Orthogonality of wave functions to negative momentum states

[0celo7]

This is a question I have about the textbook discussion, so I'll do away with the standard format.

The author of my QM book (Shankar, Principles of Quantum Mechanics) used the term "negative momentum states," all of a sudden, and I've never heard of it before. He has a little note saying that it is "left-going."

I'll quickly run through the discussion. Its the single step potential scattering problem.
Consider the (normalized) Gaussian
$$\psi_I(x)=(\pi\Delta^2)^{-1/4}e^{ik_0(x+a)}e^{-(x+a)^2/2\Delta^2}$$
This packet has mean momentum $p_0=\hbar k_0$ and mean position $\langle X\rangle=-a$. We will assume that $\Delta X$ is large and that the wave function in k space is very sharply peaked near $k_0$.
He then goes on to talk about uncertainty calculations and reflection and transmission coefficients, but I think they are irrelevant for my problem. He presents the solution to the Schrödinger equation as
$$\psi_E(x)=A\left[\left(e^{ik_1x}+\frac{B}{A}e^{-ik_1x}\right)\Theta(-x)+\frac{C}{A}e^{ik_2x}\Theta(x)\right]$$
Where A, B and C are arbitrary, $k_1=\sqrt{\frac{2mE}{\hbar^2}}$ and $k_2=\sqrt{\frac{2m(E-V_0)}{\hbar^2}}$ and $V_0$ is the potential of the step.
Since to each E there is a unique $k_1=+\sqrt{\frac{2mE}{\hbar^2}}$, we can label the eigenstates by $k_1$ instead of E. Eliminating $k_2$ in favor of $k_1$, we get
$$\psi_{k_1}(x)=A\left[\left(e^{ik_1x}+\frac{B}{A}e^{-ik_1x}\right)\Theta(-x)+\frac{C}{A}e^{ix\sqrt{k_1^2-2mV_0/\hbar^2}}\Theta(x)\right]$$
Great foresight tells us that $A=\frac{1}{\sqrt{2\pi}}$, so that our wave function is normalized. Now consider
$$a(k_1)=\langle\psi_{k_1}\vert\psi_I\rangle=\frac{1}{\sqrt{2\pi}}\left(\int_{-\infty}^{\infty}\left[e^{-ik_1x}+\frac{B^*}{A^*} e^{ik_1x}\right]\Theta(-x)\psi_I(x)dx+\int_{-\infty}^{\infty}\frac{C^*}{A^*}e^{-ik_2x}\Theta(x)\psi_I(x)dx\right)$$
This mess ought to reduce to the Fourier transform
$$a(k_1)=\frac{1}{\sqrt{2\pi}}\int e^{-ik_1x}\psi_I(x)dx$$
I get that the second integral vanishes because of the Heavyside function. My question is why the second term in the first integral vanishes. To quote Shankar, "Similarly, the second piece of the first integral also vanishes since $\psi_I$ in k space is peaked around $k=+k_0$ and is orthogonal to (left-going) negative momentum states." Does this mean that $e^{ik_1x}$ is a negative momentum state? I'm very confused by this.

Any help would be greatly appreciated.

 

[mark726]

Thank you for your question about negative momentum states. The term "negative momentum states" refers to the direction of momentum in relation to a chosen reference direction. In this case, the author is referring to "left-going" negative momentum states, which means that the momentum is directed to the left relative to the chosen reference direction. This is often used in quantum mechanics to describe the behavior of a particle in a potential, such as in the single step potential scattering problem that you mentioned.

In the context of the problem you described, the author is using the notation $k_1$ to represent the momentum of the particle. Since the potential is a step, the particle can either be reflected back with the same momentum $k_1$ or transmitted through the step with a different momentum $k_2$. The author is showing that the wave function can be written as a combination of these two possibilities, with the Heaviside function indicating the direction of the momentum (positive for transmitted and negative for reflected). The term "negative momentum states" refers to the reflected state, which is directed to the left relative to the chosen reference direction.

To answer your question about why the second term in the first integral vanishes, it is because the Gaussian wave packet given by $\psi_I(x)$ is centered around $k_0$, which is a positive momentum state. So, when you take the inner product with a negative momentum state (i.e. the reflected state), the result is zero because the two states are orthogonal. In other words, the reflected state is not a possible state for the initial Gaussian wave packet given by $\psi_I(x)$.

I hope this helps to clarify the concept of negative momentum states for you. If you have any further questions, please don't hesitate to ask. Happy studying!