Science > Physics > The =?ISO-8859-1?Q?Schr=F6dinger_Equation=2C_potential_box?==?ISO-8859-1?Q?_problem?=
| Topic: |
Science > Physics |
| User: |
"Carl" |
| Date: |
15 Sep 2006 03:57:18 PM |
| Object: |
The =?ISO-8859-1?Q?Schr=F6dinger_Equation=2C_potential_box?==?ISO-8859-1?Q?_problem?= |
Hello,
While trying to solve a physics problem relating to the Schrödinger
equation, we found something that seems unlogical. The problem given is
a real "standard" problem about separable time- and space-dependent
parts of the solution to the Schrödinger Equation.
The following is given:
An electron in the potential box
V(X)={inf for x<0, 0 for 0<=x<=L, inf for x>L}
is at t=0 described by the wave function
psi(x, 0) = 0.8psi_1(x)+0.6psi_2(x)
where the orthonormal eigenfunctions
psi_n(x) = sqrt(2/L)sin(n*pi*x/L)
and the corresponding eigenvalues
E_n = (n*pi*h~/L)^2/(2m)=n^2*E_1, n=1, 2, ...
It is asked for:
a) Make sure psi is normalized.
b) Sketch the probability of finding the electron as a function of x at
t=0, that is, P(x,0).
c) Sketch the corresponding graph at time t_1=pi*h~/(3E_1)
So, to our problem: a) is straight-forward. b) gives a sketch of an
unsymmetric function, heavy on the left side, with a peak at around
x=0.3L.
Our problems start when trying to solve c). The reason is that our
calculations and reasonings give that P(x, t_1) would be the same as
P(x, 0). Because a number of the following questions regard the
differences of P(x, 0) and P(x, t_1) and the corresponding expectation
values and so on, we feel that we must be wrong.
However, for any potential function V(x) that is not time-dependent
(V(x)=0 for all x within [0,L] clearly is time-independent), psi shall
be a time-space-dependance separable function. In those cases, denoting
the time-dependent function as phi(t), phi(t) can be written as
phi(t)=exp(-iEt/h~).
That would mean that for P(x,t),
P(x,t)=psi(x,t)*complex-conjugate(psi(x,t))=psi(x)^2 since the phi's
will multiply to unity for all t. That would mean that P(x,t) is the
same for all t.
What are we doing wrong? We just cannot believe that P(x, t) is
independent of t. We are also annoyed by the fact that P(x,t) is
unsymmetric over x (we would expect symmetry around L/2, especially
since there is no potential). If there would be a time-dependance, and
a phase velocity for P(x, t), we could still achieve symmetry if
calculating the average over time.
Any ideas are greatly appreciated!
Best regards,
Carl
--
Posted via a free Usenet account from http://www.teranews.com
.
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| User: "Jim Black" |
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| Title: =?iso-8859-1?q?Re:_The_Schr=F6dinger_Equation,_potential_box_problem?= |
15 Sep 2006 04:32:31 PM |
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Carl wrote:
Hello,
While trying to solve a physics problem relating to the Schr=F6dinger
equation, we found something that seems unlogical. The problem given is
a real "standard" problem about separable time- and space-dependent
parts of the solution to the Schr=F6dinger Equation.
The following is given:
An electron in the potential box
V(X)=3D{inf for x<0, 0 for 0<=3Dx<=3DL, inf for x>L}
is at t=3D0 described by the wave function
psi(x, 0) =3D 0.8psi_1(x)+0.6psi_2(x)
where the orthonormal eigenfunctions
psi_n(x) =3D sqrt(2/L)sin(n*pi*x/L)
and the corresponding eigenvalues
E_n =3D (n*pi*h~/L)^2/(2m)=3Dn^2*E_1, n=3D1, 2, ...
It is asked for:
a) Make sure psi is normalized.
b) Sketch the probability of finding the electron as a function of x at
t=3D0, that is, P(x,0).
c) Sketch the corresponding graph at time t_1=3Dpi*h~/(3E_1)
So, to our problem: a) is straight-forward. b) gives a sketch of an
unsymmetric function, heavy on the left side, with a peak at around
x=3D0.3L.
Our problems start when trying to solve c). The reason is that our
calculations and reasonings give that P(x, t_1) would be the same as
P(x, 0).
You are correct.
