- #1
Jan Wo
- 4
- 0
Hello
Lastly I was thinking a lot about electron density definition. It is not intuitive for me and I'm looking for any mathematical tool that could explain it to me more. My friend told me about idea to derivate it from propability density function using Dirac delta distribution. I'd like to show you this way of thinking and ask you about your opinion about it. Is it right way to think about electron density like this?
First - case of one electron. If we have only one electron, the propability density function should be at the same time its electron density function. We are looking for the operator which gives us as a mean value electron density:
##\rho(r)=\langle \psi | \hat{\rho}|\psi\rangle##
If we assume that ##\hat{\rho} \equiv \delta(r-r')## then we got electron density function for one electron from propability density.
Now let's take a N-electron case:
Let's use sum of Dirac delta operators as a electron density operator:
##\langle \psi | \sum_{i=1}^N \delta(r-r_i)|\psi\rangle = \sum_{i=1}^N \langle \psi |\delta(r-r_i)|\psi\rangle = \sum_{i=1}^N \int \psi^* (r_1, \dots , r_N) \delta (r-r_i)\psi(r_1, \dots , r_N)dr_1 \dots dr_N=*##
As we can see now, Dirac delta 'takes' one position out from this integral, let's assume that it is position number one. Then one summand should looks like: ##\int \psi^* (r, r_2, \dots , r_N)\psi (r, r_2, \dots , r_N)dr_2 \dots dr_N ## where ##r## is an position which we choose to 'check' the electron density. There is ##N## summands and the electrons are indistinguishable so we can ##N## times choose the electron number one. Then:
##*=N\int \psi^*\psi dr_2 \dots dr_N \equiv \rho(r)##
Is this right way of thinking about electron density?
Can you tell me if something is wrong in this?
Thank you in advance.
WJ
Lastly I was thinking a lot about electron density definition. It is not intuitive for me and I'm looking for any mathematical tool that could explain it to me more. My friend told me about idea to derivate it from propability density function using Dirac delta distribution. I'd like to show you this way of thinking and ask you about your opinion about it. Is it right way to think about electron density like this?
First - case of one electron. If we have only one electron, the propability density function should be at the same time its electron density function. We are looking for the operator which gives us as a mean value electron density:
##\rho(r)=\langle \psi | \hat{\rho}|\psi\rangle##
If we assume that ##\hat{\rho} \equiv \delta(r-r')## then we got electron density function for one electron from propability density.
Now let's take a N-electron case:
Let's use sum of Dirac delta operators as a electron density operator:
##\langle \psi | \sum_{i=1}^N \delta(r-r_i)|\psi\rangle = \sum_{i=1}^N \langle \psi |\delta(r-r_i)|\psi\rangle = \sum_{i=1}^N \int \psi^* (r_1, \dots , r_N) \delta (r-r_i)\psi(r_1, \dots , r_N)dr_1 \dots dr_N=*##
As we can see now, Dirac delta 'takes' one position out from this integral, let's assume that it is position number one. Then one summand should looks like: ##\int \psi^* (r, r_2, \dots , r_N)\psi (r, r_2, \dots , r_N)dr_2 \dots dr_N ## where ##r## is an position which we choose to 'check' the electron density. There is ##N## summands and the electrons are indistinguishable so we can ##N## times choose the electron number one. Then:
##*=N\int \psi^*\psi dr_2 \dots dr_N \equiv \rho(r)##
Is this right way of thinking about electron density?
Can you tell me if something is wrong in this?
Thank you in advance.
WJ
Last edited: