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 2 days ago
Qmechanic♦
99.7k121781119
asked Nov 22 at 3:35
Tonto
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up vote
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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 2 days ago
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.
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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
quantum-mechanics wavefunction units probability dimensional-analysis
edited 2 days ago
Qmechanic♦
99.7k121781119
asked Nov 22 at 3:35
Tonto
62
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edited 2 days ago
Qmechanic♦
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asked Nov 22 at 3:35
Tonto
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edited 2 days ago
Qmechanic♦
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edited 2 days ago
Qmechanic♦
99.7k121781119
edited 2 days ago
Qmechanic♦
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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
2,980615
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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 2 days ago
Thomas Jones
33118
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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
2,980615
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
2,980615
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
2,980615
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
2,980615
edited Nov 22 at 4:06
answered Nov 22 at 3:38
G. Smith
2,980615
answered Nov 22 at 3:38
G. Smith
2,980615
answered Nov 22 at 3:38
G. Smith
2,980615
2,980615
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 2 days ago
Thomas Jones
33118
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 2 days ago
Thomas Jones
33118
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 2 days ago
Thomas Jones
33118
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 2 days ago
Thomas Jones
33118
answered 2 days ago
Thomas Jones
33118
answered 2 days ago
Thomas Jones
33118
answered 2 days ago
Thomas Jones
33118
33118
add a comment |
add a comment |
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