Because a number of the following questions regard the
differences of P(x, 0) and P(x, t_1) and the corresponding expectation
values and so on, we feel that we must be wrong.
However, for any potential function V(x) that is not time-dependent
(V(x)=3D0 for all x within [0,L] clearly is time-independent), psi shall
be a time-space-dependance separable function.
Actually, psi only has to be a sum of such functions, i.e., you can
have:
psi(x,t) =3D f_1(x) g_1(t) + f_2(x) g_2(t) + ...
In those cases, denoting
the time-dependent function as phi(t), phi(t) can be written as
phi(t)=3Dexp(-iEt/h~).
That would mean that for P(x,t),
P(x,t)=3Dpsi(x,t)*complex-conjugate(psi(x,t))=3Dpsi(x)^2 since the phi's
will multiply to unity for all t. That would mean that P(x,t) is the
same for all t.
What are we doing wrong? We just cannot believe that P(x, t) is
independent of t.
Your disbelief is justified. Sampling P(x,t) at two times and finding
that it is the same at both instants does not prove that P(x,t) is
time-independent.
We are also annoyed by the fact that P(x,t) is
unsymmetric over x (we would expect symmetry around L/2, especially
since there is no potential). If there would be a time-dependance, and
a phase velocity for P(x, t), we could still achieve symmetry if
calculating the average over time.
Any ideas are greatly appreciated!
Try calculating P(x, t_1/2). This may help you to see what's going on.
.
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| User: "" |
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| Title: =?iso-8859-1?q?Re:_The_Schr=F6dinger_Equation,_potential_box_problem?= |
16 Sep 2006 01:47:02 PM |
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Thanks for the reply.
However, for any potential function V(x) that is not time-dependent
(V(x)=0 for all x within [0,L] clearly is time-independent), psi shall
be a time-space-dependance separable function.
Actually, psi only has to be a sum of such functions, i.e., you can
have:
psi(x,t) = f_1(x) g_1(t) + f_2(x) g_2(t) + ...
In those cases, denoting
the time-dependent function as phi(t), phi(t) can be written as
phi(t)=exp(-iEt/h~).
That would mean that for P(x,t),
P(x,t)=psi(x,t)*complex-conjugate(psi(x,t))=psi(x)^2 since the phi's
will multiply to unity for all t. That would mean that P(x,t) is the
same for all t.
What are we doing wrong? We just cannot believe that P(x, t) is
independent of t.
Your disbelief is justified. Sampling P(x,t) at two times and finding
that it is the same at both instants does not prove that P(x,t) is
time-independent.
We are also annoyed by the fact that P(x,t) is
unsymmetric over x (we would expect symmetry around L/2, especially
since there is no potential). If there would be a time-dependance, and
a phase velocity for P(x, t), we could still achieve symmetry if
calculating the average over time.
Any ideas are greatly appreciated!
Try calculating P(x, t_1/2). This may help you to see what's going on.
But according to our calculations and our reasoning, P(x,t)=P(x) - in
the expression for P(x,t) there simply is no t, and therefore,
inserting t_1/2 or any other t always gives the same P(x,t).
As written above, when calculating P(x,t), I get
P(x,t)=psi(x,t)*complex-conjugate(psi(x,t))=psi(x)^2
since phi(t)*complex-conjugate(phi(t))=1 (that is, for any t).
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| User: "Carl" |
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| Title: Re: The =?ISO-8859-1?Q?Schr=F6dinger_Equation=2C_potential_?==?ISO-8859-1?Q?box_problem?= |
16 Sep 2006 05:26:43 AM |
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Thanks for the reply.
However, for any potential function V(x) that is not time-dependent
(V(x)=0 for all x within [0,L] clearly is time-independent), psi shall
be a time-space-dependance separable function.
Actually, psi only has to be a sum of such functions, i.e., you can
have:
psi(x,t) = f_1(x) g_1(t) + f_2(x) g_2(t) + ...
In those cases, denoting
the time-dependent function as phi(t), phi(t) can be written as
phi(t)=exp(-iEt/h~).
That would mean that for P(x,t),
P(x,t)=psi(x,t)*complex-conjugate(psi(x,t))=psi(x)^2 since the phi's
will multiply to unity for all t. That would mean that P(x,t) is the
same for all t.
