| Topic: |
Science > Physics |
| User: |
"David Macmanus" |
| Date: |
07 Dec 2005 09:16:36 AM |
| Object: |
H-atom wavefunctions: real or imaginary? |
Are the wavefunctions of the H-atom real?
For the H-atom in spherical coordinates we separate the SWE into
the r, theta and phi parts, and the final wavefunction is thus the
product of all three of these. The phi part is 1/SQRT 2 pi (exp i m
phi), which in this form is clearly imaginary. So does it then mean that
the H-atom wavefunctions are imaginary? Or is there something going on
that means that the H-atom wfs end up real?
I'm sure I've seen the H-atom wfs described as real, so maybe people for
some reason ignore exp (phi) part.
Or maybe it's possible to swap the exp i m phi for sin or cos i m phi
and make it overall real that way???
Thanks,
David.
--
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| User: "PD" |
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| Title: Re: H-atom wavefunctions: real or imaginary? |
07 Dec 2005 09:35:10 AM |
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David Macmanus wrote:
Are the wavefunctions of the H-atom real?
For the H-atom in spherical coordinates we separate the SWE into
the r, theta and phi parts, and the final wavefunction is thus the
product of all three of these. The phi part is 1/SQRT 2 pi (exp i m
phi), which in this form is clearly imaginary. So does it then mean that
the H-atom wavefunctions are imaginary? Or is there something going on
that means that the H-atom wfs end up real?
I'm sure I've seen the H-atom wfs described as real, so maybe people for
some reason ignore exp (phi) part.
Or maybe it's possible to swap the exp i m phi for sin or cos i m phi
and make it overall real that way???
Thanks,
David.
The probability density that comes from the modulus squared of the
wavefunction is real and observable. The interference effects that come
from the superposition of two wavefunctions are real and observable.
The eigenvalues and transition probabilities that come from sandwiching
an operator between two wavefunctions are real and observable.
The wavefunction itself is not observable. Is it real? You tell me.
PD
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| User: "David Macmanus" |
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| Title: Re: H-atom wavefunctions: real or imaginary? |
07 Dec 2005 09:41:05 AM |
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"PD" <TheDraperFamily@gmail.com> wrote in message
news:1133969710.361564.268070@g14g2000cwa.googlegroups.com
The wavefunction itself is not observable. Is it real? You tell me.
Okay, I should have used the word 'complex' rather than 'imaginary'.
I take your points, but the question I'd really like answered is are the
wfs of the H-atom real or complex (mathematically speaking)?
Thanks.
--
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| User: "PD" |
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| Title: Re: H-atom wavefunctions: real or imaginary? |
07 Dec 2005 09:49:43 AM |
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David Macmanus wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message
news:1133969710.361564.268070@g14g2000cwa.googlegroups.com
The wavefunction itself is not observable. Is it real? You tell me.
Okay, I should have used the word 'complex' rather than 'imaginary'.
I take your points, but the question I'd really like answered is are the
wfs of the H-atom real or complex (mathematically speaking)?
Thanks.
The mathematical *representation* of a wavefunction is immaterial, as
long as the coupled law that governs their behavior is appropriately
matched to produce the observed outcomes. The *customary*
representation of a wavefunction is with a complex-valued function,
yes.
PD
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| User: "David Macmanus" |
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| Title: Re: H-atom wavefunctions: real or imaginary? |
07 Dec 2005 09:52:03 AM |
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"PD" <TheDraperFamily@gmail.com> wrote in message
news:1133970583.867129.320050@g14g2000cwa.googlegroups.com
The mathematical *representation* of a wavefunction is immaterial, as
long as the coupled law that governs their behavior is appropriately
matched to produce the observed outcomes. The *customary*
representation of a wavefunction is with a complex-valued function,
yes.
For the H-atom, can we exchange the exp (i m phi) for a 'real' sin or
cos function?
Thanks.
--
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| User: "PD" |
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| Title: Re: H-atom wavefunctions: real or imaginary? |
07 Dec 2005 10:02:19 AM |
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David Macmanus wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message
news:1133970583.867129.320050@g14g2000cwa.googlegroups.com
The mathematical *representation* of a wavefunction is immaterial, as
long as the coupled law that governs their behavior is appropriately
matched to produce the observed outcomes. The *customary*
representation of a wavefunction is with a complex-valued function,
yes.
