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Does the wave function \(\psi(x)=\vec{A}\left(e^{+a x}+e^{-a x}\right)\) have a well-defined momentum? Explain.

Short Answer

Expert verified
The wave function \(\psi(x)=\vec{A}\left(e^{+a x}+e^{-a x}\right)\) is not an eigenfunction of the momentum operator. Thus, it does not have a well-defined momentum.

Step by step solution

01

Identify the variables and constants

In this wave function, \(\vec{A}\) and \(a\) are constants while \(x\) is the variable.
02

Apply the momentum operator

The momentum operator is given by \(-i\hbar\frac{d}{dx}\). Apply this operator to \(\psi(x)=\vec{A}\left(e^{+a x}+e^{-a x}\right)\). This will involve taking the derivative of \(e^{+a x}+e^{-a x}\) and multiplying by \(-i\hbar\).
03

Confirm whether the resultant is a multiple of \(\psi(x)\)

If the wave function is an eigenfunction of the momentum operator, the result of Step 2 will be a multiple of \(\psi(x)\). This 'multiple' is the eigenvalue, representing the momentum.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wave Function
In quantum mechanics, a wave function is a crucial concept used to describe the quantum state of a particle. The wave function is a mathematical function that contains all the information about a system. For a single particle, it is typically written as \[\psi(x, t)\]which is dependent on both position \(x\) and time \(t\). The square of the absolute value of the wave function, \[|\psi(x, t)|^2\],is particularly important since it gives the probability density of finding the particle at position \(x\) at time \(t\).
  • A wave function must be normalized so that the total probability of finding the particle somewhere in space is 1.
  • It should be continuous and smooth to ensure proper physical interpretation.
In the given problem, the wave function is expressed as:\[\psi(x) = \vec{A}\left(e^{+a x} + e^{-a x}\right)\]Here, \( \vec{A} \) is a constant and \( a \) is a parameter. This form indicates that the wave function may consist of a superposition of exponential terms, suggesting that the particle's position description is influenced by both exponential growth and decay at rate \( a \).
Understanding wave functions is fundamental for predicting how particles behave in quantum systems, as they help provide insights into the distribution and evolution of quantum states.
Momentum Operator
The momentum operator in quantum mechanics is an essential tool for determining a particle's momentum. It is denoted by \[\hat{p} = -i\hbar \frac{d}{dx}\]where \( \hbar \) is the reduced Planck's constant and \(i\) is the imaginary unit. This operator is applied to a wave function to extract information about the momentum of the quantum state.
  • The role of the momentum operator is comparable to taking a derivative in calculus, acting to extract the rate of change with respect to position.
  • In quantum terms, it allows for the transformation of abstract wave function information into a physical quantity, the momentum.
Applying the momentum operator to the wave function \[\psi(x) = \vec{A}\left(e^{+a x} + e^{-a x}\right)\]means computing the derivative of the exponential terms and multiplying by \(-i\hbar\). Performing this action will indicate if the wave function is an eigenfunction concerning the momentum operation. If successful, the result is likely to be a scaled version of the original wave function, revealing the particle's specific momentum as the eigenvalue.
Eigenfunction
In quantum mechanics, an eigenfunction is related to an operator, such as the momentum operator, and a corresponding eigenvalue. When an operator acts on this eigenfunction, the result is simply a scaled version of the eigenfunction itself–the scaling factor being the eigenvalue. Eigenfunctions and eigenvalues are fundamental in quantum mechanics because they allow for the quantification of observable properties, such as momentum, without directly measuring them.
  • An eigenfunction provides a precise and stable description of a quantum state regarding a particular observable.
  • The eigenvalue associated with an eigenfunction of the momentum operator gives the specific momentum value.
For a wave function \[\psi(x) = \vec{A}\left(e^{+a x} + e^{-a x}\right)\],to be an eigenfunction of the momentum operator, applying \(-i\hbar \frac{d}{dx}\) should yield a multiple of itself. If the resulting differential operation simplifies back to the original form scaled by a constant, it confirms that the function is indeed an eigenfunction, with the scale factor being the system's momentum eigenvalue.
This concept is key in identifying whether wave functions reveal consistent and singular values for an observable, ensuring predictable outcomes within quantum experiments.

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Most popular questions from this chapter

Exercises \(78-88\) refer to a particle of mass \(m\) described by the wave function $$ \psi(x)=\left\\{\begin{array}{ll} 2 \sqrt{a^{3}} x e^{-a x} & x>0 \\ 0 & x<0 \end{array}\right. $$ Calculate the expectation value of the position of the particle.

In Section \(5.3\), we leamed that to be normaliable, a wave function (1) must not itself diverge and \((2)\) must fall to 0 faster than \(|\mathrm{r}|^{-1 / 2}\) as \(x\) gets lar ge. Nevertheless, we find two functions that slightly violate these requirements very useful. Consider the quantum mechanical plane wave Aelue-aul and the weird function \(\psi_{x_{1}}(x)\) pictured in Figure \(5.19\). which we here call by its preper name, the Direc delta function. (a) Which of the iwo normalizability requirements is violated by the plane wave, and which by the Dirac delta function? (b) Nomalization of the plane Wave could be accomplished if it were simply truncated, restricted to the rejion \(-b

Harmonic Oscillator: The hannonic oscillator can be solved exactly for the quantized energies, and here we compare those results with a numerical approach. Along with the values of \(m\) and \(\hbar\) discussed above, we choose our units so that the spring constant \(x\) is also 1 . The potential energy function \(U(x)\) is then simply \(\frac{1}{2} x^{2}\) For \(\Delta x\), use \(0.001\). (a) Following the above guidelines on choosing \(1 \beta(0)\) and \(\psi(\Delta x)\), test both odd and even functions at different trial values of \(E\) by finding \(\psi\) at all positive multiples of \(\Delta x\) out \(10 x=4\) and plotting the results. Note that because of the functions' symmetries there is no need to plot negative values of \(x\). Find four allowed energies. (b) What tells you that an energy is correct? (c) Compare your results with the exact values given in equation \((5-26)\).

A classical particle confined to the positive \(x\) - cxis experiences a force whose potential energy is $$ U(x)=\frac{1}{x^{2}}-\frac{2}{x}+1 \quad(\mathrm{~S} 1 \text { units }) $$ (a) By finding its minimum value and determining its behaviors at \(x=0\) and \(x=\) toe, sketch this potential energy. (b) Suppose the particle has an energy of 0.5 . Find any turning points. Would the particle be bound? (c) Suppose the particle has an energy of \(2.0 \mathrm{~J}\). Find any turning points. Would the particle be bound?

Exercise 53 outlines a procedure for predicting how the quantum-mechanically allowed energies for a harmotic oscillator should depend on a quantum number. In essence, allowed kinetic energies are the particle-in-abox energtes, except the length \(L\) is replaced by the distance between classical tuming points, expressed in terms of E. Apply this procedure to a potenu'al energy of the fonn \(U(x)=-b / x\), where \(b\) is a constant. Assume that at the origin there is an infinitely high wall, making it one turning point, and determine the other tuming pomt in terms of \(E\). For the average potential energy. use its value at half way between the tuming points. again in terms of \(E\). Find an expression for the allowed energies in terms of \(m, b\), and \(n\). (Although threedimensional. the hydrogen atom potential energy is of this form. and the allowed energy levels depend on a quantum number exactly as this simple model predicts.)

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