What are we doing wrong? We just cannot believe that P(x, t) is
independent of t.
Your disbelief is justified. Sampling P(x,t) at two times and finding
that it is the same at both instants does not prove that P(x,t) is
time-independent.
We are also annoyed by the fact that P(x,t) is
unsymmetric over x (we would expect symmetry around L/2, especially
since there is no potential). If there would be a time-dependance, and
a phase velocity for P(x, t), we could still achieve symmetry if
calculating the average over time.
Any ideas are greatly appreciated!
Try calculating P(x, t_1/2). This may help you to see what's going on.
But according to our calculations and our reasoning, P(x,t)=P(x) - in
the expression for P(x,t) there simply is no t, and therefore, inserting
t_1/2 or any other t always gives the same P(x,t).
As written above, when calculating P(x,t), I get
P(x,t)=psi(x,t)*complex-conjugate(psi(x,t))=psi(x)^2
since phi(t)*complex-conjugate(phi(t))=1 (that is, for any t).
--
Posted via a free Usenet account from http://www.teranews.com
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| User: "Greg Hansen" |
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| Title: Re: The =?ISO-8859-1?Q?Schr=F6dinger_Equation=2C_potential_?==?ISO-8859-1?Q?box_problem?= |
16 Sep 2006 09:20:11 AM |
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Carl wrote:
Hello,
While trying to solve a physics problem relating to the Schrödinger
equation, we found something that seems unlogical. The problem given is
a real "standard" problem about separable time- and space-dependent
parts of the solution to the Schrödinger Equation.
The following is given:
An electron in the potential box
V(X)={inf for x<0, 0 for 0<=x<=L, inf for x>L}
is at t=0 described by the wave function
psi(x, 0) = 0.8psi_1(x)+0.6psi_2(x)
where the orthonormal eigenfunctions
psi_n(x) = sqrt(2/L)sin(n*pi*x/L)
and the corresponding eigenvalues
E_n = (n*pi*h~/L)^2/(2m)=n^2*E_1, n=1, 2, ...
It is asked for:
a) Make sure psi is normalized.
b) Sketch the probability of finding the electron as a function of x at
t=0, that is, P(x,0).
c) Sketch the corresponding graph at time t_1=pi*h~/(3E_1)
So, to our problem: a) is straight-forward. b) gives a sketch of an
unsymmetric function, heavy on the left side, with a peak at around
x=0.3L.
Our problems start when trying to solve c). The reason is that our
calculations and reasonings give that P(x, t_1) would be the same as
P(x, 0). Because a number of the following questions regard the
differences of P(x, 0) and P(x, t_1) and the corresponding expectation
values and so on, we feel that we must be wrong.
Well, stop for a moment and think about what it *should* be. Quantum
mechanics has waves. What would you expect an asymmetric wave in a box
to do? It's heavy on the left side. Then it will slosh over to the
right side, and back to the left side, and at a particular time it will
be identical to what it was as t=0. It will slosh back and forth.
However, for any potential function V(x) that is not time-dependent
(V(x)=0 for all x within [0,L] clearly is time-independent), psi shall
be a time-space-dependance separable function. In those cases, denoting
the time-dependent function as phi(t), phi(t) can be written as
phi(t)=exp(-iEt/h~).
That would mean that for P(x,t),
P(x,t)=psi(x,t)*complex-conjugate(psi(x,t))=psi(x)^2 since the phi's
will multiply to unity for all t. That would mean that P(x,t) is the
same for all t.
What are we doing wrong? We just cannot believe that P(x, t) is
independent of t. We are also annoyed by the fact that P(x,t) is
unsymmetric over x (we would expect symmetry around L/2, especially
since there is no potential). If there would be a time-dependance, and
a phase velocity for P(x, t), we could still achieve symmetry if
calculating the average over time.
Your wavefunction is not an eigenfunction of energy. The time evolution
operator is U=exp(-iHt/hbar), so
psi(t) = U psi(0)
= U psi_1 + U psi_2 {and psi1, psi2 are energy eigenfunctions}
= exp(-i w1 t) psi_1 + exp(-i w2 t) psi_2
Find the complex conjugate of that, and the time dependence won't go away.
.
