Question

Suppose a quantum particle is in a stationary state (energy eigenstate) with a wave function \(\psi(x, t) .\) The calculation of \(\langle x\rangle,\) the expectation value of the particle's position, is shown in the text. Calculate \(d\langle x\rangle / d t(\operatorname{not}\langle d x / d t\rangle)\).

Short answer

In conclusion, for a quantum particle in a stationary state with the wave function \(\psi(x, t)\), the time derivative of the expectation value of the particle's position, \(\frac{d\langle x\rangle}{dt}\), is equal to zero. This indicates that the particle's expected position does not change over time in a stationary state.

Step-by-step solution

Formula for the expectation value of an observable

Recall that the expectation value of an observable A (in this case, the position x) for a wave function \(\psi(x, t)\) is given by the following integral: \(\langle A \rangle = \int_{-\infty}^{\infty}\psi^*(x, t)\,A\,\psi(x, t)\, dx\). Since the particle is in a stationary state (energy eigenstate), the wave function can be represented as \(\psi(x, t) = \psi(x)e^{-\frac{iE}{\hbar}t}\), where \(\psi(x)\) is the spatial part of the wave function and E is the energy of the stationary state. For the position observable x, the expectation value \(\langle x \rangle\) is given by: \(\langle x \rangle = \int_{-\infty}^{\infty}\psi^*(x, t)\,x\,\psi(x, t)\, dx\).

Time derivative of the expectation value of the position

Now, we want to find the time derivative of the expectation value of the position, i.e. \(\frac{d\langle x\rangle}{dt}\). Using the expectation value expression from Step 1, we have: \(\frac{d\langle x\rangle}{dt} = \frac{d}{dt}\left[\int_{-\infty}^{\infty}\psi^*(x, t)\,x\,\psi(x, t)\, dx\right]\). Since the spatial part of the wave function \(\psi(x)\) does not depend on time, we can take the time derivative inside the integral and apply it to the time dependence of the wave function, which comes from the term \(e^{-\frac{iE}{\hbar}t}\).

Applying the time derivative inside the integral

Applying the time derivative inside the integral, we get: \(\frac{d\langle x\rangle}{dt} = \int_{-\infty}^{\infty}\frac{d}{dt}\left[\psi^*(x, t)\,x\,\psi(x, t)\right]\, dx\). Now, we need to calculate the time derivative of \(\psi^*(x, t)\,x\,\psi(x, t)\), which can be computed using the product rule for derivatives as follows: \(\frac{d}{dt}\left[\psi^*(x, t)\,x\,\psi(x, t)\right] = \left(\frac{d\psi^*(x, t)}{dt}\right)\,x\,\psi(x, t) + \psi^*(x, t)\,x\,\frac{d\psi(x, t)}{dt}\). Since the time dependence is only due to the term \(e^{-\frac{iE}{\hbar}t}\), we have: \(\frac{d\psi^*(x, t)}{dt} = \frac{d}{dt}\left[\psi^*(x)e^{\frac{iE}{\hbar}t}\right] = \frac{iE}{\hbar}\psi^*(x)e^{\frac{iE}{\hbar}t}\), \(\frac{d\psi(x, t)}{dt} =\frac{d}{dt}\left[\psi(x)e^{-\frac{iE}{\hbar}t}\right]= -\frac{iE}{\hbar}\psi(x)e^{-\frac{iE}{\hbar}t}\).

