Issue with wavenumber in the free particle wavefunction

[shankk]

TL;DR

The stationary state solution for the free particle in one dimension is given by $\psi(x) = Ae^{\iota K x}$ where $K = \pm \frac{\sqrt{2mE}}{\hbar}$ with $E$ being the energy eigenvalue of the stationary state. Now what I'm being told is that the momentum $p = \hbar K$. De broglie relation says $p = \frac{h}{\lambda} = \hbar \frac{2\pi}{\lambda}$. This would imply that the debroglie wavenumber $\frac{\lambda}{2\pi}$ is equal to $K$. I don't understand why this would be the case.

To me, the $K$ obtained by solving the Schrödinger equation and the de broglie wavelength seem two completely unrelated quantities. Can someone explain why have we equated $K$ and $\frac{2\pi}{\lambda}$. Also, isn't writing $p = \hbar K$ implying that eigenstate of energy is also an eigenstate of the momentum operator because we have assumed a "definite value" of $p$.

 

[PeterDonis]

The stationary state solution for the free particle in one dimension is given by $\psi(x) = Ae^{\iota K x}$ where $K = \pm \frac{\sqrt{2mE}}{\hbar}$ with $E$ being the energy eigenvalue of the stationary state.

A somewhat pedantic note: the imaginary unit is just $i$; no need to use the Greek letter $\iota$.

Now what I'm being told is that the momentum $p = \hbar K$.

You can work this out for yourself. What do you get if you apply the momentum operator $\hat{p}$ to the state $\psi(x)$ you have written down?

De broglie relation says $p = \frac{h}{\lambda} = \hbar \frac{2\pi}{\lambda}$. This would imply that the debroglie wavenumber $\frac{\lambda}{2\pi}$ is equal to $K$. I don't understand why this would be the case.

The function $e^{i K x}$ describes what? A plane wave. What is the wavelength of that plane wave?

isn't writing $p = \hbar K$ implying that eigenstate of energy is also an eigenstate of the momentum operator because we have assumed a "definite value" of $p$.

You don't need to assume anything. As I noted above, you can apply the momentum operator $\hat{p}$ to the state $\psi(x)$ and check explicitly to see whether it is an eigenstate, and if so, what its eigenvalue is.

 

[PeroK]

Also, isn't writing $p = \hbar K$ implying that eigenstate of energy is also an eigenstate of the momentum operator because we have assumed a "definite value" of $p$.

No. In general, if we have a linear operator $\hat X$ with eigenvalue $\lambda$, then $\lambda^2$ is an eigenvalue of $\hat X^2$ corresponding to the same eigenvector.

That means that for the free particle in particular momentum eigenstates are also energy eigenstates.

But, it doesn't work in reverse, as we can combine vectors with eigenvalues $\pm \lambda$ to construct an eigenvector of $\hat X ^2$ that is not an eigenvector of $\hat X$. In other words:

If $v_1$ is an eigenvector of $\hat X$ with eigenvalue $\lambda$ and $v_2$ is an eigenvector of $\hat X$ with eigenvalue $-\lambda$, then $v_1 + v_2$ is an eigenvector of $\hat X^2$ with eigenvalue $\lambda^2$, but not an eigenvector of $\hat X$, as: $$\hat X (v_1 + v_2) = \lambda(v_1 - v_2)$$ Whereas: $$\hat X^2 (v_1 + v_2) = \lambda^2(v_1 + v_2)$$
PS The important point for QM is, of course, that you can have an energy eigenstate that is a superposition of momentum eigenstates with equal and opposite momenta.

 

[shankk]

A somewhat pedantic note: the imaginary unit is just $i$; no need to use the Greek letter $\iota$.
You can work this out for yourself. What do you get if you apply the momentum operator $\hat{p}$ to the state $\psi(x)$ you have written down?
The function $e^{i K x}$ describes what? A plane wave. What is the wavelength of that plane wave?
You don't need to assume anything. As I noted above, you can apply the momentum operator $\hat{p}$ to the state $\psi(x)$ and check explicitly to see whether it is an eigenstate, and if so, what its eigenvalue is.

I see, I just had to apply the momentum operator to $\psi$ and I would have received all my answers. I had thought about this problem for quite some time but this simple step just didn't cross my mind. I am a complete beginner at QM. Thanks for clearing things up.

 

[shankk]

No. In general, if we have a linear operator $\hat X$ with eigenvalue $\lambda$, then $\lambda^2$ is an eigenvalue of $\hat X^2$ corresponding to the same eigenvector.

That means that for the free particle in particular momentum eigenstates are also energy eigenstates.

But, it doesn't work in reverse, as we can combine vectors with eigenvalues $\pm \lambda$ to construct an eigenvector of $\hat X ^2$ that is not an eigenvector of $\hat X$. In other words:

If $v_1$ is an eigenvector of $\hat X$ with eigenvalue $\lambda$ and $v_2$ is an eigenvector of $\hat X$ with eigenvalue $-\lambda$, then $v_1 + v_2$ is an eigenvector of $\hat X^2$ with eigenvalue $\lambda^2$, but not an eigenvector of $\hat X$, as: $$\hat X (v_1 + v_2) = \lambda(v_1 - v_2)$$ Whereas: $$\hat X^2 (v_1 + v_2) = \lambda^2(v_1 + v_2)$$
PS The important point for QM is, of course, that you can have an energy eigenstate that is a superposition of momentum eigenstates with equal and opposite momenta.