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| User: "" |
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| Title: =?iso-8859-1?q?Re:_The_Schr=F6dinger_Equation,_potential_box_problem?= |
16 Sep 2006 01:48:30 PM |
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Greg Hansen wrote:
Carl wrote:
Hello,
While trying to solve a physics problem relating to the Schr=F6dinger
equation, we found something that seems unlogical. The problem given is
a real "standard" problem about separable time- and space-dependent
parts of the solution to the Schr=F6dinger Equation.
The following is given:
An electron in the potential box
V(X)=3D{inf for x<0, 0 for 0<=3Dx<=3DL, inf for x>L}
is at t=3D0 described by the wave function
psi(x, 0) =3D 0.8psi_1(x)+0.6psi_2(x)
where the orthonormal eigenfunctions
psi_n(x) =3D sqrt(2/L)sin(n*pi*x/L)
and the corresponding eigenvalues
E_n =3D (n*pi*h~/L)^2/(2m)=3Dn^2*E_1, n=3D1, 2, ...
It is asked for:
a) Make sure psi is normalized.
b) Sketch the probability of finding the electron as a function of x at
t=3D0, that is, P(x,0).
c) Sketch the corresponding graph at time t_1=3Dpi*h~/(3E_1)
So, to our problem: a) is straight-forward. b) gives a sketch of an
unsymmetric function, heavy on the left side, with a peak at around
x=3D0.3L.
Our problems start when trying to solve c). The reason is that our
calculations and reasonings give that P(x, t_1) would be the same as
P(x, 0). Because a number of the following questions regard the
differences of P(x, 0) and P(x, t_1) and the corresponding expectation
values and so on, we feel that we must be wrong.
Well, stop for a moment and think about what it *should* be. Quantum mec=
hanics has waves. What would you expect an asymmetric wave in a box to do?=
It's heavy on the left side. Then it will slosh over to the right side, =
and back to the left side, and at a particular time it will be identical to=
what it was as t=3D0. It will slosh back and forth.
Sorry, I don't really get your point here. If the probability function
is heavy on the left side, then the probability of finding the particle
is bigger in the left half of the box than in the right half. How could
that be? What could be the cause of that? I argue that since there is
no potential (for example an electric field) I can't see why it could
be more probable that the particle is found on the left side than on
the right.
However, for any potential function V(x) that is not time-dependent
(V(x)=3D0 for all x within [0,L] clearly is time-independent), psi shall
be a time-space-dependance separable function. In those cases, denoting
the time-dependent function as phi(t), phi(t) can be written as
phi(t)=3Dexp(-iEt/h~).
That would mean that for P(x,t),
P(x,t)=3Dpsi(x,t)*complex-conjugate(psi(x,t))=3Dpsi(x)^2 since the phi's
will multiply to unity for all t. That would mean that P(x,t) is the
same for all t.
What are we doing wrong? We just cannot believe that P(x, t) is
independent of t. We are also annoyed by the fact that P(x,t) is
unsymmetric over x (we would expect symmetry around L/2, especially
since there is no potential). If there would be a time-dependance, and
a phase velocity for P(x, t), we could still achieve symmetry if
calculating the average over time.
Your wavefunction is not an eigenfunction of energy. The time evolution =
operator is U=3Dexp(-iHt/hbar), so
psi(t) =3D U psi(0)
=3D U psi_1 + U psi_2 {and psi1, psi2 are energy eigenfunctions}
=3D exp(-i w1 t) psi_1 + exp(-i w2 t) psi_2
Find the complex conjugate of that, and the time dependence won't go away.
At first: sorry that I lack some basic understanding of these topics.
I still don't rellay get it, but seeing w1 and w2 (that is, separate
angular velocities) in what is called the U operator by you gives me a
clue.
I multiply
psi(x)=3D0.8psi_1(x) + 0.6psi_2(x) with exp(-iEt/h~) to get the time
dependent function, that is,
psi(x,t)=3D( 0.8psi_1(x) + 0.6psi_2(x) )*exp(-iEt/h~)
Is this my error? Do I have to multiply psi_1 and psi_2 separately,
substituting E in exp(-iEt/h~) to E_1 and E_2? If that is the case, it
is clear that P(x,t) will be dependet of both x and t.
.
