(Solved): quantum mechanics show that ...

quantum mechanics

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\( \left[L^{2}, r\right]=i \hbar r \times L \) 3,4 Compatibility of Observations. We have remarked before that one of the fundamental ideas of quantum mechanics is the incompatibility of making certain observations simultaneously, as summarized in Heisenberg's uncertainty relations (1.5-3), that is \( \Delta x \Delta p_{x} \sim h \). We must seek an expression of this fact in our mathematicalformalism and see how the present scheme of operators is compatible with it. The product of two operators \( A \) and \( B \) is defined as the result of successive action on a vector |\rangle (or \( \langle\mid\rangle \), that is \( (A B)|\rangle=A(B|\rangle) \). It is clear that in general \( (A B)|\rangle \neq(B A)|\rangle \) since the result depends on the order in which the operators are applied. When \( A B \neq B A \) we say that \( A \) and \( B \) do not commute, while if \( A B=B A \), the operators commute. These definitions are conveniently expressedby means of a special operator \( A B-B A=[A, B] \) called the consutator of \( A \) and \( B \). Thus if \( A \) and \( B \) commute then \( [A, B]=0 \). Since commutators will be used rather frequently in the rest of the book, it is convenient to list here some of their most important algebraic properties, all of which may be easily verified by direct expansion: (a) \( [A, B]=-[B, A] \) (b) \( [A+B, C]=[A, C]+[B, C] \) (3. 4-1) (c) \( [A B, C]=A[B, C]+[A, C] B \) We now turn to the proof of a theorem and its converse that are of fundamental importance in the formal development of quantum mechanics. If two operators have the same eigenvectors, then they commute. Let \( A \) and \( B \) be such that \( A|i\rangle=a_{1}|\rangle \) and \( B|i\rangle=b_{1}|i\rangle \); then by expanding any ket |\rangle in terms of the eigenvectors \( |t\rangle \), as |\rangle\( = \) \( =\Sigma_{1}|t\rangle\langle t \mid\rangle \), we have that \[ \begin{aligned} A B|\rangle & =A\left(\sum_{i} B|t\rangle\langle t \mid\rangle\right)=A\left(\sum_{1} b_{1}|t\rangle\langle t \mid\rangle\right) \\ & =\sum_{i}^{b} b_{1} A|t\rangle\langle t \mid\rangle=\sum_{i} b_{1} a_{1}|t\rangle\langle i \mid\rangle \end{aligned} \] \[ \begin{array}{l} B A|\rangle=B\left(\sum_{1} A|i\rangle\langle i \mid\rangle\right)=B\left(\sum_{1} a_{1}|i\rangle\langle t \mid\rangle\right) \\ =\sum_{1} a_{1} B|t\rangle\langle t \mid\rangle=\sum_{1} a_{1} b_{1}|t\rangle\langle i \mid\rangle . \\ \text { Therefore } A B|\rangle=B A|\rangle \text { or }[A, B]=0 . \end{array} \] Conversely, if two operators commute, they possess the same eigenvectors. Let \( |i\rangle \) be an eigenvector of \( A \), so that \( A|i\rangle= \) \( =a_{1}|i\rangle \). Then \( A B|i\rangle=A(B|t\rangle) \) and \( B A|i\rangle=B(A|t\rangle)=a_{1}(B|i\rangle) \). Now if \( A B=B A \), we find that \( A(B|t\rangle)=a_{1}(B|i\rangle) \) so that \( B|i\rangle \) is an eigenvector of \( A \) with eigenvalue \( a_{1} \). This can be true, however, only if \( B|i\rangle \) is proportional to \( |i\rangle \) or \( B|t\rangle=b_{1}|t\rangle \); consequently \( A \) and \( B \) have the same eigenvectors. Although we have implicity assumed no degeneracy in the above arguments, the theorem is valid in all cases. Thus we can state that the condition for the compatibility of a simultaneous measurement of two quantities is that their corresponding operators conmute. Thisfollows directly from the fact that if a system is in a state described by an eigenvector \( |i\rangle \) of both \( A \) and \( B \), then a measurement of these observables is certain to lead to \( d_{1} \) and \( b_{1} \) respectively, regardless of the time or order of measurement. We shall now see that the value of the commutator of two operators is directly related to a quantitative expression of the uncertainty principle. The uncertainty \( \Delta \Omega \) in a dynamical quantity \( A \) is defined by \[ \Delta a=\left[\left\langle(A-\langle A\rangle)^{a}\right\rangle\right]^{\frac{2}{2}}=\left[\left\langle