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Kinetic energy of the hydrogen atom in its ground state

I saw another post about this but i didn't quite find what i was looking for there so i thought i'd give it a go instead with a thread.

1. Homework Statement
Calculate the exact value of the kinetic energy of the hydrogen atom in its ground state. No more information is given, we are referred to litterature and stuff.

2. Homework Equations
I am thinking that my approach is wrong in the sense that i am trying to do something in a way that it's not allowed to. We have the hamiltonian [tex]\hat{H}\psi=E\psi[/tex] which in this case is [tex](-\frac{\hbar^2}{2\mu}\nabla^2-\frac{Ze^2}{4\pi\epsilon_0r})\psi(r,\theta,\phi)=E\psi(r,\theta,\phi)=-13.6eV\psi(r,\theta,\phi)[/tex] and the last term is because we know that it is in its groundstate.

3. The Attempt at a Solution
What i thought that i'd do that i now suspect is not "allowed" mathematically was to let the [itex]\psi(r,\theta,\phi)[/itex]s cancel themselves out and then since the first term on the left side represents the kinetic energy, just move the right term over to the right side and simply calculate the value. The value i get is incredibly small, since it is the hydrogenatom in its groundstate we can interpret the function with the values [tex]Z=1[/tex][tex]e=1.6*10^{-19}[/tex][tex]\epsilon_0=8.8*10^{-12}[/tex] and [tex]r=0.52Å=0.52*10^{-10}[/tex] and if i proceed to do as i intended the value for the kinetic energy i get is 13.6eV or so which is wrong, i've been nudge to believe that the kinetic energy should be [itex]\frac{1}{2}[/itex][tex]kinetic energy=-13.6+\frac{(1.6*10^{-19})^2}{(4\pi8.8*10^{-12}*0.52*10^{-10})}=-13.599999eV[/tex]

Should my approach instead be to try to calculate [tex](-\frac{\hbar^2}{2\mu}\nabla^2)\frac{1}{\sqrt{\pi a_0}}e^{-\frac{r}{a_0}}[/tex] given that [tex]\psi(r,\theta,\phi)_{100}=\frac{1}{\sqrt{\pi a_0}}e^{-\frac{r}{a_0}}[/tex] How would i go about and do this?
 

blue_leaf77

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When it comes to quantum mechanics, one can only calculate the average of an observable to describe the theoretical result of a measurement on that observable. So, the word "exact" in the question is supposed to be understood as the "expression" of the average or expectation value of kinetic energy.
By the way, although your approach is indeed wrong, I need to mention that you cannot add two quantities with different units. That's why you get that 13.599999.
 
When it comes to quantum mechanics, one can only calculate the average of an observable to describe the theoretical result of a measurement on that observable. So, the word "exact" in the question is supposed to be understood as the "expression" of the average or expectation value of kinetic energy.
By the way, although your approach is indeed wrong, I need to mention that you cannot add two quantities with different units. That's why you get that 13.599999.
The question is indeed to calculate the exact value. This is the 2nd of 2 questions, where the first one is the calculate the value with the Thomas-Fermi approach and Hartree units, which gave me the value 0.3, and then the 2nd question, the one i am adressing above, is to instead calculate the exact value.
 

PeroK

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The question is indeed to calculate the exact value. This is the 2nd of 2 questions, where the first one is the calculate the value with the Thomas-Fermi approach and Hartree units, which gave me the value 0.3, and then the 2nd question, the one i am adressing above, is to instead calculate the exact value.
You know the total energy. Maybe if you could calculate the expected value of the potential energy, then you could get the expected value of the kinetic energy from that?
 
You know the total energy. Maybe if you could calculate the expected value of the potential energy, then you could get the expected value of the kinetic energy from that?
Hmm, that approach might actually work! Of course i would do this like [tex]\langle\psi_{100}|potentialenergy|\psi_{100}\rangle[/tex] right? That's how i've calculated expected values before. I've just finished lunch, but i'll give that a try after and i'll get back to you guys, thanks for the help! :)
 

PeroK

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I am thinking that my approach is wrong in the sense that i am trying to do something in a way that it's not allowed to. We have the hamiltonian [tex]\hat{H}\psi=E\psi[/tex] which in this case is [tex](-\frac{\hbar^2}{2\mu}\nabla^2-\frac{Ze^2}{4\pi\epsilon_0r})\psi(r,\theta,\phi)=E\psi(r,\theta,\phi)=-13.6eV\psi(r,\theta,\phi)[/tex] and the last term is because we know that it is in its groundstate.
What you have here is essentially an operator equation:

