2}x)}$). Assume that there is a function $p$ that depends only on $U$ and its derivatives, and does not depend on $omega$. In this case, $P(x,t)$ is also a function of $x-vt$: $P(x,t)=F(x-vt)$.
As a consequence:
$$frac{D_t P}{D_x P}=-v=-sqrt{frac{hbaromega}{2m}},$$
which contradicts the assumption that $P(x,t)$ does not depend on $omega$.
edited Nov 3 at 20:31
Maurizio Barbato
1,239620
1,239620
answered Nov 2 at 15:45
atarasenko
2764
2764
Thank you very much for having answered my question. Unfortunately, this argument does not work. To see why note that $p$ is a function also of the spatial derivatives of $U$ and of $frac{partial U}{partial t}$. Such spatial derivatives are zero for the special solution you chose. So you cannot conclude that $p$ is a constant function.
– Maurizio Barbato
Nov 2 at 16:24
Dear @atarasenko, your argument about the second formulation is interesting, but only proves the following fact. Consider $P(x,t)$ for the special solution $U(x,t)=cosleft(sqrt{frac{2m omega}{hbar}}x-omega t right)$. We have $P(x,t)=F(x-vt)$, with $v=[hbar omega/(2m)]^{1/2}$. Under the $mathbf{hypothesis}$ that $F$ is not constant, then $P(x,t)$ must depend on $omega$. We can say it in another equivalent way: if $P(x,t)$ turns out to be independent of $omega$, then it is a constant function.
– Maurizio Barbato
Nov 3 at 10:06
Note that this is not the thesis I formulated in the post, since you can find some non-constant function $p$ such that $P(x,t)$ turns out to be constant for the special solution $U(x,t)=cosleft(sqrt{frac{2m omega}{hbar}}x-omega t right)$ (think of appropriate ratios of partial derivatives). The point to prove (which is difficult) is that such functions cannot satisfy a continuity equation.
– Maurizio Barbato
Nov 3 at 10:13
I have posted the question also on Math OverFlow: see Conserved Positive Charge for a PDE, but no one has given any answer up to now. So it is for sure a difficult problem, and frankly I do not imagine it can be given a simple and clear solution.
– Maurizio Barbato
Nov 3 at 10:22
add a comment |
Thank you very much for having answered my question. Unfortunately, this argument does not work. To see why note that $p$ is a function also of the spatial derivatives of $U$ and of $frac{partial U}{partial t}$. Such spatial derivatives are zero for the special solution you chose. So you cannot conclude that $p$ is a constant function.
– Maurizio Barbato
Nov 2 at 16:24
Dear @atarasenko, your argument about the second formulation is interesting, but only proves the following fact. Consider $P(x,t)$ for the special solution $U(x,t)=cosleft(sqrt{frac{2m omega}{hbar}}x-omega t right)$. We have $P(x,t)=F(x-vt)$, with $v=[hbar omega/(2m)]^{1/2}$. Under the $mathbf{hypothesis}$ that $F$ is not constant, then $P(x,t)$ must depend on $omega$. We can say it in another equivalent way: if $P(x,t)$ turns out to be independent of $omega$, then it is a constant function.
– Maurizio Barbato
Nov 3 at 10:06
Note that this is not the thesis I formulated in the post, since you can find some non-constant function $p$ such that $P(x,t)$ turns out to be constant for the special solution $U(x,t)=cosleft(sqrt{frac{2m omega}{hbar}}x-omega t right)$ (think of appropriate ratios of partial derivatives). The point to prove (which is difficult) is that such functions cannot satisfy a continuity equation.
– Maurizio Barbato
Nov 3 at 10:13
I have posted the question also on Math OverFlow: see Conserved Positive Charge for a PDE, but no one has given any answer up to now. So it is for sure a difficult problem, and frankly I do not imagine it can be given a simple and clear solution.
– Maurizio Barbato
Nov 3 at 10:22
Thank you very much for having answered my question. Unfortunately, this argument does not work. To see why note that $p$ is a function also of the spatial derivatives of $U$ and of $frac{partial U}{partial t}$. Such spatial derivatives are zero for the special solution you chose. So you cannot conclude that $p$ is a constant function.
– Maurizio Barbato
Nov 2 at 16:24
Thank you very much for having answered my question. Unfortunately, this argument does not work. To see why note that $p$ is a function also of the spatial derivatives of $U$ and of $frac{partial U}{partial t}$. Such spatial derivatives are zero for the special solution you chose. So you cannot conclude that $p$ is a constant function.
