The normalized wave functions for the ground state, ψ0(x), and the first excited state, ψ1(x), o

The normalized wave functions for the ground state, ψ0(x),...

The normalized wave functions for the ground state, $\psi_{0}(x),$ and the first excited state, $\psi_{1}(x),$ of a quantum harmonic oscillator are $\psi_{0}(x)=\left(\frac{a}{\pi}\right)^{1 / 4} e^{-a x^{2} / 2} \quad \psi_{1}(x)=\left(\frac{4 a^{3}}{\pi}\right)^{1 / 4} x e^{-a x^{2} / 2}$ where $a=m \omega / \hbar$ . A mixed state, $\psi_{01}(x),$ is constructed from these states: $$ \psi_{01}(x)=\frac{1}{\sqrt{2}}\left[\psi_{0}(x)+\psi_{1}(x)\right] $$ $\psi_{0}(x),$ and the first excited state, $\psi_{1}(x),$ of a quantum harmonic oscillator are $\psi_{0}(x)=\left(\frac{a}{\pi}\right)^{1 / 4} e^{-a x^{2} / 2} \quad \psi_{1}(x)=\left(\frac{4 a^{3}}{\pi}\right)^{1 / 4} x e^{-a x^{2} / 2}$ where $a=m \omega / \hbar$ . A mixed state, $\psi_{01}(x),$ is constructed from these states: $$ \psi_{01}(x)=\frac{1}{\sqrt{2}}\left[\psi_{0}(x)+\psi_{1}(x)\right] $$ The symbol $\langle q\rangle_{s}$ denotes the expectation value of the quantity $q$ for the state $\psi_{s}(x) .$ Calculate the expectation values (a) $\langle x\rangle_{0},$ (b) $\langle x\rangle_{1},$ and $(\mathrm{c})\langle x\rangle_{01} .$

The normalized wave functions for the ground state, $\psi_{0}(x),$ and the first excited state, $\psi_{1}(x),$ of a quantum harmonic oscillator are $\psi_{0}(x)=\left(\frac{a}{\pi}\right)^{1 / 4} e^{-a x^{2} / 2} \quad \psi_{1}(x)=\left(\frac{4 a^{3}}{\pi}\right)^{1 / 4} x e^{-a x^{2} / 2}$ where $a=m \omega / \hbar$ . A mixed state, $\psi_{01}(x),$ is constructed from these states:
$$
\psi_{01}(x)=\frac{1}{\sqrt{2}}\left[\psi_{0}(x)+\psi_{1}(x)\right]
$$
$\psi_{0}(x),$ and the first excited state, $\psi_{1}(x),$ of a quantum harmonic oscillator are
$\psi_{0}(x)=\left(\frac{a}{\pi}\right)^{1 / 4} e^{-a x^{2} / 2} \quad \psi_{1}(x)=\left(\frac{4 a^{3}}{\pi}\right)^{1 / 4} x e^{-a x^{2} / 2}$ where $a=m \omega / \hbar$ . A mixed state, $\psi_{01}(x),$ is constructed from these states:
$$
\psi_{01}(x)=\frac{1}{\sqrt{2}}\left[\psi_{0}(x)+\psi_{1}(x)\right]
$$
The symbol $\langle q\rangle_{s}$ denotes the expectation value of the quantity $q$ for the state $\psi_{s}(x) .$ Calculate the expectation values (a) $\langle x\rangle_{0},$ (b) $\langle x\rangle_{1},$ and $(\mathrm{c})\langle x\rangle_{01} .$

Key Concepts

Eigenstates

Eigenstates are the specific solutions to the time-independent Schrödinger equation for a quantum system, each corresponding to a definite energy eigenvalue. In the context of the harmonic oscillator, these states are represented by wave functions involving Hermite polynomials and Gaussian factors.

Superposition States

Superposition states are formed by the linear combination of two or more eigenstates, resulting in a new valid state of the system. Their construction leads to interference effects in observables, and the calculation of expectation values must take into account the cross terms that arise from the combination of different eigenstates.

Parity

Parity refers to the symmetry of a function or system under spatial inversion (x ? -x). In the case of the quantum harmonic oscillator, the ground state is even (symmetric) and the first excited state is odd (antisymmetric), which has implications for the evaluation of certain integrals such as expectation values.

Quantum Harmonic Oscillator

This is a fundamental model in quantum mechanics that describes a particle subject to a restoring force proportional to its displacement. Its mathematical framework leads to quantized energy levels and eigenfunctions that are solutions to the Schrödinger equation with a quadratic potential.

Wavefunction Normalization

Normalization is the process of scaling the wave function so that the total probability of finding the particle somewhere in space is unity. This condition ensures that the integral of the absolute square of the wave function over all space equals one.

Expectation Value

The expectation value of an observable in quantum mechanics represents the average outcome of measurements of that observable on a large number of identically prepared systems. For a given operator, it is calculated using the integral of the product of the complex conjugate of the wave function, the operator, and the wave function itself.

Transcript

00:01 Given the wave functions, including the mixed state wave function, we're going to start with the ground state, and we're going to take the expectation value for the position operator.

00:13 And proceeding, we get the following integral.

00:19 And we see that this is going to be an integral over an even or symmetric interval, meaning that it's symmetric about the vertical axis, the y -axis.

00:31 In other words, it's an even interval or symmetric interval about the origin from minus infinity to infinity.

00:38 But this is an integral of an odd function.

00:41 So from that observation, we can already conclude that this integral should be zero.

00:47 You could also look this up in an integral table.

00:50 Or real quick, you can actually do the integral by making a quick change of variables, as i'll show you.

00:59 So there's a real quick way to do that.

01:01 You can write this as two integrals.

01:04 So an integral from minus infinity to zero and an integral from zero to infinity.

01:09 And you'll perform a change of variables on just the first integral.

01:14 So we basically change x to minus x.

01:18 That's the change of variables.

01:20 And when we do that, we get an integral from infinity to zero.

01:24 We leave the second integral alone.

01:27 And this is the form that we now have.

01:33 So in doing that, we come to this next form.

01:41 We flip the integral on the first integral itself, and we have a minus sign.

01:47 Now notice that the minus sign on negative x right here has canceled with the minus sign in d negative x.

01:57 And the minus sign in negative x up here has gone away because it's negative x squared.

02:02 So that leaves us with x dx in the next line, and we just have e to the negative a x squared.

02:10 So we just have a minus sign right here simply because we flipped this integral right here.

02:16 So now this integral will cancel with the integral that we left alone on the right, and we therefore have zero.

02:31 So the expectation value for x in the ground state is zero.

02:35 Next, we continue, and we calculate the expectation value for x in the first excited state...

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