I Is the potential energy always negative in the ground state of a hydrogen atom?

[LagrangeEuler]

Why energy of the electron in ground state of hydrogen atom is negative $E_1=-13,6 \rm{eV}$? I am confused because energy is sum of kinetic and potential energy. Kinetic energy is always positive. How do you know that potential energy is negative in this problem?

 

[PeroK]

Why energy of the electron in ground state of hydrogen atom is negative $E_1=-13,6 \rm{eV}$? I am confused because energy is sum of kinetic and potential energy. Kinetic energy is always positive. How do you know that potential energy is negative in this problem?

That value assumes the standard convention that the PE at infinity is zero. In any case, the point at which energy is zero is arbitrary.

 

[Dale]

Why energy of the electron in ground state of hydrogen atom is negative $E_1=-13,6 \rm{eV}$? I am confused because energy is sum of kinetic and potential energy. Kinetic energy is always positive. How do you know that potential energy is negative in this problem?

As @PeroK said, this refers to the convention that the energy is 0 when the electron is far from the proton. When the electron starts far away (unbound) and then goes to the ground state it releases 13.6 eV of energy. Conversely, if the electron starts in the ground state and then is pulled far away from the proton it requires 13.6 eV of energy. This indicates that the energy of the ground state is 13.6 eV lower than the energy of the unbound state. So if the latter is by convention set to 0 then the former must be -13.6 eV

 

[hilbert2]

<edited>

 

[DrClaude]

And, any potential well with $V(\mathbf{r})<0$, no matter how shallow, will have at least one bound state with a negative energy eigenvalue.

Really?

 

[hilbert2]

Really?

If you create potential energy functions like

$V(x)=-C\exp\left(-kx^2 \right)$

or

$V(x,y)=-C\exp\left(-k_x x^2 - k_y y^2 \right)$

with $C>0$, $k_x >0$ and $k_y >0$, which have a negative value for any $x$ or $(x,y)$, they should have a negative-energy bound state no matter how small $C$ is. Isn't this what is meant in the linked article?

https://www.researchgate.net/publication/253032187_Criteria_for_bound-state_solutions_in_quantum_mechanics

 

[DrClaude]

If you create potential energy functions like

$V(x)=-C\exp\left(-kx^2 \right)$

or

$V(x,y)=-C\exp\left(-k_x x^2 - k_y y^2 \right)$

with $C>0$, $k_x >0$ and $k_y >0$, which have a negative value for any $x$ or $(x,y)$, they should have a negative-energy bound state no matter how small $C$ is. Isn't this what is meant in the linked article?

https://www.researchgate.net/publication/253032187_Criteria_for_bound-state_solutions_in_quantum_mechanics

I haven't checked the article, but the statement above is markedly different from the one you made in post #4:

And, any potential well with $V(\mathbf{r})<0$, no matter how shallow, will have at least one bound state with a negative energy eigenvalue.

There are plenty of potentials well shapes for which there is a minimum depth below which there is no bound state.

 

[hilbert2]

Then, I should have stated it as

$V(\mathbf{r}) < 0$ for all $\mathbf{r}=(x_1 ,x_2 ,\dots ,x_n ) \in \mathbb{R}^n$

Or maybe you mean a potential that has the kind of negative infinity somewhere that causes a "falling to center" problem?

 

[DrClaude]

Then, I should have stated it as

$V(\mathbf{r}) < 0$ for all $\mathbf{r}=(x_1 ,x_2 ,\dots ,x_n ) \in \mathbb{R}^n$

Or maybe you mean a potential that has the kind of negative infinity somewhere that causes a "falling to center" problem?

I don't see what you are getting at. While one-dimensional attractive potentials always have at least one bound state, this is not true in higher dimensions. For instance, one can show that the spherical well
$$
V(r) = \left\{ \begin{array}{ll} -V_0 & r < a \\ 0 & r>a \end{array} \right.
$$
has no bound states for $V_0 a^2 < \pi^2 \hbar􏰠^2/8m$.

 

[hilbert2]

I don't see what you are getting at. While one-dimensional attractive potentials always have at least one bound state, this is not true in higher dimensions. For instance, one can show that the spherical well
$$
V(r) = \left\{ \begin{array}{ll} -V_0 & r < a \\ 0 & r>a \end{array} \right.
$$
has no bound states for $V_0 a^2 < \pi^2 \hbar􏰠^2/8m$.

Can you give a reference where this is proved? Sorry for causing confusion if I gave incorrect information.

Edit: Ok, it was as easy to find as this: https://quantummechanics.ucsd.edu/ph130a/130_notes/node227.html

 

[vanhees71]

Really?

I don't remember by heart now, how it is in 3 dimensions. I think, there's an answer in the textbook by Messiah, but I don't have the book at hand now either :-(.

 

[hilbert2]

I don't remember by heart now, how it is in 3 dimensions. I think, there's an answer in the textbook by Messiah, but I don't have the book at hand now either :-(.

Some old edition of that book is readable in Archive.org

https://archive.org/details/QuantumMechanicsVolumeI/mode/2up

https://archive.org/details/QuantumMechanicsVolumeIi/mode/2up

 

[hilbert2]

So, I guess that an n-dimensional hypercube potential well, with $V(x_1 ,\dots ,x_n )=V_0$ when any of $|x_k |$ is greater than $L/2$ and otherwise $V (x_1 ,\dots ,x_n )=0$, always has a bound state no matter how close to zero the positive number $V_0$ is. This is because the 3D eigenfunctions are just products of the solutions of the equivalent 1-dimensional problem.

A fun problem to consider would be to make a finite potential well where the region with $V(x_1 ,\dots ,x_n ) = 0$ is a hypercube with rounded corners, i.e. an intersection of a hypercube with side length $L$ and a sphere with radius $L/2 \leq R \leq \sqrt{n}L/2$.