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What are the units of probability density?
Units of probability density? If bound electron is thought of as a cloud of charge, and it's charge density is proportional to the probability density. Then coulombs /m3 proportional to?
quantum-mechanics wavefunction units probability dimensional-analysis
edited Nov 22 '18 at 5:53
Qmechanic♦
102k121831161
asked Nov 22 '18 at 3:35
TontoTonto
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Units of probability density? If bound electron is thought of as a cloud of charge, and it's charge density is proportional to the probability density. Then coulombs /m3 proportional to?
quantum-mechanics wavefunction units probability dimensional-analysis
edited Nov 22 '18 at 5:53
Qmechanic♦
102k121831161
asked Nov 22 '18 at 3:35
TontoTonto
62
add a comment |
Units of probability density? If bound electron is thought of as a cloud of charge, and it's charge density is proportional to the probability density. Then coulombs /m3 proportional to?
quantum-mechanics wavefunction units probability dimensional-analysis
edited Nov 22 '18 at 5:53
Qmechanic♦
102k121831161
asked Nov 22 '18 at 3:35
TontoTonto
62
Units of probability density? If bound electron is thought of as a cloud of charge, and it's charge density is proportional to the probability density. Then coulombs /m3 proportional to?
quantum-mechanics wavefunction units probability dimensional-analysis
quantum-mechanics wavefunction units probability dimensional-analysis
edited Nov 22 '18 at 5:53
Qmechanic♦
102k121831161
asked Nov 22 '18 at 3:35
TontoTonto
62
edited Nov 22 '18 at 5:53
Qmechanic♦
102k121831161
asked Nov 22 '18 at 3:35
TontoTonto
62
edited Nov 22 '18 at 5:53
Qmechanic♦
102k121831161
edited Nov 22 '18 at 5:53
Qmechanic♦
102k121831161
edited Nov 22 '18 at 5:53
Qmechanic♦
102k121831161
102k121831161
asked Nov 22 '18 at 3:35
TontoTonto
62
asked Nov 22 '18 at 3:35
TontoTonto
62
asked Nov 22 '18 at 3:35
TontoTonto
62
62
add a comment |
add a comment |
2 Answers
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The units of probability density in three-dimensional space are inverse volume, $[L]^{-3}$. This is because probability itself is a dimensionless number, such as 0.5 for a probability of 50%.
A normalized wavefunction $Psi(vec{r},t)$ has dimensions $[L]^{-3/2}$ because the square of its complex magnitude, integrated over all space, must be 1. There is 100% probability that the electron is somewhere.
The charge density at a point is the particle’s charge times the probability density at that point.
When the wavefunction is for an electron, the charge density is $rho(vec{r},t)=-e|Psi(vec{r},t)|^2$, which could be expressed in units such as Coulombs per cubic meter ($text{C/m}^3$).
answered Nov 22 '18 at 3:38
G. SmithG. Smith
4,9131021
add a comment |
You can think about it by looking at what quantities you get from probability density. Namely, probability is dimensionless (it's just a number between 0 and 1), and given by the integral of probability density, so $P = int rho text{ d}V$.
The volume element has dimensions $ [L]^3$, so the probability density must have dimensions $[L]^{-3}$.
answered Nov 22 '18 at 9:30
Thomas JonesThomas Jones
36618
add a comment |
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2 Answers
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2 Answers
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The units of probability density in three-dimensional space are inverse volume, $[L]^{-3}$. This is because probability itself is a dimensionless number, such as 0.5 for a probability of 50%.
A normalized wavefunction $Psi(vec{r},t)$ has dimensions $[L]^{-3/2}$ because the square of its complex magnitude, integrated over all space, must be 1. There is 100% probability that the electron is somewhere.
The charge density at a point is the particle’s charge times the probability density at that point.
When the wavefunction is for an electron, the charge density is $rho(vec{r},t)=-e|Psi(vec{r},t)|^2$, which could be expressed in units such as Coulombs per cubic meter ($text{C/m}^3$).
answered Nov 22 '18 at 3:38
G. SmithG. Smith
4,9131021
add a comment |
The units of probability density in three-dimensional space are inverse volume, $[L]^{-3}$. This is because probability itself is a dimensionless number, such as 0.5 for a probability of 50%.