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| User: "Greg Hansen" |
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| Title: Re: The =?ISO-8859-1?Q?Schr=F6dinger_Equation=2C_potential_?==?ISO-8859-1?Q?box_problem?= |
16 Sep 2006 09:52:59 PM |
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wrote:
Greg Hansen wrote:
Carl wrote:
Hello,
While trying to solve a physics problem relating to the Schrödinger
equation, we found something that seems unlogical. The problem given is
a real "standard" problem about separable time- and space-dependent
parts of the solution to the Schrödinger Equation.
The following is given:
An electron in the potential box
V(X)={inf for x<0, 0 for 0<=x<=L, inf for x>L}
is at t=0 described by the wave function
psi(x, 0) = 0.8psi_1(x)+0.6psi_2(x)
where the orthonormal eigenfunctions
psi_n(x) = sqrt(2/L)sin(n*pi*x/L)
and the corresponding eigenvalues
E_n = (n*pi*h~/L)^2/(2m)=n^2*E_1, n=1, 2, ...
It is asked for:
a) Make sure psi is normalized.
b) Sketch the probability of finding the electron as a function of x at
t=0, that is, P(x,0).
c) Sketch the corresponding graph at time t_1=pi*h~/(3E_1)
So, to our problem: a) is straight-forward. b) gives a sketch of an
unsymmetric function, heavy on the left side, with a peak at around
x=0.3L.
Our problems start when trying to solve c). The reason is that our
calculations and reasonings give that P(x, t_1) would be the same as
P(x, 0). Because a number of the following questions regard the
differences of P(x, 0) and P(x, t_1) and the corresponding expectation
values and so on, we feel that we must be wrong.
Well, stop for a moment and think about what it *should* be. Quantum mechanics has waves. What would you expect an asymmetric wave in a box to do? It's heavy on the left side. Then it will slosh over to the right side, and back to the left side, and at a particular time it will be identical to what it was as t=0. It will slosh back and forth.
Sorry, I don't really get your point here. If the probability function
is heavy on the left side, then the probability of finding the particle
is bigger in the left half of the box than in the right half. How could
that be? What could be the cause of that? I argue that since there is
no potential (for example an electric field) I can't see why it could
be more probable that the particle is found on the left side than on
the right.
The system was prepared that way. An electron might have been fired at
it. Or a particular potential used to shape the wavefunction, and then
reduced to zero. The problem is artificially simple so that you can
solve it, but a more realistic one would be to fire an electron at an atom.
However, for any potential function V(x) that is not time-dependent
(V(x)=0 for all x within [0,L] clearly is time-independent), psi shall
be a time-space-dependance separable function. In those cases, denoting
the time-dependent function as phi(t), phi(t) can be written as
phi(t)=exp(-iEt/h~).
That would mean that for P(x,t),
P(x,t)=psi(x,t)*complex-conjugate(psi(x,t))=psi(x)^2 since the phi's
will multiply to unity for all t. That would mean that P(x,t) is the
same for all t.
What are we doing wrong? We just cannot believe that P(x, t) is
independent of t. We are also annoyed by the fact that P(x,t) is
unsymmetric over x (we would expect symmetry around L/2, especially
since there is no potential). If there would be a time-dependance, and
a phase velocity for P(x, t), we could still achieve symmetry if
calculating the average over time.
Your wavefunction is not an eigenfunction of energy. The time evolution operator is U=exp(-iHt/hbar), so
psi(t) = U psi(0)
= U psi_1 + U psi_2 {and psi1, psi2 are energy eigenfunctions}
= exp(-i w1 t) psi_1 + exp(-i w2 t) psi_2
Find the complex conjugate of that, and the time dependence won't go away.
At first: sorry that I lack some basic understanding of these topics.
I still don't rellay get it, but seeing w1 and w2 (that is, separate
angular velocities) in what is called the U operator by you gives me a
clue.
I multiply
psi(x)=0.8psi_1(x) + 0.6psi_2(x) with exp(-iEt/h~) to get the time
dependent function, that is,
psi(x,t)=( 0.8psi_1(x) + 0.6psi_2(x) )*exp(-iEt/h~)
Is this my error? Do I have to multiply psi_1 and psi_2 separately,
substituting E in exp(-iEt/h~) to E_1 and E_2? If that is the case, it
is clear that P(x,t) will be dependet of both x and t.
