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[Solved]: quantum mechanics   1. Find the e
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(Solved): quantum mechanics   1. Find the eigenvalu ...


quantum mechanics

1. Find the eigenvalues of \( p \) and the normalization constant \( C \) in (3.7-11) in the case where the particle is confi???????

\( 3.7 \) The \( x \)-representation, Up until now we have spoken of the states of a physical system being completely describ

The representation in which \( x_{\text {or }} \) is diagonal and the \( |x\rangle \) are the base vectors is called the \( x

\[ \begin{aligned} p|\rangle & =\int p|x\rangle \mathrm{d} x\langle x \mid\rangle=i \hbar \int\left(\frac{\mathrm{d}}{\mathrm

\[ H(r,-\imath \hbar \nabla)\langle\mathrm{r} \mid E\rangle=E\langle\mathrm{r} \mid E\rangle \] which may be written in full ???????

 

1. Find the eigenvalues of \( p \) and the normalization constant \( C \) in (3.7-11) in the case where the particle is confined to move in the region \( |x| each of which can be expressed in terms of a complete set of basis vectors \( |i\rangle \) spanning the space. The particular choice of basis vectors (i.e., of the representation) was not specified. For any given problem there is often one representation which more conveniently expresses the physical behavior of the system under consideration than any other. In this The \( x-r^{2} \) presentation 93 and the following selections we shall examine some of the most frequently used representations. Let us momentarily designate by \( x_{0 p} \) the operator corresponding to the \( x \) coordinate. Then \[ x_{o p}|x\rangle=x|x\rangle, \quad\langle x| x_{0 p}=\langle x| x_{0} \] From physical experience as well as from the proof given at the end of Section 3.6, we know that any value of the coordinate is always possible, at least within certain ranges, Therefore the spectrum of \( x \) is continuous and the orthogonality and normalization condition can be written as \[ \left\langle x^{\prime} \mid x^{\prime \prime}\right\rangle=\delta\left(x^{\prime}-x^{\prime \prime}\right) \text {. } \] For an operator \( f\left(x_{o p}\right) \) which is a function of \( x_{o p} \), we have, according to \( (3.4-10) \), \[ f\left(x_{\mathrm{op}}\right)|x\rangle=f(x)|x\rangle . \] The representation in which \( x_{\text {or }} \) is diagonal and the \( |x\rangle \) are the base vectors is called the \( x \)-representation or the configuration representation. To find the representative of the momentum op- The representation in which \( x_{\text {or }} \) is diagonal and the \( |x\rangle \) are the base vectors is called the \( x \)-representation or the configuration representation. To find the representative of the momentum operator \( p \) in the \( x \)-representation we make use of the commutation relation (3.4-8), \[ x_{o p} p-p x_{o p}=t A . \] Multiplying on opposite sides by \( \left\langle x^{\prime}\right| \) and \( \left|x^{\prime \prime}\right\rangle \), and making use of \( (3.7-1) \) and \( (3.7-2) \), we find that \[ \left\langle x^{1}\left|x_{o p} p\right| x^{\prime \prime}\right\rangle=\left\langle x^{\prime}\left|p x_{o p}\right| x^{\prime \prime}\right\rangle=i \text { h }\left\langle x^{\prime \prime} \mid x^{11}\right\rangle \] or \[ \left(x^{\prime}-x^{\prime \prime}\right)\left\langle x^{\prime}|p| x^{\prime \prime}\right\rangle=\operatorname{t\hbar \delta }\left(x^{\prime}-x^{\prime \prime}\right) . \] A comparison with \( [3.5-5(b)] \) then shows that \[ \left\langle x^{\prime}|p| x^{\prime \prime}\right\rangle=-i \hbar^{2}\left(x^{\prime}-x^{\prime \prime}\right) . \] However, we have seen that \[ \delta^{\prime}\left(x^{\prime}-x^{\prime \prime}\right)=\frac{\mathrm{d}}{\mathrm{d} x^{\prime \prime}}\left\langle x^{\prime} \mid x^{\prime \prime}\right\rangle=-\frac{\mathrm{d}}{\mathrm{d} x^{\prime 1}}\left\langle x^{\prime} \mid x^{\prime \prime}\right\rangle \] so that the previous expression is equivalent to \[ \left\langle x^{\prime}|p| x^{\prime \prime}\right\rangle=-i \hbar \frac{\mathrm{d}}{\mathrm{d} x^{\prime}}\left\langle x^{\prime} \mid x^{\prime \prime}\right\rangle=\left\lfloor A \frac{\mathrm{d}}{\mathrm{d} x^{\prime \prime}}\left\langle x^{\prime} \mid x^{\prime \prime}\right\rangle .