# Orthogonality of wave functions to negative momentum states

• 0celo7
In summary: Your Name]In summary, the author of the QM textbook discussed the concept of "negative momentum states" in the context of a single step potential scattering problem. These states refer to the direction of momentum relative to a chosen reference direction, with "left-going" negative momentum states indicating that the momentum is directed to the left. The author showed that the wave function can be written as a combination of reflected and transmitted states, with the Heaviside function indicating the direction of the momentum. The term "negative momentum states" specifically refers to the reflected state, which is orthogonal to the initial Gaussian wave packet centered around a positive momentum state.
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.

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!

## What is the concept of orthogonality in wave functions?

The concept of orthogonality in wave functions refers to the mathematical property of two wave functions being perpendicular to each other. This means that the integral of the product of the two wave functions over all space is equal to zero. In other words, the two wave functions have no overlap or interference with each other.

## What is the importance of orthogonality in quantum mechanics?

Orthogonality is a fundamental concept in quantum mechanics as it allows for the description of the state of a system in terms of its wave function. This allows for the calculation of various physical properties and observables of the system, such as energy levels and probabilities, through the use of mathematical operations on the wave function.

## What does it mean for a wave function to be orthogonal to negative momentum states?

For a wave function to be orthogonal to negative momentum states means that the wave function has no overlap or interference with states that have negative momentum values. This is important in quantum mechanics as it allows for the description of a particle's momentum state and its associated wave function in a mathematical framework.

## How does the orthogonality of wave functions to negative momentum states affect the uncertainty principle?

The orthogonality of wave functions to negative momentum states plays a crucial role in the Heisenberg uncertainty principle. This principle states that the more precisely we know the position of a particle, the less precisely we can know its momentum, and vice versa. The orthogonality of wave functions restricts the possible values of a particle's momentum, thus contributing to the uncertainty in its measurement.

## What are some applications of the orthogonality of wave functions to negative momentum states?

The concept of orthogonality of wave functions to negative momentum states has several applications in quantum mechanics. It is used in the calculation of energy levels and probabilities of a system, as well as in the development of quantum algorithms for various computational tasks. It also plays a role in the study of quantum entanglement and the behavior of particles in wave-particle duality.

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