For the H-atom, can we exchange the exp (i m phi) for a 'real' sin or
cos function?
Thanks.
For some things, like probability density, yes. The reason is that the
phase information that is contained in the (imaginary) argument of a
complex wavefunction, can be also represented as a phase in a sine or
cosine.
PD
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| User: "David Macmanus" |
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| Title: Re: H-atom wavefunctions: real or imaginary? |
07 Dec 2005 10:13:43 AM |
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"PD" <TheDraperFamily@gmail.com> wrote in message
news:1133971338.914956.73160@g47g2000cwa.googlegroups.com
For some things, like probability density, yes. The reason is that the
phase information that is contained in the (imaginary) argument of a
complex wavefunction, can be also represented as a phase in a sine or
cosine.
Thanks, but I don't quite understand this. Could you explain a little
more? Are you saying that exp(-i m phi) times exp(i m phi) is the same
as sin^2 or something??
Thanks,
David.
--
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| User: "PD" |
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| Title: Re: H-atom wavefunctions: real or imaginary? |
07 Dec 2005 10:34:12 AM |
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David Macmanus wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message
news:1133971338.914956.73160@g47g2000cwa.googlegroups.com
For some things, like probability density, yes. The reason is that the
phase information that is contained in the (imaginary) argument of a
complex wavefunction, can be also represented as a phase in a sine or
cosine.
Thanks, but I don't quite understand this. Could you explain a little
more? Are you saying that exp(-i m phi) times exp(i m phi) is the same
as sin^2 or something??
Thanks,
David.
A exp(i m phi) = A [cos(m phi) + i sin (m phi)]
Now take the modulus of this number.
Any online tutorial about complex numbers will show how this works.
PD
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| User: "David Macmanus" |
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| Title: Re: H-atom wavefunctions: real or imaginary? |
07 Dec 2005 10:41:50 AM |
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"PD" <TheDraperFamily@gmail.com> wrote in message
news:1133973252.556655.225850@o13g2000cwo.googlegroups.com
A exp(i m phi) = A [cos(m phi) + i sin (m phi)]
Now take the modulus of this number.
Yes, agreed. But doesn't this mean that if we exchange the exp for cos
OR sin we lose something when calculating probabilies?
I mean, exp{ }^2 = cos{ }^2 + sin{ }^2.
So if we replace exp{ } with just sin{ } then we end up with a
probability of just sin{ }^2, which is not unity.
So how can we swap exp for just sin OR just cos and get the right
probabilities? Don't we still need both sin and cos?
Thanks.
--
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| User: "PD" |
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| Title: Re: H-atom wavefunctions: real or imaginary? |
07 Dec 2005 11:07:22 AM |
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David Macmanus wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message
news:1133973252.556655.225850@o13g2000cwo.googlegroups.com
A exp(i m phi) = A [cos(m phi) + i sin (m phi)]
Now take the modulus of this number.
Yes, agreed. But doesn't this mean that if we exchange the exp for cos
OR sin we lose something when calculating probabilies?
I mean, exp{ }^2 = cos{ }^2 + sin{ }^2.
So if we replace exp{ } with just sin{ } then we end up with a
probability of just sin{ }^2, which is not unity.
So how can we swap exp for just sin OR just cos and get the right
probabilities? Don't we still need both sin and cos?
Thanks.
I didn't suggest just replacing exp( ) with sin ( ). That would be
dangerous, as you note.
But observe that cos (m phi) = sin (m phi + pi/2), so that
|psi|^2 = [A exp (i m phi)] [A exp (-i m phi)] = A^2 [sin^2(m phi) +
cos^2(m phi)]
= A^2 [sin^2(m phi) + sin^2(m phi + pi/2)] = A^2.
