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I want to show that $$K=K(x,x',t-t')=\sum_{\beta}\exp\left[\frac{-iE_{\beta}}{\hbar}(t-t')\right]$$ satisfies the Schrödinger equation $$ H|\psi\rangle = i\hbar\partial_t|\psi\rangle$$ with respect to $x$ and $t$, where the $\beta$'s are the Eigenstates of the Hamiltonian and satisfy $$\sum |\beta\rangle\langle\beta|=1.$$ So I started calculating and got $$ i\hbar\partial_t K =...= \langle x\vert\exp\left[\frac{-iE_{\beta}}{\hbar}(t-t')\right]\vert x'\rangle. $$ But here I am stuck since I am not that familiar with QM and bra-ket notation/relations. What is the next step or which fact do I have to use to show the claim?

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  • $\begingroup$ You forget the eigenstate $\:|\beta\rangle \:$ and a coefficient $\:c_{\beta} \:$ under the sum : $$ K=K(x,x',t-t')=\sum_{\beta}c_{\beta}\cdot e^{\frac{-iE_{\beta}}{\hbar}(t-t')} \cdot |\beta\rangle $$ The coefficient $\:c_{\beta} \:$ is $\:\langle\beta |(x')\:$ $\endgroup$ – Frobenius Jun 20 '18 at 19:07
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Hint :

In my opinion you must deal with the inverse problem : that of finding the propagator $\:K(\mathbf{x},t\;;\;\mathbf{x}_{0},t_{0})\:$ from the non-relativistic Schr$\ddot{\rm{o}}$dinger equation, so to understand why it is the transition probability amplitude for a particle present at position $\:\mathbf{x}_{0}\:$ on time $\:t_{0}\:$ to propagate at position $\:\mathbf{x}\:$ on time $\:t\:$ and how to use it for the solutions of the equation.

If by your efforts you could answer your question directly without dealing with the inverse problem then you would gain much on mathematics but lose much on physics.


In the following I sketch the main points of the inverse problem :
1. Let the non-relativistic Schr$\ddot{\rm{o}}$dinger equation \begin{equation} i\hbar\;\dfrac{\partial \psi(\mathbf{x},t)}{\partial t}=H(\mathbf{x},t)\;\psi(\mathbf{x},t) \tag{sk-01} \end{equation} where $\:H(\mathbf{x},t)\:$ the Hamiltonian, an hermitian operator.

2. Let $\:H\:$ be dependent only on the position coordinate $\:\mathbf{x}\:$ but not on time, that is $\:H=H(\mathbf{x})$.

3. Now we suppose that $\:H\:$ has (real) eigenvalues $\:E_{\jmath}\:$ with corresponding eigenfunctions $\:u_{\jmath}(\mathbf{x})\:$ \begin{equation} H(\mathbf{x})u_{\jmath}(\mathbf{x})=E_{\jmath}\;u_{\jmath}(\mathbf{x}) \qquad \jmath=1,2,\cdots \tag{sk-02} \end{equation} where the index $\:\jmath\:$ takes discrete or continuous values. For simplicity we suppose that $\:\jmath\:$ takes discrete values and that the set of eigenfunctions $\:\lbrace u_{\jmath} \rbrace, \jmath=1,2,...\:$ is a complete basis of the Hilbert space of states $\:\psi $. So, any state $\:\psi(\mathbf{x},t)\:$ can be expressed as \begin{equation} \psi(\mathbf{x},t)=\sum_{\jmath}a_{\jmath}(t) \; u_{\jmath}(\mathbf{x}) \tag{sk-03} \end{equation}

4. So \begin{equation} \sum_{\jmath}\left[ i\hbar\;\dfrac{da_{\jmath}(t)}{dt} -E_{\jmath}a_{\jmath}(t)\right] u_{\jmath}(\mathbf{x})=\mathbf{0} \tag{sk-04} \end{equation} and from the completeness of the basis $\:\lbrace u_{\jmath}\rbrace\:$ \begin{equation} a_{\jmath}(t)=a_{\jmath}(t_{0})\; \exp \left[ -i \frac{E_{\jmath}}{\hbar}(t-t_{0})\right] \qquad \jmath=1,2,\cdots \tag{sk-05} \end{equation}