##\hat{H}\psi = E \psi##

or

##\hat{T}\psi + \hat{V} \psi = E\psi##

What you can't do is cancel the ##\psi##. What you could do is take the inner product with ##\psi## to get an equation involving expected values:

##\langle \psi | \hat{T}\psi + \hat{V} \psi \rangle = \langle \psi | E\psi \rangle##

##\langle T \rangle + \langle V \rangle = E##
 
Last edited:
What you have here is essentially an operator equation:

##\hat{H}\psi = E \psi##

or

##\hat{T}\psi + \hat{V} \psi = E\psi##

What you can't do is cancel the ##\psi##. What you could do is take the inner product with ##\psi## to get an equation involving expected values:

##<\psi | \hat{T}\psi + \hat{V} \psi> = <\psi | E\psi>##

##<T> + <V> = E##
I'll get to that right now :) Get back to you in a second
 
I still don't get it quite right it looks like. [tex]\langle\psi|V(r)|\psi\rangle=\langle\psi|-\frac{k}{r}|\psi\rangle=\frac{-k4\pi}{\pi a_0}\int_{0}^{\infty} re^{\frac{2r}{a_0}} dr[/tex] which i get to become [tex]-ka_0^2=-\frac{-Ze^2a_0}{4\pi\epsilon_0}=\frac{-(1.6*10^{-10})^20.52*10^{-10}}{4\pi8.8*10^{-12}}=-1.2*10^{-38}[/tex] I must have done something wrong somewhere. The 4pi early in the numerator came from the [itex]sin(\theta)d\theta d\phi[/itex] in the spherical coordinates integral.
 

PeroK

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I still don't get it quite right it looks like. [tex]\langle\psi|V(r)|\psi\rangle=\langle\psi|-\frac{k}{r}|\psi\rangle=\frac{-k4\pi}{\pi a_0}\int_{0}^{\infty} re^{\frac{2r}{a_0}} dr[/tex] which i get to become [tex]-ka_0^2=-\frac{-Ze^2a_0}{4\pi\epsilon_0}=\frac{-(1.6*10^{-10})^20.52*10^{-10}}{4\pi8.8*10^{-12}}=-1.2*10^{-38}[/tex] I must have done something wrong somewhere. The 4pi early in the numerator came from the [itex]sin(\theta)d\theta d\phi[/itex] in the spherical coordinates integral.
What wave function are you using? Also, I would break this problem down and first calculate:

##\langle \frac{1}{r} \rangle##

Whay do you get for that?
 
What wave function are you using? Also, I would break this problem down and first calculate:

##\langle \frac{1}{r} \rangle##

Whay do you get for that?
The wavefunction i am using is the one for hydrigen in its m=0, l=0 and n=1-state, that is [itex]\frac{1}{\sqrt{\pi a_0}}e^{\frac{-r}{a_0}}[/itex]
 

PeroK

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The wavefunction i am using is the one for hydrigen in its m=0, l=0 and n=1-state, that is [itex]\frac{1}{\sqrt{\pi a_0}}e^{\frac{-r}{a_0}}[/itex]
It should be:

[itex]\frac{1}{\sqrt{\pi a_0^3}}e^{\frac{-r}{a_0}}[/itex]

By the way, you may not need any numbers for this problem at all.
 
It should be:

[itex]\frac{1}{\sqrt{\pi a_0^3}}e^{\frac{-r}{a_0}}[/itex]

By the way, you may not need any numbers of this problem at all.
Now i see it should ba 3 there.... this vastly changes things.. ! I'll do the math once again for the expected potential and i'll get back!
 
What wave function are you using? Also, I would break this problem down and first calculate:

##\langle \frac{1}{r} \rangle##

Whay do you get for that?
For this i get [tex]\frac{1}{a_0}=1.92*10^{10}[/tex] which if i later also add in the rest of the equation for the potential energy gives me that the expected value for the potential energy should be [itex]-4.45*10^{-18}[/itex] which is wrong i guess.. :/
 

PeroK

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For this i get [tex]\frac{1}{a_0}=1.92*10^{10}[/tex] which if i later also add in the rest of the equation for the potential energy gives me that the expected value for the potential energy should be [itex]-4.45*10^{-18}[/itex] which is wrong i guess.. :/
No numbers allowed! How is ##a_0## defined?
 
Yes, okay, but how is it related to the potential ##V(r)##?
The way i'd like to see it be is that the potential at the inner most radious for the electron (ground state) should be 0 because it can't jump back to a lower state but i guess this is not the case.. :/
 

PeroK

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The way i'd like to see it be is that the potential at the inner most radious for the electron (ground state) should be 0 because it can't jump back to a lower state but i guess this is not the case.. :/

##a_0 = \frac{4\pi \epsilon_0 \hbar^2}{me^2}##

You could have looked that up online.