\right. \] Now, since \( \left|x^{\prime}\right\rangle \) and \( \left|x^{\prime \prime}\right\rangle \) are arbitrary, 94 General Principles \[ p\left|x^{\prime \prime}\right\rangle=\ell \hbar \frac{\mathrm{d}}{\mathrm{d} x^{\prime 1}}\left|x^{\prime 1}\right\rangle, \quad\left\langle x^{1}\right| p=-t \hbar \frac{\mathrm{d}}{\mathrm{d} x^{1}}\left\langle x^{\prime}\right| . \] If we apply the former equation to an arbitrary ket vector, and integrate by parts, we have \[ \begin{aligned} p|\rangle & =\int p|x\rangle \mathrm{d} x\langle x \mid\rangle=i \hbar \int\left(\frac{\mathrm{d}}{\mathrm{d} x}|x\rangle\right) \mathrm{d} x\langle x \mid\rangle \\ & =-i \hbar \int|x\rangle \mathrm{d} x \frac{\mathrm{d}}{\mathrm{d} x}\langle x \mid\rangle \end{aligned} \] from which it follows that \[ p^{\mathrm{n}}|\rangle=(-\iota \hbar)^{\mathrm{n}} \int|x\rangle \mathrm{d} x\left(\frac{\mathrm{d}}{\mathrm{d} x}\right)^{\mathrm{n}}\langle x \mid\rangle, \] or, for any function \( f(p) \) expandable in a power series, \[ f(p)|\rangle=\int|x\rangle \mathrm{d} x f\left(-\ell \hbar \frac{d}{d x}\right)\langle x \mid\rangle . \] Similar results are, of course, obtained for operation on a bra vector ?l. Suppose now that we have an operator \( A(p, x) \), involving both \( p \) and \( x \), and we wish to find its eigenvalues, i, e., to solve the equation \( A|a\rangle=a|a\rangle \). Using \( (3.7-4) \) we observe that this is equivalent to solving the equation \[ \int\left|x^{\prime}\right\rangle \mathrm{d} x^{\prime} A\left(x^{\prime},-i \hbar \mathrm{d} / \mathrm{d} x^{\prime}\right)\left\langle x^{\prime} \mid a\right\rangle=a|a\rangle . \] Multiplication on the left by \( \langle x| \) and use of (3.7-2) gives us \[ A(x,-i \hbar \mathrm{d} / \mathrm{d} x)\langle x \mid a\rangle=a\langle x \mid a\rangle, \] or in the three-dimensional case, \[ A(\mathbf{r},-i \hbar \nabla)\langle\mathbf{r} \mid a\rangle=a\langle\mathbf{r} \mid a\rangle . \] This important result tells us that we may find the eigenvalues of an operator in the \( x \)-representation by solving a differential equation like \( (3.7-4) \) in which the substitutions \[ \mathrm{r} \rightarrow \mathrm{r}, \mathrm{p} \rightarrow-i \hbar \nabla, \quad|a\rangle \rightarrow\langle\mathrm{r} \mid a\rangle, \] have been made in the eigenvalue operator equation. In particular if \( A(\mathrm{r}, \mathrm{p}) \) is the Hamiltonian operator \( H(\mathrm{r}, \mathrm{p})=\mathrm{p}^{2} / 2 m+V(\mathbf{r}) \), then in the x-representation it becomes the differential operator \[ H(\mathrm{r},-i \hbar \nabla)=\left(-\hbar^{2} / 2 m\right) \nabla^{2}+V(\mathrm{r}) \text {. } \] Consequently, the equation which must be used to find the allowed values \( E \) of the energy is \[ H(r,-\imath \hbar \nabla)\langle\mathrm{r} \mid E\rangle=E\langle\mathrm{r} \mid E\rangle \] which may be written in full as \[ \left[-\frac{A^{3}}{2 m} \nabla^{3}+V(r)\right]\langle r \mid E\rangle=E\langle r \mid E\rangle . \] Identifying \( \langle r \mid E\rangle \) with \( \psi_{E}(r) \) we see that (3.7-10) leads directly to the Schroedinger time-independent equation (1.9-3). That the identification of \( \langle r \mid \vec{E}\rangle \) with \( \hbar_{E}(r) \) is correct can be seen by an application of Eq. (3.5-15). This shows that \[ P(r, E) \mathrm{d}^{3} \mathrm{r}=|\langle\mathrm{r} \mid E\rangle|^{8} \mathrm{~d}^{3} \mathrm{r}=\left|\psi_{E}(\mathrm{r})\right|^{3} \mathrm{~d}^{3} \mathrm{r} \] is just the probability for finding the particle within \( \mathrm{d}^{3} \mathrm{r} \) at \( r \) when the energy is \( \bar{B} \), in accordance with \( (1.9-4) \). Thus we see that the discussions of chapters I and II correspond to using a particular representation ( \( x \)-representation) contained within the more general theory. In the special case of the momentum, the equation \( p_{o p}|p\rangle=p|p\rangle \) becomes simply \[ -\mathcal{L} \hbar \frac{\mathrm{d}}{\mathrm{d} x}\langle x \mid p\rangle=p\langle x \mid p\rangle, \] whose solution is just a plane wave \[ \langle x \mid p\rangle=C \exp t p x / \hbar, \] indicating that the spectrum of eigenvalues of \( P \) is continuous, if the entire space from \( -\infty \) to \( +\infty \) is available to the particle. The constant \( C \) is found from the normalization condition, \[ \begin{aligned} \delta\left(p^{\prime}-p^{\prime \prime}\right) & =\left\langle p^{\prime} \mid p^{\prime \prime}\right\rangle=\int\left\langle p^{\prime} \mid x\right\rangle \mathrm{d} x\left\langle x \mid p^{\prime \prime}\right\rangle \\ & \left.=|C|^{3}\right\rangle \exp i\left(p^{\prime}-p^{\prime \prime}\right) x / \hbar \mathrm{d} x . \end{aligned} \] A comparison with \( (3,5-3) \) shows that \( Q=(2 \pi \wedge)^{-\frac{1}{7}} \) and hence that \[ \langle x \mid p\rangle=(2 \pi \hbar)^{-\frac{1}{2}} \exp \hat{p} x / \hbar ; \] or, in three dimensions \[ \langle\mathrm{r} \mid \mathrm{p}\rangle=(2 \pi)^{-3 / \mathrm{a}} \exp \mathcal{p} \cdot r / \hbar, \] which are the eigenfunctions of the momentum operator in the x-representation.


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When the particle in a potential box `abs(x) V(x)=? xa . So the Schrodinger equation in the region -a ?¯h22md2?dx2+V(x)?=E? ........
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