– Maurizio Barbato
Nov 2 at 16:24
Dear @atarasenko, your argument about the second formulation is interesting, but only proves the following fact. Consider $P(x,t)$ for the special solution $U(x,t)=cosleft(sqrt{frac{2m omega}{hbar}}x-omega t right)$. We have $P(x,t)=F(x-vt)$, with $v=[hbar omega/(2m)]^{1/2}$. Under the $mathbf{hypothesis}$ that $F$ is not constant, then $P(x,t)$ must depend on $omega$. We can say it in another equivalent way: if $P(x,t)$ turns out to be independent of $omega$, then it is a constant function.
– Maurizio Barbato
Nov 3 at 10:06
Dear @atarasenko, your argument about the second formulation is interesting, but only proves the following fact. Consider $P(x,t)$ for the special solution $U(x,t)=cosleft(sqrt{frac{2m omega}{hbar}}x-omega t right)$. We have $P(x,t)=F(x-vt)$, with $v=[hbar omega/(2m)]^{1/2}$. Under the $mathbf{hypothesis}$ that $F$ is not constant, then $P(x,t)$ must depend on $omega$. We can say it in another equivalent way: if $P(x,t)$ turns out to be independent of $omega$, then it is a constant function.
– Maurizio Barbato
Nov 3 at 10:06
Note that this is not the thesis I formulated in the post, since you can find some non-constant function $p$ such that $P(x,t)$ turns out to be constant for the special solution $U(x,t)=cosleft(sqrt{frac{2m omega}{hbar}}x-omega t right)$ (think of appropriate ratios of partial derivatives). The point to prove (which is difficult) is that such functions cannot satisfy a continuity equation.
– Maurizio Barbato
Nov 3 at 10:13
Note that this is not the thesis I formulated in the post, since you can find some non-constant function $p$ such that $P(x,t)$ turns out to be constant for the special solution $U(x,t)=cosleft(sqrt{frac{2m omega}{hbar}}x-omega t right)$ (think of appropriate ratios of partial derivatives). The point to prove (which is difficult) is that such functions cannot satisfy a continuity equation.
– Maurizio Barbato
Nov 3 at 10:13
I have posted the question also on Math OverFlow: see Conserved Positive Charge for a PDE, but no one has given any answer up to now. So it is for sure a difficult problem, and frankly I do not imagine it can be given a simple and clear solution.
– Maurizio Barbato
Nov 3 at 10:22
I have posted the question also on Math OverFlow: see Conserved Positive Charge for a PDE, but no one has given any answer up to now. So it is for sure a difficult problem, and frankly I do not imagine it can be given a simple and clear solution.
– Maurizio Barbato
Nov 3 at 10:22
add a comment |
After some careful thought I must conclude that the mathematical formulation of Bohm's statement I have given in my post above is wrong for the reason that atarasenko identified in his comment.
Indeed, if we start from the usual on-dimensional Schrödinger equation for a particle in the case of a potential $W(x)$:
begin{equation}
i hbar frac{partial psi}{partial t}(x,t) = -frac{hbar^2}{2m} frac{partial^2}{partial x^2}psi(x,t) + W(x) psi(x,t),
end{equation}
and we change $W(x)$ by adding a constant $W_0$, the new solution $psi$ corresponding to the initial data $x mapsto psi(x,0)$ is $tilde{psi}(x,t)=psi(x,t) exp(-iW_0 t/ hbar)$. If $U$ and $V$ are respectively the real and imaginary part of $psi$, and $tilde{U}$ and $tilde{V}$ are respectively the real and imaginary part of $tilde{psi}$, then from
begin{equation}
frac{partial tilde{psi}}{partial t} = exp(-iW_0t/ hbar) left[ frac{partial psi}{partial t} - frac{i W_0}{hbar} psi right],
end{equation}
we get
begin{equation}
frac{partial tilde{U}}{partial t}(x,0)= frac{partial U}{partial t}(x,0) + frac{W_0}{hbar} V(x,0),
end{equation}
so $x mapsto frac{partial tilde{U}}{partial t}(x,0)$ and $x mapsto frac{partial U}{partial t}(x,0)$ do not coincide. Since Bohm requires that the usual theory based on the complex-valued wave function $psi$ and the (eventual) theory based on the real-valued wave function $U$ should be equivalent, we conclude that the mathematical formulation I have given above is incorrect.