The next-to-last step isn't really necessary, but it's more
illuminating when you are doing a transition state instead, like
<psi2 | psi1> = [A exp (i m phi)] [B exp (-i n phi)]
PD
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| User: "David Macmanus" |
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| Title: Re: H-atom wavefunctions: real or imaginary? |
07 Dec 2005 11:28:31 AM |
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"PD" <TheDraperFamily@gmail.com> wrote in message
news:1133975242.003902.10520@g44g2000cwa.googlegroups.com
I didn't suggest just replacing exp( ) with sin ( ). That would be
dangerous, as you note.
But observe that cos (m phi) = sin (m phi + pi/2), so that
|psi|^2 = [A exp (i m phi)] [A exp (-i m phi)] = A^2 [sin^2(m phi) +
cos^2(m phi)]
= A^2 [sin^2(m phi) + sin^2(m phi + pi/2)] = A^2.
Right. Then I think I've got it.
Thanks very much.
David.
--
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| User: "Bruce Scott TOK" |
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| Title: Re: H-atom wavefunctions: real or imaginary? |
07 Dec 2005 10:49:59 AM |
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David Macmanus wrote:
Yes, agreed. But doesn't this mean that if we exchange the exp for cos
OR sin we lose something when calculating probabilies?
I mean, exp{ }^2 = cos{ }^2 + sin{ }^2.
So if we replace exp{ } with just sin{ } then we end up with a
probability of just sin{ }^2, which is not unity.
So how can we swap exp for just sin OR just cos and get the right
probabilities? Don't we still need both sin and cos?
Thanks.
No it just means you keep both cos and sin parts.
Note exp{ }^2 = cos{ }^2 + sin{ }^2 is not correct. You want
exp(i a) times exp (-i a) = 1
--
ciao,
Bruce
drift wave turbulence: http://www.rzg.mpg.de/~bds/
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| User: "David Macmanus" |
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| Title: Re: H-atom wavefunctions: real or imaginary? |
07 Dec 2005 10:53:58 AM |
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"Bruce Scott TOK" <Use-Author-Supplied-Address-Header@[127.1]> wrote in
message news:200512071649.jB7GnxbZ010882@ipp.mpg.de
No it just means you keep both cos and sin parts.
Okay, so if we have to keep both cos and sin then we have to keep 'i' as
well, so we can't make the wavefunctions overall real after all! Is that
right?
Note exp{ }^2 = cos{ }^2 + sin{ }^2 is not correct. You want
exp(i a) times exp (-i a) = 1
Granted!
Thanks.
--
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| User: "Dirk Van de moortel" |
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| Title: Re: H-atom wavefunctions: real or imaginary? |
07 Dec 2005 12:11:09 PM |
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"David Macmanus" <macmanus@tripos.com> wrote in message news:c7543386c2ebe387c711daa7f7640bcb.35661@mygate.mailgate.org...
"Bruce Scott TOK" <Use-Author-Supplied-Address-Header@[127.1]> wrote in
message news:200512071649.jB7GnxbZ010882@ipp.mpg.de
No it just means you keep both cos and sin parts.
Okay, so if we have to keep both cos and sin then we have to keep 'i' as
well, so we can't make the wavefunctions overall real after all! Is that
right?
If you don't like complex numbers to represent the wavefunction,
you can always represent it with pairs of real numbers.
You can always write
(x,y) in stead of x + i y ,
and x^2+y^2 as the modulus of the number.
Then you could define addition and multiplication of such "number
pairs" and you could write all the quantities you are dealing with
as number pairs in stead of complex numbers. All the results
would be the same, but you would never deal with "i".
Exercises:
(1) How would you represent the compex number
A exp(i m phi)
with such a real number pair?
(2) How would you define the "sum" of the pairs
(x1,y1) and (x2,y2)
?
(3) How would you define the "product" of the pairs
(x1,y1) and (x2,y2)
?
(4) To which number pair would the complex unit i correspond?
(you know, i, the thing that satisfies i^2 = -1)
(5) To which number pairs would the ordinary real numbers
correspond?
Dirk Vdm
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| User: "Josef Matz" |
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| Title: Re: H-atom wavefunctions: real or imaginary? |
07 Dec 2005 12:25:50 PM |
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"Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com> schrieb
im Newsbeitrag news:1BFlf.67456$mT6.4097071@phobos.telenet-ops.be...
