Science > Physics > Simultaneous vanishing of a wave function and its derivative?
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Science > Physics |
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"" |
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31 Aug 2006 03:21:32 PM |
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Simultaneous vanishing of a wave function and its derivative? |
Confer a one-dimensional one-particle wave function of a bound
state. Obviously, it is easily possible that the wave function and its
second derivative vanish simultaneously, e.g., at the zero-crossings
if the wave function is sinusoidal.
But what's about points where the wave function and its first
derivative vanish both? Apparently, this happens in regions of
infinite potential, where the wave function is equivalent to zero, and
in the infinity where a bound state vanishes also.
My question is: Is it possible that both the wave function and its
first derivative vanish at a certain point though the potential is
finite there? In other words: is it thinkable that within a small
region the amplitude is zero just at one point without having a node
there? How can the answer be deduced from the time independent
Schroedinger equation H psi(x) = E psi(x) with the Hamiltonian H(x) =
(-i hbar d/dx)^2/2m + V(x)?
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| User: "" |
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| Title: Re: Simultaneous vanishing of a wave function and its derivative? |
31 Aug 2006 04:49:12 PM |
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In article <1157055692.590203.259320@h48g2000cwc.googlegroups.com>, writes:
Confer a one-dimensional one-particle wave function of a bound
state. Obviously, it is easily possible that the wave function and its
second derivative vanish simultaneously, e.g., at the zero-crossings
if the wave function is sinusoidal.
But what's about points where the wave function and its first
derivative vanish both? Apparently, this happens in regions of
infinite potential, where the wave function is equivalent to zero, and
in the infinity where a bound state vanishes also.
My question is: Is it possible that both the wave function and its
first derivative vanish at a certain point though the potential is
finite there? In other words: is it thinkable that within a small
region the amplitude is zero just at one point without having a node
there? How can the answer be deduced from the time independent
Schroedinger equation H psi(x) = E psi(x) with the Hamiltonian H(x) =
(-i hbar d/dx)^2/2m + V(x)?
I would think that, at least for a well behaved potential, the answer
is no. You've a second order equation which, together with
appropriate initial conditions (values of function and derivative at a
given point, for example) has a unique solution. But there is already
a solution for the case where both the function and its derivative are
zero at the same point, i.e. the solution psi = 0. so, by uniqueness,
that's the only solution.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "" |
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| Title: Re: Simultaneous vanishing of a wave function and its derivative? |
01 Sep 2006 05:15:01 PM |
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wrote:
I would think that, at least for a well behaved potential, the answer
is no. You've a second order equation which, together with
appropriate initial conditions (values of function and derivative at a
given point, for example) has a unique solution. But there is already
a solution for the case where both the function and its derivative are
zero at the same point, i.e. the solution psi = 0. so, by uniqueness,
that's the only solution.
With your help I found the right theorem in a textbook about
analysis. Shortened repetition: If the coefficient functions of a
linear differential equation of n-th order are continuous in [a,b],
then there exists for each x0 in [a,b] an isomorphism between the
n-dimensional solution space for psi(x) and the tuple of initial
values {psi(x0), psi'(x0), ..., psi^(n-1)(x0)} where psi^(n) denotes
the n-th derivative.
So a well-behaved potential has to be continuous, especially it must
not be infinite at any point. The mathematics is clear now but how can
I understand intuitively from the quantum mechanics point of view
that psi and psi' never vanish simultaneously? (In a real system the
potential is always finite.)
Martina
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| User: "" |
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| Title: Re: Simultaneous vanishing of a wave function and its derivative? |
01 Sep 2006 07:32:38 PM |
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In article <1157148901.289931.251750@e3g2000cwe.googlegroups.com>, writes:
mmeron@cars3.uchicago.edu wrote:
I would think that, at least for a well behaved potential, the answer
is no. You've a second order equation which, together with
appropriate initial conditions (values of function and derivative at a
given point, for example) has a unique solution. But there is already
a solution for the case where both the function and its derivative are
zero at the same point, i.e. the solution psi = 0. so, by uniqueness,
that's the only solution.
With your help I found the right theorem in a textbook about
analysis. Shortened repetition: If the coefficient functions of a
linear differential equation of n-th order are continuous in [a,b],
then there exists for each x0 in [a,b] an isomorphism between the
n-dimensional solution space for psi(x) and the tuple of initial
values {psi(x0), psi'(x0), ..., psi^(n-1)(x0)} where psi^(n) denotes
the n-th derivative.
Yes, that's the ticket.
So a well-behaved potential has to be continuous, especially it must
not be infinite at any point.
Indeed. Though (my memory is a bit hazy here) it is possible that
"piecewise continuous" is still OK. But certainly not infinite.
The mathematics is clear now but how can
I understand intuitively from the quantum mechanics point of view
that psi and psi' never vanish simultaneously? (In a real system the
potential is always finite.)
Hmm, that's a good question. I'll get back to you on this if
something occurs to me.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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