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Consider a particle in an infinite potential well, If we solve for the energy using Schrodinger's equation we get the solutions, we get sine functions for wavefunction and Energy in a state is proportional to $n^2$. My questions are:

(i) What exactly is the Energy of the particle, in this case since there is no potential energy, is it just Kinetic Energy?

(ii) We know that $\psi(x)$ denotes the probability density of finding a particle at $x$, similarly there will be $\psi(p)$ (momentum) right? and $\psi(0)$ may not be zero. Now my doubt is how can the particle have variable momentum if it has fixed energy. Because $\mathrm{energy} = KE = p^2/2m$. What am I missing?

Qmechanic
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Saivardhan Annam
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2 Answers2

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The sharp walls or delta potentials in QM are best interpreted as approximations to a smooth, potentials, which are however difficult to deal with mathematically. e.g., $U(x)=U_0\tanh(\alpha x)$ is a step potential, whereas two such steps may form a nearly squared well, which, for sufficiently large $U_0$ would be well approximated by an infinite well (for low energy states.)

If you have no problem with a square well, you may just think of a very deep square well. The solutions in the infinite well provide an estimate to what extent this approximation is valid: $E_n\ll U_0$.

Then the Hamiltonian is usual $H=K+U$, which commutes with neither position, nor momentum operators, which means that position and momentum are uncertain in the energy eigenstates.

Roger V.
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Consider a particle in an infinite potential well...

(i) What exactly is the Energy of the particle, in this case since there is no potential energy, is it just Kinetic Energy?

Yes, it is just kinetic energy.

(ii) We know that $\psi(x)$ denotes the probability density of finding a particle at $x$, similarly there will be $\psi(p)$ (momentum) right?

I would encourage you to use a different symbol for probability density in position space ($\psi(x)$) versus probability density in momentum space ($\tilde \psi(p)$), since when you use the same symbol in the same sentence, it would normally be interpreted as the same function, which $\psi(x)$ and $\tilde \psi(p)$ are not.

and $\psi(0)$ may not be zero.

See? I have no idea if you mean the probability density in space at $x=0$, or in momentum space at $p=0$. So, this part of your question is unclear.

Now my doubt is how can the particle have variable momentum if it has fixed energy. Because $\mathrm{energy} = KE = p^2/2m$. What am I missing?

The length of the momentum will be fixed in an energy eigenstate, but the direction (or sign in 1D) will not necessarily be fixed.

[from comments] but how can the particle have zero momentum when the total energy is non zero and potential energy is zero...

Your phrase "have zero momentum," is not really meaningful, unless the state is a momentum eigenstate, which an energy eigenstate is not. You can really only ask about the expectation value of the momentum in an energy eigenstate.

Energy eigenstates are standing waves $$ \psi(x)\propto \sin(p_n x) =\frac{1}{2i}\left(e^{ip_n x} - e^{-ip_n x}\right)\;, $$ which you can see are linear combinations of "forward traveling" waves $$ e^{ip_n x} $$ and "backward travelling" waves $$ e^{-ip_n x}\;, $$ with coefficients such that the overall wave function is real.

For such a standing wave the expectation value of $p$ is zero: $$ \langle \hat p \rangle \propto \int dx \sin(p_n x)\cos(p_n x) = 0 $$ but the expectation value of $p^2$ is not zero: $$ \langle \hat p^2 \rangle = \frac{n^2\pi^2}{L^2} $$

hft
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