As for the second mathematical formulation of Bohm's statement I gave in my NOTE (2) above, I have discussed it at some length in my post on MathOverflow Conserved Positive Charge for a PDE, but no one could give some answer. This problem seems not trivial at all, and maybe it could have a positive answer, in the sense that there exist some functions $p geq 0$ and $j$ which satisfy properties (i),(ii) and (iii) I listed in NOTE (2) above. For this reason, I suspect that this mathematical formulation is flawed too, this time probably because Bohm had in his mind some kind of invariance argument based on physical considerations (hidden in Condition (3) required for the probability function in Section (4.2) of his book Quantum Theory) which we do not know and which would make the non-existence result transparent.
add a comment |
After some careful thought I must conclude that the mathematical formulation of Bohm's statement I have given in my post above is wrong for the reason that atarasenko identified in his comment.
Indeed, if we start from the usual on-dimensional Schrödinger equation for a particle in the case of a potential $W(x)$:
begin{equation}
i hbar frac{partial psi}{partial t}(x,t) = -frac{hbar^2}{2m} frac{partial^2}{partial x^2}psi(x,t) + W(x) psi(x,t),
end{equation}
and we change $W(x)$ by adding a constant $W_0$, the new solution $psi$ corresponding to the initial data $x mapsto psi(x,0)$ is $tilde{psi}(x,t)=psi(x,t) exp(-iW_0 t/ hbar)$. If $U$ and $V$ are respectively the real and imaginary part of $psi$, and $tilde{U}$ and $tilde{V}$ are respectively the real and imaginary part of $tilde{psi}$, then from
begin{equation}
frac{partial tilde{psi}}{partial t} = exp(-iW_0t/ hbar) left[ frac{partial psi}{partial t} - frac{i W_0}{hbar} psi right],
end{equation}
we get
begin{equation}
frac{partial tilde{U}}{partial t}(x,0)= frac{partial U}{partial t}(x,0) + frac{W_0}{hbar} V(x,0),
end{equation}
so $x mapsto frac{partial tilde{U}}{partial t}(x,0)$ and $x mapsto frac{partial U}{partial t}(x,0)$ do not coincide. Since Bohm requires that the usual theory based on the complex-valued wave function $psi$ and the (eventual) theory based on the real-valued wave function $U$ should be equivalent, we conclude that the mathematical formulation I have given above is incorrect.
As for the second mathematical formulation of Bohm's statement I gave in my NOTE (2) above, I have discussed it at some length in my post on MathOverflow Conserved Positive Charge for a PDE, but no one could give some answer. This problem seems not trivial at all, and maybe it could have a positive answer, in the sense that there exist some functions $p geq 0$ and $j$ which satisfy properties (i),(ii) and (iii) I listed in NOTE (2) above. For this reason, I suspect that this mathematical formulation is flawed too, this time probably because Bohm had in his mind some kind of invariance argument based on physical considerations (hidden in Condition (3) required for the probability function in Section (4.2) of his book Quantum Theory) which we do not know and which would make the non-existence result transparent.
add a comment |
After some careful thought I must conclude that the mathematical formulation of Bohm's statement I have given in my post above is wrong for the reason that atarasenko identified in his comment.
Indeed, if we start from the usual on-dimensional Schrödinger equation for a particle in the case of a potential $W(x)$:
begin{equation}
i hbar frac{partial psi}{partial t}(x,t) = -frac{hbar^2}{2m} frac{partial^2}{partial x^2}psi(x,t) + W(x) psi(x,t),
end{equation}
and we change $W(x)$ by adding a constant $W_0$, the new solution $psi$ corresponding to the initial data $x mapsto psi(x,0)$ is $tilde{psi}(x,t)=psi(x,t) exp(-iW_0 t/ hbar)$. If $U$ and $V$ are respectively the real and imaginary part of $psi$, and $tilde{U}$ and $tilde{V}$ are respectively the real and imaginary part of $tilde{psi}$, then from
begin{equation}
frac{partial tilde{psi}}{partial t} = exp(-iW_0t/ hbar) left[ frac{partial psi}{partial t} - frac{i W_0}{hbar} psi right],
end{equation}
we get
begin{equation}
frac{partial tilde{U}}{partial t}(x,0)= frac{partial U}{partial t}(x,0) + frac{W_0}{hbar} V(x,0),
end{equation}
so $x mapsto frac{partial tilde{U}}{partial t}(x,0)$ and $x mapsto frac{partial U}{partial t}(x,0)$ do not coincide. Since Bohm requires that the usual theory based on the complex-valued wave function $psi$ and the (eventual) theory based on the real-valued wave function $U$ should be equivalent, we conclude that the mathematical formulation I have given above is incorrect.