Okay, but for the free particle case, we first obtained the energy eigenstate and found out that it is also an eigenstate of the momentum operator. So, for this particular case of a free particle, every eigenstate of energy is also an eigenstate of momentum, right?

 

[PeroK]

Okay, but for the free particle case, we first obtained the energy eigenstate and found out that it is also an eigenstate of the momentum operator. So, for this particular case of a free particle, every eigenstate of energy is also an eigenstate of momentum, right?

No. See above. Every momentum eigenstate is an energy eigenstate but not vice versa.

 

[shankk]

No. See above. Every momentum eigenstate is an energy eigenstate but not vice versa.

An energy eigenstate is a wavefunction $\psi(x)$ for which $\hat{H}\psi = E\psi$. On solving for free particle in one dimension, we get $\psi(x) = Ae^{iKx}$ with $K = \pm \frac{\sqrt{2mE}}{\hbar}$. Now, we apply the momentum operator $-i\hbar \frac{\partial}{\partial x} (Ae^{iKx}) = \hbar K(Ae^{iKx})$. And so, $\psi$ is also an eigenstate of $\hat{p}$ with eigenvalue $\hbar K$. And so, $\psi$ is also a momentum eigenstate.

 

[PeroK]

An energy eigenstate is a wavefunction $\psi(x)$ for which $\hat{H}\psi = E\psi$. On solving for free particle in one dimension, we get $\psi(x) = Ae^{iKx}$ with $K = \pm \frac{\sqrt{2mE}}{\hbar}$. Now, we apply the momentum operator $-i\hbar \frac{\partial}{\partial x} (Ae^{iKx}) = \hbar K(Ae^{iKx})$. And so, $\psi$ is also an eigenstate of $\hat{p}$ with eigenvalue $\hbar K$. And so, $\psi$ is also a momentum eigenstate.

You need to take more care with the mathematics here. To be more precise, we have:
$$\psi(x) = Ae^{iKx} + Be^{-iKx}, \ \ \text{with} \ \ K = \frac{\sqrt{2mE}}{\hbar}$$ which is only a momentum eigenstate if $A = 0$ or $B = 0$.

 

[PeterDonis]

we first obtained the energy eigenstate

No, you obtained an energy eigenstate. It is not the only possible one. @PeroK showed you another one. If you're wondering where his came from, try just considering $B e^{-i K x}$ by itself, and check that it is both an energy eigenstate and a momentum eigenstate--but with a different eigenvalue for momentum (flipped sign) from the state you wrote down in the OP.

 

[vanhees71]

What you are struggling here is what's named "degeneracy". In your example of a free particle moving in one dimension with the Hamiltonian
$$\hat{H}=\frac{1}{2m} \hat{p}^2$$
the energy eigenstates are determined by the time-independent Schrödinger equation
$$\hat{H} u_E(x)=E u_E(x),$$
where $E$ is an energy eigenvalue and $u_E$ an energy eigenfunction for this eigenvalue. Now use $\hat{p}=-\mathrm{i} \hbar \partial_x$. Then you get the equation
$$-\frac{\hbar^2}{2m} \partial_x^2 u_E(x)=E u_E(x).$$
The general solution of this 2nd-order differential equation is
$$u_E(x)=A \exp(\mathrm{i} k x) + B \exp(-\mathrm{i} k x),$$
where $k=\sqrt{2mE}/\hbar$.

Obviously you must have $k \in \mathbb{R}$, such that the $u_E$ are at least "normalizable to a $\delta$ distribution", i.e., you must have $E \geq 0$, and only for $E=0$ you'd get a unique eigenstate. For $E>0$ there are always two linearly independent eigenstates, namely
$$u_{E,R}=A \exp(\mathrm{i} k x), \quad u_{E,L} = A \exp(-\mathrm{i} k x),$$
where $A$ is some constant to normalize the states (but that's not so important here).

It's also easy to understand, why you have this two-fold degeneracy: The energy of the free particle is determined by its momentum, and $E_p=p^2/(2m)$. Now there are two possibilities for a free particle to have this energy, namely to move with a momentum $p>0$ (i.e., to the right) or $-p<0$ (i.e., to the left).

So in this case it's much more convenient to use the momentum eigenstates as a complete set of (generalized) eigenfunctions, which are not degenerate and use the fact that any momentum eigenstate is also an energy eigenstate. As stressed already in this thread, of course not any energy eigenstate is also a momentum eigenstate because of the degeneracy of the energy eigenstates: If you have a superposition of a right and a left-moving energy eigenstate with the same energy eigenvalue you get again an energy eigenstate with that energy but it's no longer a momentum eigenstate.