"David Macmanus" <macmanus@tripos.com> wrote in message
news:c7543386c2ebe387c711daa7f7640bcb.35661@mygate.mailgate.org...
"Bruce Scott TOK" <Use-Author-Supplied-Address-Header@[127.1]> wrote in
message news:200512071649.jB7GnxbZ010882@ipp.mpg.de
No it just means you keep both cos and sin parts.
Okay, so if we have to keep both cos and sin then we have to keep 'i' as
well, so we can't make the wavefunctions overall real after all! Is that
right?
If you don't like complex numbers to represent the wavefunction,
you can always represent it with pairs of real numbers.
You can always write
(x,y) in stead of x + i y ,
and x^2+y^2 as the modulus of the number.
Then you could define addition and multiplication of such "number
pairs" and you could write all the quantities you are dealing with
as number pairs in stead of complex numbers. All the results
would be the same, but you would never deal with "i".
Exercises:
(1) How would you represent the compex number
A exp(i m phi)
with such a real number pair?
(2) How would you define the "sum" of the pairs
(x1,y1) and (x2,y2)
?
(3) How would you define the "product" of the pairs
(x1,y1) and (x2,y2)
?
(4) To which number pair would the complex unit i correspond?
(you know, i, the thing that satisfies i^2 = -1)
(5) To which number pairs would the ordinary real numbers
correspond?
Dirk Vdm
But this vectorroom is isomorh with complex numers and the rules are
identical.
Therefore ists the same. To use the i is just the nicest and easiest
representation of
the complex things indeed. And if you use the real valued description does
not mean
to have someting real in the complex vectorroom. So one has to be careful.
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| User: "Dirk Van de moortel" |
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| Title: Re: H-atom wavefunctions: real or imaginary? |
07 Dec 2005 12:28:04 PM |
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"Josef Matz" <josefmatz@arcor.de> wrote in message news:439727f3$0$9643$9b4e6d93@newsread2.arcor-online.net...
"Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com> schrieb
im Newsbeitrag news:1BFlf.67456$mT6.4097071@phobos.telenet-ops.be...
"David Macmanus" <macmanus@tripos.com> wrote in message
news:c7543386c2ebe387c711daa7f7640bcb.35661@mygate.mailgate.org...
"Bruce Scott TOK" <Use-Author-Supplied-Address-Header@[127.1]> wrote in
message news:200512071649.jB7GnxbZ010882@ipp.mpg.de
No it just means you keep both cos and sin parts.
Okay, so if we have to keep both cos and sin then we have to keep 'i' as
well, so we can't make the wavefunctions overall real after all! Is that
right?
If you don't like complex numbers to represent the wavefunction,
you can always represent it with pairs of real numbers.
You can always write
(x,y) in stead of x + i y ,
and x^2+y^2 as the modulus of the number.
Then you could define addition and multiplication of such "number
pairs" and you could write all the quantities you are dealing with
as number pairs in stead of complex numbers. All the results
would be the same, but you would never deal with "i".
Exercises:
(1) How would you represent the compex number
A exp(i m phi)
with such a real number pair?
(2) How would you define the "sum" of the pairs
(x1,y1) and (x2,y2)
?
(3) How would you define the "product" of the pairs
(x1,y1) and (x2,y2)
?
(4) To which number pair would the complex unit i correspond?
(you know, i, the thing that satisfies i^2 = -1)
(5) To which number pairs would the ordinary real numbers
correspond?
Dirk Vdm
But this vectorroom is isomorh with complex numers and the rules are
identical.
Therefore ists the same. To use the i is just the nicest and easiest
representation of
the complex things indeed.
Of course.
Some people seem to be allergic to complex numbers,
specially when they are used to describe part of nature.
This is a nice way out for them :-)
And if you use the real valued description does
not mean
to have someting real in the complex vectorroom. So one has to be careful.
One always has to be careful...
Dirk Vdm
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| User: "Ben Rudiak-Gould" |
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| Title: Re: H-atom wavefunctions: real or imaginary? |
07 Dec 2005 10:42:42 AM |
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David Macmanus wrote:
For the H-atom in spherical coordinates we separate the SWE into
the r, theta and phi parts, and the final wavefunction is thus the
product of all three of these. The phi part is 1/SQRT 2 pi (exp i m
phi), which in this form is clearly imaginary. So does it then mean that
the H-atom wavefunctions are imaginary? Or is there something going on
that means that the H-atom wfs end up real?