A^{3}\right\rangle-\langle A\rangle^{2}\right]^{\frac{1}{2}} . \] Clearly \( \Delta a \) is the root-mean-square deviation of the expectation of \( A \). Then the following important relation holds: \[ \Delta a \Delta b \geq \frac{2}{2}|\langle[A, B]\rangle| \text {. (3.4-3) } \] To prove this, we first derive the Cauchy-Schwarz inequality. \( { }^{(3)} \) Let \( |u\rangle \) and \( \mid \) v) be any two vectors and \( \lambda \) an arbitrary real number. Then, by the definition of the magnitude of a vector, \[ \{\langle u|+\lambda\langle v|)(|u\rangle+\lambda|v\rangle) \geq 0 \] or \[ \langle u \mid u\rangle+\lambda(\langle v \mid u\rangle+\langle u \mid v\rangle)+\lambda^{2}\langle v \mid v\rangle \geq 0 \text {. } \] This in turn requires that \[ \begin{array}{l} \qquad\langle u \mid u\rangle+\langle v \mid v\rangle \geq \frac{1}{4}|\langle u \mid v\rangle+\langle v \mid u\rangle|^{2} \quad(3.4-4) \\ \text { For }|u\rangle=(A-\langle A\rangle\rangle|\rangle \text { and }|v\rangle=t(B-\langle B\rangle)|\rangle \text {, where } A \text { and } E \text { are } \\ \text { hermitian, one obtains } \end{array} \] \[ \begin{aligned} \langle u \mid u\rangle & =\left\langle\left|(A-\langle A\rangle)^{3}\right|\right\rangle=\langle\Delta a)^{2},\langle v \mid v\rangle=\left\langle\left|(B-\langle B\rangle)^{a}\right|\right\rangle=(\Delta b)^{a} \\ \langle u \mid v\rangle & =t\langle|(A-\langle A\rangle)(B-\langle B\rangle)|\rangle \\ & =t\langle|A B-\langle A\rangle B-\langle B\rangle A+\langle A\rangle\langle B\rangle|\rangle \\ \langle v \mid u\rangle & =-t\langle|(B-\langle B\rangle)(A-\langle A\rangle)|\rangle \\ & =-i\langle|B A-\langle A\rangle B-\langle B\rangle A+\langle B\rangle\langle A\rangle|\rangle . \end{aligned} \] By substituting these values into \( (3,4-4) \), we obtain the inequality (3. 4-3). Before making full use of (3.4-3) we prove yet another result: \[ (A B)+=B+A+\text {. } \] (3.4-5) To show this, we note that if \( A B|m\rangle=|n\rangle \), then \( \langle m|(A B)+=\langle n| \). Hence, if \( B|m\rangle=|s\rangle \) we have that \( A|s\rangle=|n\rangle \). The adjoint of this last equation is \( \langle\Omega| A t=\langle n| \), and since \( \langle g|=\langle m| B t \), we have \( \langle m| B+A \dagger=\langle n| \). By comparison we conclude that \( (3,4-5\rangle \) is cor rect. Using the same reasoning we are led to the more general result \( (A B \ldots M N)+=N+M t \ldots B+A+ \). \( (3.4-6) \) If \( A \) and \( B \) are hermitian, then \( [A, B]+=-[A, B] \), so that \( [A, B] \) is antihermitian, and we can always write \[ [A, B]=+i C \] where \( C \) is hermitian (see Problem 3.2-2). Going back to (3.4-2) let us suppose that \( A \) corresponds to the \( x \) coordinate and \( B \) to its conjugate momentum \( P_{*} \). Then we know that according to the uncertainty principle \( \Delta x \Delta p_{x} \neq 0 \) so that also \( \left[x, p_{x}\right] \neq 0 \). This leads us to state the third principle of quantum mechanics: III. To satisfy the unoertainty prinoiple the commutator of canoniealiy conjugate obeervables must be different from zero. More preaisely we must have \[ \left[q_{1}, p_{1}\right]=i h_{1} \] where \( q_{1} \) is one of the coordinates \( x, y, z \), and \( p_{1} \) the corresponding conjugate momentum. On the other hand \( \left[q_{t}, p_{1}\right]=0 ;\left[q_{s}, q_{1}\right]=0 ;\left[p_{1}, p_{2}\right]=0 \), \( (3.4-9) \) since there is no limit to the precision with which these variables can be measured simultaneously. In this form, Eq. \( \{3.4-8\} \) may be considered as the definition of \( \hbar \). According to \( (3.4-7) \), \( \$ \) should be a hermitian operator, but we further assume that it is simply a multiple of the idemfactor, so that strictly speaking, we should write \( (3,4-7) \) as \( \left[q_{1}, p_{1}\right]=t \hbar T \) where now \( \hbar \) is a scalar quantity. However, since the meaning is clear, we shall omit the factor \( I \). Then by using \( (3.4-2) \) the uncertainty relations become \( \Delta q_{1} \Delta p_{1} z \hbar / 2 \) where \( \Delta q_{1} \) and \( \Delta p_{1} \) are defined according to \( (3.4-2) \). Given a function \( f(\xi) \) that is expandable in a power series, \( f(\xi)=\Sigma_{\mathrm{n}} c_{\mathrm{n}} 5^{\mathrm{n}} \), we define a function \( f(A) \) of an operator \( A \) by the operator \( f(A)=\sum_{n} c_{r r} A^{n} \) so that \[ f(A)|a\rangle=f(a)|a\rangle . \] Obviously \( [f(A), A]=0 \). Now by using the results of Problem 3. 