5. The coefficients $\:a_{\jmath}(t_{0}) \:$ are determined from the initial state of the system \begin{equation} \psi_{0}(\mathbf{x}) \equiv \psi(\mathbf{x},t_{0})=\sum_{\jmath}a_{\jmath}(t_{0}) \; u_{\jmath}(\mathbf{x}) \tag{sk-06} \end{equation}

6. Under these conditions we have the following general solution of the Schr$\ddot{\rm{o}}$dinger equation (sk-01) \begin{equation} \psi(\mathbf{x},t)=\sum_{\jmath}a_{\jmath}(t_{0})\; \exp \left[ -i \frac{E_{\jmath}}{\hbar}(t-t_{0})\right] \; u_{\jmath}(\mathbf{x}) \tag{sk-07} \end{equation}

7. Now, equation (sk-01) is linear, that is, if to two initial states $\:\psi_{0}^{(1)}(\mathbf{x}) \:$ and $\:\psi_{0}^{(2)}(\mathbf{x}) \:$ at time $\: t_{0}\:$ there correspond the solutions $\:\psi^{(1)}(\mathbf{x},t) \:$ and $\:\psi^{(2)}(\mathbf{x},t) \:$ respectively, then to the initial state $\:\psi_{0}=a_{1}\psi_{0}^{(1)}+a_{2}\psi_{0}^{(2)}\:$, where $\:a_{1},a_{2}\:$ are complex numbers, there corresponds the solution $\:\psi=a_{1}\psi^{(1)}+a_{2}\psi^{(2)}$. Any initial state $\:\psi_{0}(\mathbf{x})\:$ may be considered as the superposition (linear combination) of impulses expressed via the Dirac function $\:\delta^{3}(\mathbf{x})\:$ as \begin{equation} \psi_{0}(\mathbf{x}) = \int \limits_{\mathbf{x}_{0}} \delta^{3}(\mathbf{x}_{0}-\mathbf{x}) \;\psi_{0}(\mathbf{x}_{0}) \; d^{3}\mathbf{x}_{0} \tag{sk-08} \end{equation}

8. If our system is a single point particle and this particle is appeared suddenly at point $\:\mathbf{x}_{0}\:$ on time $\: t_{0}\:$, then the expression $\:\delta^{3}(\mathbf{x}_{0}-\mathbf{x})\:$ under the integral represents the state (wave) function of this particle. If to this initial state there corresponds the solution \begin{align} K(\mathbf{x},t \; \boldsymbol{;} \;\mathbf{x}_{0},t_{0})\equiv \: & \text{the state function at point $\:\mathbf{x}\:$ on time $\:t\:$} \nonumber\\ &\text{corresponding to the appearance of the} \nonumber\\ &\text{particle at point $\:\mathbf{x}_{0}\:$ on time $\:t_{0}\:$} \tag{sk-09} \end{align} then, because of the linearity discussed in the previous paragraph, the solution $\psi(\mathbf{x},t)$ corresponding to the initial state $\psi_{0}(\mathbf{x})$ of equation (sk-08) would be \begin{equation} \psi(\mathbf{x},t) = \int \limits_{\mathbf{x}_{0}}K(\mathbf{x},t \; \boldsymbol{;} \;\mathbf{x}_{0},t_{0}) \; \psi_{0}(\mathbf{x}_{0}) \; d^{3}\mathbf{x}_{0} \tag{sk-10} \end{equation}

9. In mathematical terms $\:K(\mathbf{x},t \; ; \;\mathbf{x}_{0},t_{0})\:$ is called kernel while in physical terms is called propagator since it is the transition probability amplitude for a particle present at position $\:\mathbf{x}_{0}\:$ on time $\:t_{0}\:$ to propagate at position $\:\mathbf{x}\:$ on time $\:t$.

10. We'll determine $\:K(\mathbf{x},t \; ; \;\mathbf{x}_{0},t_{0})\:$ from equation (sk-07) \begin{equation} K(\mathbf{x},t \; \boldsymbol{;} \;\mathbf{x}_{0},t_{0})=\sum_{\jmath}c_{\jmath}\; \exp \left[ -i \frac{E_{\jmath}}{\hbar}(t-t_{0})\right] u_{\jmath}(\mathbf{x}) \tag{sk-11} \end{equation} where $\:c_{\jmath}\:$ are the "coordinates" of the initial state $\:\delta^{3}(\mathbf{x}_{0}-\mathbf{x})\:$ relatively to the complete basis $\:\lbrace u_{\jmath} \rbrace, \jmath=1,2,...\:$ according to equation (sk-06)
\begin{equation} \delta^{3}(\mathbf{x}_{0}-\mathbf{x}) =\sum_{\jmath}c_{\jmath}u_{\jmath}(\mathbf{x}) \tag{sk-12} \end{equation}