##V(r)## can be expressed, therefore, in terms of ##a_0##

Also, you'll need ##E_0## in terms of ##a_0##.
 
##a_0 = \frac{4\pi \epsilon_0 \hbar^2}{me^2}##

You could have looked that up online.

##V(r)## can be expressed, therefore, in terms of ##a_0##

Also, you'll need ##E_0## in terms of ##a_0##.
Oh i didn't understand your question earlier, let me dig into this for a while, i'll get back to you
 

vela

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The value for the kinetic energy I get is 13.6 eV or so which is wrong.

That's actually the right answer, but that's not the answer you got. You got -13.6 eV, which is wrong because the kinetic energy shouldn't be negative. You can't just erase the minus sign because you don't want it there.
 
Guys the mystery has just solved itself.. as i said in my OP, i was nudged to believe that the answer was 1/2... the source that told me this just clarified that it was 1/2 hartree units.. which translates to 13.6eV. The problem is solved, thank you guys for your help!
 

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Guys the mystery has just solved itself.. as i said in my OP, i was nudged to believe that the answer was 1/2... the source that told me this just clarified that it was 1/2 hartree units.. which translates to 13.6eV. The problem is solved, thank you guys for your help!
It's good that you got the answer but this problem ought to make you think about your mathematical techniques. First, it is easier to break these problems down. If you have a potential based on ##\frac{1}{r}## or ##\frac{1}{r^2}## then it is a good idea to calculate ##\langle \frac{1}{r} \rangle## or ##\langle \frac{1}{r^2} \rangle## separately. You can then use this result to calculate ##\langle V \rangle## and you also have it up your sleeve, so you don't end up doing the same integration several times.

Sticking with the algebra not only makes the problem easier to solve and makes it much easier to spot an error, but lets you actually see the physics. In this case:

##E_0 = -\frac{\hbar^2}{2ma_0^2}##, ##V(r) = -\frac{\hbar^2}{ma_0 r}##, hence ##\langle V \rangle = -\frac{\hbar^2}{ma_0^2} = 2E_0##

This gives you ##\langle T \rangle = \frac{\hbar^2}{2ma_0^2} = -E_0## without any arithmetic at all.

So, you never had to go back to the values for ##m_e## and ##Z## etc. All these numbers were encapsulated in the Bohr radius ##a_0##.

If you tried the same problem for higher energy states, you'll see a mathematical pattern emerge, which is much harder to see if you just plug numbers into equations. In this problem, you weren't really sure whether the answer was right or wrong or out by a factor of ##10^8##. It was just some number that emerged from a complicated numerical expression.

My advice would be to think about these things. You are doing maths and physics at a level now where the "plug and chug" approach is already letting you down.
 
It's good that you got the answer but this problem ought to make you think about your mathematical techniques. First, it is easier to break these problems down. If you have a potential based on ##\frac{1}{r}## or ##\frac{1}{r^2}## then it is a good idea to calculate ##\langle \frac{1}{r} \rangle## or ##\langle \frac{1}{r^2} \rangle## separately. You can then use this result to calculate ##\langle V \rangle## and you also have it up your sleeve, so you don't end up doing the same integration several times.

Sticking with the algebra not only makes the problem easier to solve and makes it much easier to spot an error, but lets you actually see the physics. In this case:

##E_0 = -\frac{\hbar^2}{2ma_0^2}##, ##V(r) = -\frac{\hbar^2}{ma_0 r}##, hence ##\langle V \rangle = -\frac{\hbar^2}{ma_0^2} = 2E_0##

This gives you ##\langle T \rangle = \frac{\hbar^2}{2ma_0^2} = -E_0## without any arithmetic at all.

So, you never had to go back to the values for ##m_e## and ##Z## etc. All these numbers were encapsulated in the Bohr radius ##a_0##.

If you tried the same problem for higher energy states, you'll see a mathematical pattern emerge, which is much harder to see if you just plug numbers into equations. In this problem, you weren't really sure whether the answer was right or wrong or out by a factor of ##10^8##. It was just some number that emerged from a complicated numerical expression.

My advice would be to think about these things. You are doing maths and physics at a level now where the "plug and chug" approach is already letting you down.
I see your point. I am taking this class in addition to the classes i currently have, i am majoring in an electrical engineering programme with emphasis on physics and i maybe want to take a masters in physics here that's given to the physics students. With that said i don't have the same background as the others taking this class and i guess that alters my way to solve stuff. The way i normally solve stuff in electrical engineering is to mash in the numbers and see what i get
 

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