As for the second mathematical formulation of Bohm's statement I gave in my NOTE (2) above, I have discussed it at some length in my post on MathOverflow Conserved Positive Charge for a PDE, but no one could give some answer. This problem seems not trivial at all, and maybe it could have a positive answer, in the sense that there exist some functions $p geq 0$ and $j$ which satisfy properties (i),(ii) and (iii) I listed in NOTE (2) above. For this reason, I suspect that this mathematical formulation is flawed too, this time probably because Bohm had in his mind some kind of invariance argument based on physical considerations (hidden in Condition (3) required for the probability function in Section (4.2) of his book Quantum Theory) which we do not know and which would make the non-existence result transparent.
After some careful thought I must conclude that the mathematical formulation of Bohm's statement I have given in my post above is wrong for the reason that atarasenko identified in his comment.
Indeed, if we start from the usual on-dimensional Schrödinger equation for a particle in the case of a potential $W(x)$:
begin{equation}
i hbar frac{partial psi}{partial t}(x,t) = -frac{hbar^2}{2m} frac{partial^2}{partial x^2}psi(x,t) + W(x) psi(x,t),
end{equation}
and we change $W(x)$ by adding a constant $W_0$, the new solution $psi$ corresponding to the initial data $x mapsto psi(x,0)$ is $tilde{psi}(x,t)=psi(x,t) exp(-iW_0 t/ hbar)$. If $U$ and $V$ are respectively the real and imaginary part of $psi$, and $tilde{U}$ and $tilde{V}$ are respectively the real and imaginary part of $tilde{psi}$, then from
begin{equation}
frac{partial tilde{psi}}{partial t} = exp(-iW_0t/ hbar) left[ frac{partial psi}{partial t} - frac{i W_0}{hbar} psi right],
end{equation}
we get
begin{equation}
frac{partial tilde{U}}{partial t}(x,0)= frac{partial U}{partial t}(x,0) + frac{W_0}{hbar} V(x,0),
end{equation}
so $x mapsto frac{partial tilde{U}}{partial t}(x,0)$ and $x mapsto frac{partial U}{partial t}(x,0)$ do not coincide. Since Bohm requires that the usual theory based on the complex-valued wave function $psi$ and the (eventual) theory based on the real-valued wave function $U$ should be equivalent, we conclude that the mathematical formulation I have given above is incorrect.
As for the second mathematical formulation of Bohm's statement I gave in my NOTE (2) above, I have discussed it at some length in my post on MathOverflow Conserved Positive Charge for a PDE, but no one could give some answer. This problem seems not trivial at all, and maybe it could have a positive answer, in the sense that there exist some functions $p geq 0$ and $j$ which satisfy properties (i),(ii) and (iii) I listed in NOTE (2) above. For this reason, I suspect that this mathematical formulation is flawed too, this time probably because Bohm had in his mind some kind of invariance argument based on physical considerations (hidden in Condition (3) required for the probability function in Section (4.2) of his book Quantum Theory) which we do not know and which would make the non-existence result transparent.
edited Nov 29 at 20:29
answered Nov 29 at 19:33
Maurizio Barbato
1,239620
1,239620
add a comment |
add a comment |
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If a constant $c$ is added to $W$, the wave function $psi$ is multiplied by $exp(-ict/hbar)$. The real part of the wave function $U$ changes in a nontrivial way: is becomes a linear combination of $U$ and $V$. Therefore it is not clear what you mean by same initial conditions in property (i).
– atarasenko
Nov 2 at 13:05
Also an interesting point: if $W(x)$ is real, then there always exists a solution $psi=iV(t,x)$, with $U=0$. In this case any function $p$ that depends only on $U$ and its derivatives is constant. But from the physical point of view solutions $psi=U(t,x)$ and $psi=iU(t,x)$ are equivalent, because the wave functions for them differ only by a constant phase.
– atarasenko
Nov 2 at 13:24
@atarasenko As for the first observation, I mean exactly the same $U(x,0)$ and $frac{partial U}{partial t}(x,0)$. As for the second observation it is simply wrong. Write down the Schrödinger equation and you will see that $i V(x,t)$ cannot be a solution.
– Maurizio Barbato
Nov 2 at 13:44
If $psi=U$ is a solution, then $ipsi$ is a solution as well. It is due to linearity of the Schrödinger equation.
– atarasenko
Nov 2 at 13:53
Regarding the first observation - if the initial conditions $U(0,x)$ and $D_tU(0,x)$ are the same but functions $W$ differ by a constant, then the wave functions $psi$ will be completely different (not just by a phase factor $exp(−ict/hbar)$). And there is no reason for $p$ to be the same for 2 physically different solutions.
– atarasenko
Nov 2 at 13:56