A normalized wavefunction $Psi(vec{r},t)$ has dimensions $[L]^{-3/2}$ because the square of its complex magnitude, integrated over all space, must be 1. There is 100% probability that the electron is somewhere.
The charge density at a point is the particle’s charge times the probability density at that point.
When the wavefunction is for an electron, the charge density is $rho(vec{r},t)=-e|Psi(vec{r},t)|^2$, which could be expressed in units such as Coulombs per cubic meter ($text{C/m}^3$).
answered Nov 22 '18 at 3:38
G. SmithG. Smith
4,9131021
add a comment |
The units of probability density in three-dimensional space are inverse volume, $[L]^{-3}$. This is because probability itself is a dimensionless number, such as 0.5 for a probability of 50%.
A normalized wavefunction $Psi(vec{r},t)$ has dimensions $[L]^{-3/2}$ because the square of its complex magnitude, integrated over all space, must be 1. There is 100% probability that the electron is somewhere.
The charge density at a point is the particle’s charge times the probability density at that point.
When the wavefunction is for an electron, the charge density is $rho(vec{r},t)=-e|Psi(vec{r},t)|^2$, which could be expressed in units such as Coulombs per cubic meter ($text{C/m}^3$).
answered Nov 22 '18 at 3:38
G. SmithG. Smith
4,9131021
The units of probability density in three-dimensional space are inverse volume, $[L]^{-3}$. This is because probability itself is a dimensionless number, such as 0.5 for a probability of 50%.
A normalized wavefunction $Psi(vec{r},t)$ has dimensions $[L]^{-3/2}$ because the square of its complex magnitude, integrated over all space, must be 1. There is 100% probability that the electron is somewhere.
The charge density at a point is the particle’s charge times the probability density at that point.
When the wavefunction is for an electron, the charge density is $rho(vec{r},t)=-e|Psi(vec{r},t)|^2$, which could be expressed in units such as Coulombs per cubic meter ($text{C/m}^3$).
answered Nov 22 '18 at 3:38
G. SmithG. Smith
4,9131021
edited Nov 22 '18 at 4:06
answered Nov 22 '18 at 3:38
G. SmithG. Smith
4,9131021
answered Nov 22 '18 at 3:38
G. SmithG. Smith
4,9131021
answered Nov 22 '18 at 3:38
G. SmithG. Smith
4,9131021
4,9131021
add a comment |
add a comment |
You can think about it by looking at what quantities you get from probability density. Namely, probability is dimensionless (it's just a number between 0 and 1), and given by the integral of probability density, so $P = int rho text{ d}V$.
The volume element has dimensions $ [L]^3$, so the probability density must have dimensions $[L]^{-3}$.
answered Nov 22 '18 at 9:30
Thomas JonesThomas Jones
36618
add a comment |
You can think about it by looking at what quantities you get from probability density. Namely, probability is dimensionless (it's just a number between 0 and 1), and given by the integral of probability density, so $P = int rho text{ d}V$.
The volume element has dimensions $ [L]^3$, so the probability density must have dimensions $[L]^{-3}$.
answered Nov 22 '18 at 9:30
Thomas JonesThomas Jones
36618
add a comment |
You can think about it by looking at what quantities you get from probability density. Namely, probability is dimensionless (it's just a number between 0 and 1), and given by the integral of probability density, so $P = int rho text{ d}V$.
The volume element has dimensions $ [L]^3$, so the probability density must have dimensions $[L]^{-3}$.
answered Nov 22 '18 at 9:30
Thomas JonesThomas Jones
36618
You can think about it by looking at what quantities you get from probability density. Namely, probability is dimensionless (it's just a number between 0 and 1), and given by the integral of probability density, so $P = int rho text{ d}V$.
The volume element has dimensions $ [L]^3$, so the probability density must have dimensions $[L]^{-3}$.
answered Nov 22 '18 at 9:30
Thomas JonesThomas Jones
36618
answered Nov 22 '18 at 9:30
Thomas JonesThomas Jones
36618
answered Nov 22 '18 at 9:30
Thomas JonesThomas Jones
36618
answered Nov 22 '18 at 9:30
Thomas JonesThomas Jones
36618
36618
add a comment |
add a comment |
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