Yes, that is your problem. The energy operator is not exp(-iEt/hbar),
it's exp(-iHt/hbar). There's a difference. There is no single E that
characterizes your wavefunction because it's not an eigenfunction of the
energy. But the professor has kindly expressed it as a superposition of
energy eigenfunctions.
U psi(x) = exp(-iHt/hbar) * [0.8 psi_1(x) + 0.6 psi_2(x)]
= 0.8 exp(-iHt/hbar) psi_1(x) + 0.6 exp(-iHt/hbar) psi_2(x)
= 0.8 exp(-i E1 t/hbar) psi_1(x) + 0.6 exp(-i E2 t/hbar) psi_2(x)
H, the Hamiltonian, is an operator, and E1 and E2 are numbers, the
energy eigenvalues of psi_1 and psi_2.
When you find the squared magnitude, the cross-terms will remain,
oscillating with a beat frequency of (E1-E2)/hbar.
.
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| User: "Jim Black" |
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| Title: =?iso-8859-1?q?Re:_The_Schr=F6dinger_Equation,_potential_box_problem?= |
16 Sep 2006 05:30:39 PM |
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wrote:
Greg Hansen wrote:
Your wavefunction is not an eigenfunction of energy. The time evolution operator is U=exp(-iHt/hbar), so
psi(t) = U psi(0)
= U psi_1 + U psi_2 {and psi1, psi2 are energy eigenfunctions}
= exp(-i w1 t) psi_1 + exp(-i w2 t) psi_2
Find the complex conjugate of that, and the time dependence won't go away.
At first: sorry that I lack some basic understanding of these topics.
I still don't rellay get it, but seeing w1 and w2 (that is, separate
angular velocities) in what is called the U operator by you gives me a
clue.
I multiply
psi(x)=0.8psi_1(x) + 0.6psi_2(x) with exp(-iEt/h~) to get the time
dependent function, that is,
psi(x,t)=( 0.8psi_1(x) + 0.6psi_2(x) )*exp(-iEt/h~)
Is this my error?
Yes. Try plugging 0.8psi_1(x) + 0.6psi_2(x) into the time-independent
Schrodinger equation; you'll find that it is not a solution, for any
value of E. In general, the solution to a differential equation does
not have to be a product like f(x)*g(t). But that's okay, because if
the equation is linear (and the Schrodinger equation is), we add such
solutions together to find the one we need.
Do I have to multiply psi_1 and psi_2 separately,
substituting E in exp(-iEt/h~) to E_1 and E_2?
Yes. The separation-of-variables technique tells us that
psi(x,t) = psi_1(x) * exp(- i E_1 t / hbar)
is a solution of the Schrodinger equation and that
psi(x,t) = psi_2(x) * exp(- i E_2 t / hbar)
is also a solution. Since the Schrodinger equation is linear, we can
deduce that
psi(x,t) = 0.8 psi_1(x) * exp(- i E_1 t / hbar)
+ 0.6 psi_2(x) * exp(- i E_2 t / hbar)
is a solution, and this solution matches the initial condition given.
If that is the case, it
is clear that P(x,t) will be dependet of both x and t.
.
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| User: "Jim Black" |
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| Title: =?iso-8859-1?q?Re:_The_Schr=F6dinger_Equation,_potential_box_problem?= |
16 Sep 2006 06:22:26 PM |
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wrote:
Greg Hansen wrote:
Your wavefunction is not an eigenfunction of energy. The time evolution operator is U=exp(-iHt/hbar), so
psi(t) = U psi(0)
= U psi_1 + U psi_2 {and psi1, psi2 are energy eigenfunctions}
= exp(-i w1 t) psi_1 + exp(-i w2 t) psi_2
Find the complex conjugate of that, and the time dependence won't go away.
At first: sorry that I lack some basic understanding of these topics.
I still don't rellay get it, but seeing w1 and w2 (that is, separate
angular velocities) in what is called the U operator by you gives me a
clue.
Perhaps you have yet to learn about operators in detail. The "H" in
U=exp(-iHt/hbar) is not a number, but an operator called the
Hamiltonian operator. For a one-dimensional-system, the Hamiltonian
operator is:
H = - (hbar^2 / 2m) d^2/dx^2 + V(x)
[Here, "d" is used to represent partial differentiation because the
character set is restricted.]
A simpler example of an operator is the derivative (e.g. d/dx). Also,
multiplying by some function f(x) is an operator.