If m=0 then the phi part is a constant real number. I can't remember whether
the radial and orbital parts are always real independent of the associated
quantum numbers, but I know that some of the eigenstates are real and others
aren't. (Of course, by "real" I mean "can be made real by an appropriate
choice of the arbitrary phase factor".)
Or maybe it's possible to swap the exp i m phi for sin or cos i m phi
and make it overall real that way???
Well, Phi(m) and Phi(-m) are complex conjugates, so assuming R(r) and
Theta(theta) are real, 1/sqrt(2) R(r)Theta(theta)(Phi_m(phi) + Phi_-m(phi))
should be real, and it will have the form you describe since e^ix + e^-ix =
2 cos x. This will even be an energy eigenstate unless there's an external
magnetic field. I may be wrong about all of this, though -- it's been a
while since I took quantum mechanics.
-- Ben
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| User: "Josef Matz" |
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| Title: Re: H-atom wavefunctions: real or imaginary? |
07 Dec 2005 09:28:18 AM |
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Wave functions are complex.
"David Macmanus" <macmanus@tripos.com> schrieb im Newsbeitrag
news:488e29a4475d42342622baaaa0a4bbdf.35661@mygate.mailgate.org...
Are the wavefunctions of the H-atom real?
For the H-atom in spherical coordinates we separate the SWE into
the r, theta and phi parts, and the final wavefunction is thus the
product of all three of these. The phi part is 1/SQRT 2 pi (exp i m
phi), which in this form is clearly imaginary. So does it then mean that
the H-atom wavefunctions are imaginary? Or is there something going on
that means that the H-atom wfs end up real?
I'm sure I've seen the H-atom wfs described as real, so maybe people for
some reason ignore exp (phi) part.
Or maybe it's possible to swap the exp i m phi for sin or cos i m phi
and make it overall real that way???
Thanks,
David.
--
Posted via Mailgate.ORG Server - http://www.Mailgate.ORG
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| User: "David Macmanus" |
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| Title: Re: H-atom wavefunctions: real or imaginary? |
07 Dec 2005 09:34:30 AM |
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"Josef Matz" <josefmatz@arcor.de> wrote in message
news:4396fe58$0$27883$9b4e6d93@newsread4.arcor-online.net
Wave functions are complex.
So you are saying that the eigenstates of the Hamiltonian are never real
for the H-atom?
--
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| User: "" |
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| Title: Re: H-atom wavefunctions: real or imaginary? |
07 Dec 2005 11:57:02 AM |
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In article <32e0f963f6a1aa4edcb8308442ec1b2b.35661@mygate.mailgate.org>, "David Macmanus" <macmanus@tripos.com> writes:
"Josef Matz" <josefmatz@arcor.de> wrote in message
news:4396fe58$0$27883$9b4e6d93@newsread4.arcor-online.net
Wave functions are complex.
So you are saying that the eigenstates of the Hamiltonian are never real
for the H-atom?
Is this a problem? In what way?
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: "Josef Matz" |
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| Title: Re: H-atom wavefunctions: real or imaginary? |
07 Dec 2005 11:37:10 AM |
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Yes of shure because you also have the time propagator exp(iwt), a complex
phase factor
variing with time and w = E / hbar where E is the bond energy of the
eigenstate.
The wave function therefore is only real sometimes on a certain place on a
certain time, not
more as for any other complex function.
"David Macmanus" <macmanus@tripos.com> schrieb im Newsbeitrag
news:32e0f963f6a1aa4edcb8308442ec1b2b.35661@mygate.mailgate.org...
"Josef Matz" <josefmatz@arcor.de> wrote in message
news:4396fe58$0$27883$9b4e6d93@newsread4.arcor-online.net
Wave functions are complex.
So you are saying that the eigenstates of the Hamiltonian are never real
for the H-atom?
--
Posted via Mailgate.ORG Server - http://www.Mailgate.ORG
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