Why is the KE operator negative in QM?

[sungholee]

In the Hamilonian for an H₂⁺, the kinetic energy of the electron (KE of nucleus ignored due to born-oppenheimer approximation) has a negative sign in front of it.

I understand the signs for the potential energy operators but not for the KE apart from the strictly mathematical point of view. Can someone explain this please?

 

[blue_leaf77]

In the Hamilonian for an H2+, the kinetic energy of the electron (KE of nucleus ignored due to born-oppenheimer approximation) has a negative sign in front of it.

Do you mean the minus sign in $p = -\frac{\hbar^2}{2m} \frac{d}{dx}$? It is a general momentum operator, it looks like this for all quantum systems, not only restricted to H₂⁺. Speaking of the minus sign, $p$ is an operator, it's not a number, it's not yet a measured physical quantity. Therefore, whether it contains negative signs or imaginary number does not make it peculiar.

 

[sungholee]

Oh right. Then I assume there is no other explanation except for the mathematical derivation, as it's not technically got a physical meaning until the operator has been "operated" on a function, is that right?

Thanks so much

 

[blue_leaf77]

as it's not technically got a physical meaning until the operator has been "operated" on a function, is that right?

Yes, an operator becomes physical if it's measured. Only then, a question of whether negative signs or complex number are allowed make sense.

 

[sungholee]

Great. Thank you!

 

[Steven Hanna]

Do you mean the minus sign in $p = -\frac{\hbar^2}{2m} \frac{d}{dx}$? It is a general momentum operator, it looks like this for all quantum systems, not only restricted to H₂⁺. Speaking of the minus sign, $p$ is an operator, it's not a number, it's not yet a measured physical quantity. Therefore, whether it contains negative signs or imaginary number does not make it peculiar.

I too was wondering about this. The equation T = -(ħ/2m)∇^2 seems to imply that the energy of a system decreases with ∇^2, which is counter intuitive. As the second spatial derivative is increased, the spatial frequency of a wavefunction should increase.

 

[blue_leaf77]

the spatial frequency of a wavefunction

What is that?

The equation T = -(ħ/2m)∇^2 seems to imply that the energy of a system increases with ∇^2, which is counter intuitive.

Applying the kinetic energy operator on a wavefunction does not give a value of kinetic energy, be it the measured one or the average one. Instead, it will just give you another wavefunction. If you want to calculate the quantum mechanical kinetic energy for a given wavefunction, you should calculate the expectation value $\langle \psi |P^2/(2m)| \psi \rangle$. This will in general give you the "simulated" value of the average kinetic energy had the measurement is repeated infinite of times.

 

[Steven Hanna]

What is that?

Applying the kinetic energy operator on a wavefunction does not give a value of kinetic energy, be it the measured one or the average one. Instead, it will just give you another wavefunction. If you want to calculate the quantum mechanical kinetic energy for a given wavefunction, you should calculate the expectation value $\langle \psi |P^2/(2m)| \psi \rangle$. This will in general give you the "simulated" value of the average kinetic energy had the measurement is repeated infinite of times.

Thanks for your quick reply! First, I edited my original post to say that the negative sign seems to imply that energy decreases with ∇². So if you act the kinetic energy operator on an eigenfunction, wouldn't the eigenvalue be equal to the kinetic energy? If so, wouldn't this mean that energy decreases with ∇²?. Also, why is there no negative sign next to P²/(2m) in the expected value formula?

By spatial frequency I mean the number of cycles undergone by the wavefunction per unit distance. I believe this should increase with the second spatial derivative, e.g. sin(2x) vs. sin(x).

 

[blue_leaf77]

why is there no negative sign next to P2/(2m) in the expected value formula?

The minus sign appear when you subtitute $\mathbf{P}=-i\hbar \nabla$.

the negative sign seems to imply that energy decreases with ∇2

Mathematically, saying that something is decreasing or increasing with an operator, in this case the nabla, does not bear any well-defined meaning. You have to make it act on something before saying about a particular behavior of the resulting function. Moreover, for a wavefunction which goes to zero as it approaches infinity and is square-integrable, the expectation value of kinetic energy is always real and positive,
$$
\langle \psi |P^2/(2m)| \psi \rangle = -\frac{\hbar^2}{2m} \int \psi^*(x) \frac{d^2}{dx^2} \psi(x) dx \\
= -\frac{\hbar^2}{2m} \left( \psi^*(x)\frac{d}{dx} \psi(x)\Big|_{-\infty}^\infty - \int \frac{d}{dx} \psi(x) \frac{d}{dx} \psi^*(x) dx \right) \\
= \frac{\hbar^2}{2m} \int \Big|\frac{d}{dx} \psi(x)\Big|^2 dx \geq 0
$$

 

[Steven Hanna]

The minus sign appear when you subtitute $\mathbf{P}=-i\hbar \nabla$.

Mathematically, saying that something is decreasing or increasing with an operator, in this case the nabla, does not any well-defined meaning. You have to make it act on something before saying about a particular behavior of the resulting function. Moreover, for a wavefunction which goes to zero as it approaches infinity and is square-integrable, the expectation value of kinetic energy is always real and positive,
$$
\langle \psi |P^2/(2m)| \psi \rangle = -\frac{\hbar^2}{2m} \int \psi^*(x) \frac{d^2}{dx^2} \psi(x) dx \\
= -\frac{\hbar^2}{2m} \left( \psi^*(x)\frac{d}{dx} \psi(x)\Big|_{-\infty}^\infty - \int \frac{d}{dx} \psi(x) \frac{d}{dx} \psi^*(x) dx \right) \\
= \frac{\hbar^2}{2m} \int \Big|\frac{d}{dx} \psi(x)\Big|^2 dx \geq 0
$$

thanks so much! I think I get it now: the expected value of kinetic energy is never negative, and the negative sign in the KE operator is there because (-i)*(-i) = -1.

 

[blue_leaf77]

thanks so much! I think I get it now: the expected value of kinetic energy is never negative, and the negative sign in the KE operator is there because (-i)*(-i) = -1.

Yes exactly.