11. Taking the inner product of above equation with $\:u_{k}(\mathbf{x})\:$ we have \begin{equation} \left\langle \delta^{3}(\mathbf{x}_{0}-\mathbf{x}),u_{k}(\mathbf{x})\right \rangle =\sum_{\jmath}c_{\jmath}\left\langle u_{\jmath}(\mathbf{x}),u_{k}(\mathbf{x})\right \rangle \tag{sk-13} \end{equation} The complete basis $\:\lbrace u_{\jmath} \rbrace, \jmath=1,2,...\:$ is orthogonal since its members are eigenfunctions of the hermitian operator $\:H$. Without loss of generality we suppose that these eigenfunctions are normalized (unit norm) so that
\begin{equation} \left\langle u_{\jmath}(\mathbf{x}),u_{k}(\mathbf{x})\right \rangle = \int \limits_{\mathbf{x}} u_{\jmath}(\mathbf{x}) \;u_{k}^{\boldsymbol{*}}(\mathbf{x})\;d^{3}\mathbf{x}=\delta_{\jmath k} \tag{sk-14} \end{equation} By equations (sk-14), (sk-13) \begin{equation} c_{k}=\left\langle \delta^{3}(\mathbf{x}_{0}-\mathbf{x}),u_{k}(\mathbf{x})\right \rangle = \int \limits_{\mathbf{x}} \delta^{3}(\mathbf{x}_{0}-\mathbf{x}) \;u_{k}^{\boldsymbol{*}}(\mathbf{x})\;d^{3}\mathbf{x}=u_{k}^{\boldsymbol{*}}(\mathbf{x}_{0}) \tag{sk-15} \end{equation} that is \begin{equation} c_{\jmath}=u_{\jmath}^{\boldsymbol{*}}(\mathbf{x}_{0}) \tag{sk-16} \end{equation}
so equations (sk-12) and (sk-11) yield respectively \begin{equation} \delta^{3}(\mathbf{x}_{0}-\mathbf{x}) =\sum_{\jmath}u_{\jmath}^{\boldsymbol{*}}(\mathbf{x}_{0})\;u_{\jmath}(\mathbf{x}) \tag{sk-17} \end{equation} and \begin{equation} K(\mathbf{x},t \; \boldsymbol{;} \;\mathbf{x}_{0},t_{0})=\sum_{\jmath} u_{\jmath}^{\boldsymbol{*}}(\mathbf{x}_{0})\;\exp \left[ -i \frac{E_{\jmath}}{\hbar}(t-t_{0})\right] u_{\jmath}(\mathbf{x}) \tag{sk-18} \end{equation}

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  • $\begingroup$ Thank you for the hint. I really appreciate it. I will try to solve the inverse of the problem and get back to this post if I have a solution or if there is a problem. Thanks. Nonetheless, I still don't know what to make of the expression $|e^{...}|x'\rangle$ in my initial post. $\endgroup$ – laguna Jun 21 '18 at 11:06
  • $\begingroup$ @laguna : Ok, I am waiting to see your (successful I wish) effort and help you where you would stuck. On the beginning don't try to use the bra-ket notation, leave it for the future after solving the inverse problem. $\endgroup$ – Frobenius Jun 21 '18 at 11:18
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Note that, in bra-ket notation, if the $\{|\beta\rangle\}$ are the eigenstates of $\hat{H}$,

$$ \begin{aligned} \hat{K}(t-t') &= \exp\left[-\tfrac{i}{\hbar}\hat{H}(t-t')\right] \\ &= \sum_\beta |\beta\rangle \exp\left[-\tfrac{i}{\hbar}E_\beta(t-t')\right] \langle \beta|. \end{aligned} $$

Therefore, in the position representation,

$$ \begin{aligned} K(x,x',t-t') &= \langle x| \hat{K}(t-t') |x' \rangle \\ &= \sum_\beta \exp\left[-\tfrac{i}{\hbar}E_\beta(t-t')\right] \beta(x)\beta^*(x'). \end{aligned} $$

Can you see that both forms satisfy the SE?

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