4-2, it follows that , \[ \left[x^{\mathrm{n}}, p_{x}\right]=\operatorname{tAn}_{x}^{n-1},\left[x, p_{x}^{\mathrm{n}}\right]=t_{A m p}{ }^{\mathrm{n}-1}-\quad(3.4-11) \] We note that when \( x \) refers to a scalar quantity, \( n x^{n-1} \) is just the \( x \)-derivative of \( x^{\mathrm{n}} \). Therefore it seems logical for us to adopt (3.4-11) as the definition of the corresponding derivatives of an operator of the form \( x^{\mathrm{n}} \) or \( p_{x}{ }^{\mathrm{n}} \). These results can be employed to extend our notion of the derivative of an operator. By using (3.4-10) and (3.4-11) we see that \[ \left[f(x), p_{\mathrm{x}}\right]=\sum_{\mathrm{n}} b_{\mathrm{n}}\left[x^{\mathrm{n}}, p_{\mathrm{x}}\right]=t \hbar \sum_{\mathrm{a}} n \mathrm{o}_{\mathrm{n}} x^{\mathrm{n}-\mathbf{1}} \] The right-hand side is formally identical to \( \partial f / \partial x \) times \( t \hbar \). Hence, we shall adopt as our definition of the derivative of a function of an operator the relation \[ \begin{array}{l} \partial f / \partial x=(\downarrow \hbar)^{-1}\left[f(x), p_{x}\right],(3.4-12) \\ \text { and, if } f=f\left(p_{x}\right), \\ \partial f / \partial p_{x}=\left(\llcorner\hbar)^{-1}\left[x, f\left(p_{x}\right)\right] .\right. \end{array} \] A very important observable in quantum as well as in classical mechanics is theorbital angular momentum \( L=r \times p \), with components \( L_{z}=x F_{y}-y p_{x} \), etc. By using \( (3.4-8) \) and the properties (3. 4-1) of the commutator, it is easily verified that, if \( L^{3}=L_{\pi}^{3}+ \) \( +L_{y}{ }^{3}++L_{z}{ }^{3} \), then \[ \left[\mathrm{L}^{3}, L_{z}\right]=0,\left[L_{x}, L_{y}\right]=i n L_{2} \] with similar relations for the other components. The relations (3.4-14) can be expressed most concisely in vector form by \( L \times L=\mathcal{L} L,\left[L^{2}, L\right]=0 \). \( (3,4-15) \) Here then, we see that it is impossible to determine simultaneously any two components of the angular momentum, although it is possible to make such a determination of the magnitude of the angular momentum, and one of its components. This is equiv alent to saying that it is impossible to precisely fix the direction of the angular momentum in space. Indeed the uncertainty in two components, say \( x \) and \( y \), of \( L \) is such that \( \Delta L_{x} \Delta L_{y} \geq \frac{1}{2} \hbar_{2}\left|\left\langle L_{z}\right\rangle\right| \). This is a significant point to which we shall return later on. Let us consider the Hamiltonian operator of a particle, that is \[ H=\mathrm{p}^{3} / 2 m+V(\mathrm{r}) \] where \( V(r) \) is the potential energy, \( L_{z} \), and hence also \( L^{2} \), commute with \( \mathrm{p}^{3} \), but they donot commute with \( V(r) \) (Problem 3.4-5). Thus in general, it is not possible to know exactly the energy and either one component or the magnitude of the angular momentum of a particle (unless the latter vanishes). However, if the forces are central so that the potential energy depends only on the magnitude of \( \mathbf{r} \), i.e., \( V(r) \), then we have, according to Problem 3.4-5, that \[ \left[L_{z}, V(r)\right]=0,\left[\mathrm{~L}^{3}, V(r)\right]=0 \text {, } \] and therefore both \( L_{2} \) and \( L^{2} \) commute with the Hamiltonian, \[ \left[L_{\mathbf{z}}, H\right]=0, \quad\left[\mathrm{~L}^{3}, H\right]=0 . \] Therefore in the case of central forces it is possible to know simultaneously the energy, the magnitude of the angular momentum and one of its components. This just reflects the conservation of orbital angular momentum for central forces, as we shall see later (Sect. 4.2).