An operator can be applied to a function like so:
H psi(x)
= [- (hbar^2 / 2m) d^2/dx^2 + V(x)] psi(x)
= - (hbar^2 / 2m) d^2(psi(x))/dx^2 + V(x) * psi(x)
d/dx psi(x)
= d(psi(x))/dx
f(x) psi(x)
= f(x) * psi(x)
You can exponentiate (raise to a power) an operator by applying it
multiple times; for example:
(d/dx)^2 psi(x)
= (d/dx) [(d/dx) psi(x)]
= d^2(psi(x))/dx^2
For the multiply-by-a-function operator, exponentiation is equivalent
to normal exponentiation:
(f(x))^2 psi(x)
= f(x) * [f(x) * psi(x)]
= f(x)^2 * psi(x)
You may be wondering what it means to raise e to the power of an
operator. Well, consider what it means to raise e to the power of a
number:
e^x = lim_{n -> infinity} (1 + x/n)^n
This is what you'd end up with if you put one dollar in a bank account
for a year at n*100% interest per year, and the interest was compounded
not yearly, quarterly, or monthly, but continuously.
Or, for negative x, it's the proportion of water you'd have left in a
container after one minute if the water drained out at the rate
(current amount of water) * x / minute.
The definition e^x = lim_{n -> infinity} (1 + x/n)^n also makes sense
for operators.
Now, something we'll need to use is the fact that H is a linear
operator. That means that:
H(a*psi_1 + b*psi_2) = a*H psi_1 + b*H psi_2
All of the other operators I've mentioned above are linear operators,
as you can probably verify. In fact, all of the operators we use in
quantum mechanics are linear.
Now if H is a linear operator, it's pretty simple to confirm that
(1 - iHt/hbar/n) is also a linear operator, and also that
(1 - iHt/hbar/n)^n is a linear operator. You can prove the last one by
induction if you want to. If we were mathematicians, we'd then want a
proof that the limit of a series of linear operators is still linear,
possibly with some conditions attached, but let's hand-wave over that
step. We've got:
exp(-iHt/hbar) (psi_1 + psi_2)
= lim_{n -> infinity} (1 - iHt/hbar/n)^n (psi_1 + psi_2)
= lim_{n -> infinity} (1 - iHt/hbar/n)^n psi_1
+ lim_{n -> infinity} (1 - iHt/hbar/n)^n psi_2
Now we know that:
H psi_1 = E_1 * psi_1
H psi_2 = E_2 * psi_2
[These are just the time-independent Schrodinger equation, in
shorthand.]
So when we apply (1 - iHt/hbar/n) to psi_1 (or, by linearity, to a
multiple of psi_1):
(1 - iHt/hbar/n) psi_1
= psi_1 - i t/hbar/n (H psi_1)
= psi_1 - i t/hbar/n (E_1 psi_1)
= (1 - i E_1 t / hbar / n) psi_1
It's the same as multiplying it by (1 - i E_1 t / hbar / n). Applying
(1 - iHt/hbar/n) to psi_2 multiplies it by a different number,
(1 - i E_2 t / hbar / n).
So, continuing:
exp(-iHt/hbar) (psi_1 + psi_2)
= lim_{n -> infinity} (1 - iHt/hbar/n)^n (psi_1 + psi_2)
= lim_{n -> infinity} (1 - iHt/hbar/n)^n psi_1
+ lim_{n -> infinity} (1 - iHt/hbar/n)^n psi_2
= lim_{n -> infinity} (1 - i E_1 t / hbar / n)^n psi_1
+ lim_{n -> infinity} (1 - i E_2 t / hbar / n)^n psi_2
= exp(- i E_1 t / hbar) psi_1
+ exp(- i E_2 t / hbar) psi_2
Q.E.D.
Hopefully, you can make some sense out of this operator stuff.
Apologies if I've made no sense or confused the issue further!
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| User: "Edward Green" |
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| Title: =?iso-8859-1?q?Re:_The_Schr=F6dinger_Equation,_potential_box_problem?= |
16 Sep 2006 05:45:24 PM |
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wrote:
Sorry, I don't really get your point here. If the probability function
is heavy on the left side, then the probability of finding the particle
is bigger in the left half of the box than in the right half. How could
that be? What could be the cause of that? I argue that since there is
no potential (for example an electric field) I can't see why it could
be more probable that the particle is found on the left side than on
the right.
