What are the units of probability density?
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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
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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 at 5:53
Qmechanic♦
99.7k121781119
asked Nov 22 at 3:35
Tonto
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Tonto is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
1
down vote
favorite
up vote
1
down vote
favorite
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 at 5:53
Qmechanic♦
99.7k121781119
asked Nov 22 at 3:35
Tonto
62
New contributor
Tonto is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 at 5:53
Qmechanic♦
99.7k121781119
asked Nov 22 at 3:35
Tonto
62
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Tonto is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited Nov 22 at 5:53
Qmechanic♦
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asked Nov 22 at 3:35
Tonto
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edited Nov 22 at 5:53
Qmechanic♦
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edited Nov 22 at 5:53
Qmechanic♦
99.7k121781119
edited Nov 22 at 5:53
Qmechanic♦
99.7k121781119
99.7k121781119
asked Nov 22 at 3:35
Tonto
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asked Nov 22 at 3:35
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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 at 3:38
G. Smith
3,063615
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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 at 9:30
Thomas Jones
35618
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
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 at 3:38
G. Smith
3,063615
add a comment |
up vote
6
down vote
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 at 3:38
G. Smith
3,063615
add a comment |
up vote
6
down vote
up vote
6
down vote
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 at 3:38
G. Smith
3,063615
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 at 3:38
G. Smith
3,063615
edited Nov 22 at 4:06
answered Nov 22 at 3:38
G. Smith
3,063615
answered Nov 22 at 3:38
G. Smith
3,063615
answered Nov 22 at 3:38
G. Smith
3,063615
3,063615
add a comment |
add a comment |
up vote
1
down vote
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 at 9:30
Thomas Jones
35618
add a comment |
up vote
1
down vote
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 at 9:30
Thomas Jones
35618
add a comment |
up vote
1
down vote
up vote
1
down vote
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 at 9:30
Thomas Jones
35618
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 at 9:30
Thomas Jones
35618
answered Nov 22 at 9:30
Thomas Jones
35618
answered Nov 22 at 9:30
Thomas Jones
35618
answered Nov 22 at 9:30
Thomas Jones
35618
35618
add a comment |
add a comment |
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