Evaluating the time derivative of the expectation value of the position

Substituting the time derivatives back into the expression for \(\frac{d\langle x\rangle}{dt}\), we have: \(\frac{d\langle x\rangle}{dt} = \int_{-\infty}^{\infty}\left[\left(\frac{iE}{\hbar}\psi^*(x)e^{\frac{iE}{\hbar}t}\right)\,x\,\psi(x)e^{-\frac{iE}{\hbar}t} + \psi^*(x)e^{\frac{iE}{\hbar}t}\,x\,\left(-\frac{iE}{\hbar}\psi(x)e^{-\frac{iE}{\hbar}t}\right)\right]\, dx\). Simplifying, we notice that the two terms inside the integral sum to zero: \(\frac{d\langle x\rangle}{dt} = \int_{-\infty}^{\infty}\left[\frac{iEx}{\hbar}\psi^*(x)\psi(x) - \frac{iEx}{\hbar}\psi^*(x)\psi(x)\right]dx = 0\). So, the time derivative of the expectation value of the particle's position in a stationary state is equal to zero, i.e. \(\frac{d\langle x\rangle}{dt} = 0\).

Key concepts

Expectation Value

When studying quantum mechanics, the concept of an expectation value plays a critical role in understanding the typical behavior of a quantum system. In essence, the expectation value is a statistical mean of an observable quantity—such as position, momentum, or energy—over all the possible outcomes of a given wave function.

Mathematically, for a particle with a wave function \(\psi(x, t)\), the expectation value of an observable \(A\) is given by the integral \(\langle A \rangle = \int_{-\infty}^{\infty}\psi^*(x, t) A \psi(x, t) dx\), where \(\psi^*(x, t)\) is the complex conjugate of \(\psi(x, t)\). This operation blends quantum probabilities with the classical concept of an average, offering a bridge between the two worlds.

Stationary State

A stationary state is a quantum state with all its properties unchanging in time. This term is frequently used interchangeably with energy eigenstate. In a quantum context, if a particle is in a stationary state, it means that the probability density associated with its wave function does not evolve with time.

The stationary states are significant because they represent the allowed energy levels of a quantum system and are described by wave functions of the form \(\psi(x, t) = \psi(x)e^{-\frac{iE}{\hbar}t}\), where \(\psi(x)\) is the spatial aspect and does not depend on time. Here \(E\) is the energy associated with that state, making the time evolution purely exponential with no change in the wave function's magnitude.

Wave Function

The wave function, denoted by \(\psi(x, t)\), is a fundamental mathematical object in quantum mechanics. It contains all the information about a quantum system. This complex-valued function can describe particles and their quantum states, and its absolute square \(|\psi(x, t)|^2\) provides the probability density of finding a particle at position \(x\) at time \(t\).

It's critical to acknowledge that the wave function alone doesn't have a direct physical interpretation until it's linked with an observable through the expectation value. The behavior of \(\psi(x, t)\) underlies much of quantum mechanics, giving rise to phenomena like interference and superposition.

Time Derivative

Within quantum mechanics, the time derivative is a way to express how a quantum state or an observable changes over time. Taking the time derivative of quantities like the expectation value of position provides insight into their dynamic properties.

For a stationary state, one might expect that since the state 'does not change' over time, the expectation value of an observable wouldn't either. The calculation of the time derivative involves utilizing the product rule and recognizing the explicit time dependence that comes from the exponential term \(e^{-\frac{iE}{\hbar}t}\). In the particular situation of a stationary state, when you work through the math, the dynamic particulars cancel each other out, leading to a zero time derivative for the expectation value, highlighting the 'stationary' aspect of such states.

Energy Eigenstate

Quantum particles possess discrete energy levels which are referred to as energy eigenstates. These are special states of a quantum system where the energy operator (Hamiltonian) acting on the wave function returns the same wave function times a constant (the energy eigenvalue). This corresponds to one of the quantum system's possible energy levels.

The term 'eigenstate' originates from the German word 'eigen,' meaning 'own' or 'characteristic.' In a stationary or energy eigenstate, as expressed in the wave function \(\psi(x, t) = \psi(x)e^{-\frac{iE}{\hbar}t}\), the space-dependent part \(\psi(x)\) holds the spatial information and the exponential term captures the time evolution. This separation is at the heart of the concept of an energy eigenstate, where \(E\) represents an eigenvalue corresponding to the system's energy when in that eigenstate.