Asymmetric initial conditions?
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| User: "Carl" |
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| Title: Re: The =?ISO-8859-1?Q?Schr=F6dinger_Equation=2C_potential_?==?ISO-8859-1?Q?box_problem?= |
16 Sep 2006 10:25:54 AM |
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Greg Hansen wrote:
Carl wrote:
Hello,
While trying to solve a physics problem relating to the Schrödinger
equation, we found something that seems unlogical. The problem given is
a real "standard" problem about separable time- and space-dependent
parts of the solution to the Schrödinger Equation.
The following is given:
An electron in the potential box
V(X)={inf for x<0, 0 for 0<=x<=L, inf for x>L}
is at t=0 described by the wave function
psi(x, 0) = 0.8psi_1(x)+0.6psi_2(x)
where the orthonormal eigenfunctions
psi_n(x) = sqrt(2/L)sin(n*pi*x/L)
and the corresponding eigenvalues
E_n = (n*pi*h~/L)^2/(2m)=n^2*E_1, n=1, 2, ...
It is asked for:
a) Make sure psi is normalized.
b) Sketch the probability of finding the electron as a function of x at
t=0, that is, P(x,0).
c) Sketch the corresponding graph at time t_1=pi*h~/(3E_1)
So, to our problem: a) is straight-forward. b) gives a sketch of an
unsymmetric function, heavy on the left side, with a peak at around
x=0.3L.
Our problems start when trying to solve c). The reason is that our
calculations and reasonings give that P(x, t_1) would be the same as
P(x, 0). Because a number of the following questions regard the
differences of P(x, 0) and P(x, t_1) and the corresponding expectation
values and so on, we feel that we must be wrong.
Well, stop for a moment and think about what it *should* be. Quantum
mechanics has waves. What would you expect an asymmetric wave in a box
to do? It's heavy on the left side. Then it will slosh over to the
right side, and back to the left side, and at a particular time it will
be identical to what it was as t=0. It will slosh back and forth.
Sorry, I don't really get your point here. If the probability function
is heavy on the left side, then the probability of finding the particle
is bigger in the left half of the box than in the right half. How could
that be? What could be the cause of that? I argue that since there is no
potential (for example an electric field) I can't see why it could be
more probable that the particle is found on the left side than on the right.
However, for any potential function V(x) that is not time-dependent
(V(x)=0 for all x within [0,L] clearly is time-independent), psi shall
be a time-space-dependance separable function. In those cases, denoting
the time-dependent function as phi(t), phi(t) can be written as
phi(t)=exp(-iEt/h~).
That would mean that for P(x,t),
P(x,t)=psi(x,t)*complex-conjugate(psi(x,t))=psi(x)^2 since the phi's
will multiply to unity for all t. That would mean that P(x,t) is the
same for all t.
What are we doing wrong? We just cannot believe that P(x, t) is
independent of t. We are also annoyed by the fact that P(x,t) is
unsymmetric over x (we would expect symmetry around L/2, especially
since there is no potential). If there would be a time-dependance, and
a phase velocity for P(x, t), we could still achieve symmetry if
calculating the average over time.
Your wavefunction is not an eigenfunction of energy. The time evolution
operator is U=exp(-iHt/hbar), so
psi(t) = U psi(0)
= U psi_1 + U psi_2 {and psi1, psi2 are energy eigenfunctions}
= exp(-i w1 t) psi_1 + exp(-i w2 t) psi_2
Find the complex conjugate of that, and the time dependence won't go away.
At first: sorry that I lack some basic understanding of these topics.
I still don't rellay get it, but seeing w1 and w2 (that is, separate
angular velocities) in what is called the U operator by you gives me a
clue.
I multiply
psi(x)=0.8psi_1(x) + 0.6psi_2(x) with exp(-iEt/h~) to get the time
dependent function, that is,
psi(x,t)=( 0.8psi_1(x) + 0.6psi_2(x) )*exp(-iEt/h~)
Is this my error? Do I have to multiply psi_1 and psi_2 separately,
substituting E in exp(-iEt/h~) to E_1 and E_2? If that is the case, it
is clear that P(x,t) will be dependet of both x and t.
--
Posted via a free Usenet account from http://www.teranews.com
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