The following question is from David Griffiths' Introduction to Quantum Mechanics:

Problem 2.13 A particle in the harmonic oscillator potential starts out in the state $$\Psi(x,0) = A[3 \psi_0(x) + 4 \psi_1(x)]\,.$$ (a) Find $A$.

(b) Construct $\Psi(x,t)$ and $|\Psi(x,t)|^2$.

Part (b) asks to construct the general solution from the instance at time $t = 0$.

The strategy given in the book is to add the time dependence to each stationary state, forming a general solution:

To fit $\Psi(x,0)$ you write down the general linear combination of these solutions: $$ \Psi(x,0)=\sum_{n=1}^\infty c_n\psi_n(x):\tag{2.16} $$ the miracle is that you can *always match the specified initial state by appropriate choice of the constants $c_1,c_2,c_3,\ldots$. To construct $\Psi(x,t)$ you simply tack onto each term its characteristic time dependence, $e^{-iE_nt/\hbar}$: $$ \Psi(x,t)=\sum_{n=1}^\infty c_n\psi_n(x) e^{-iE_nt/\hbar}=\sum_{n=1}^\infty c_n\Psi_n(x,t).\tag{2.17} $$

I'm not sure, however, about how we can guarantee that two solutions $ {\psi_1}$ and $ {\psi_2}$ to the time-dependent equation don't have $ {\psi_1(x,0)} = {\psi_2(x,0)}$. If we can't guarantee this, then how do we know that the solution found by Griffiths' method is unique?


3 Answers 3


But how can we guarantee that two solutions $\boldsymbol {\psi_1}$ and $\boldsymbol {\psi_2}$ to the time-dependent equation don't have $\boldsymbol {\psi_1(x,0)} = \boldsymbol {\psi_2(x,0)}$. If we can't guarantee this, then how do we know that the solution found by Griffith's method is unique?

I interpret that your question basically asks how do we know that the time-dependent Schrödinger equation with an initial condition has a unique solution. To show this, we use a common method for showing the uniqueness of a solution to a linear differential equation: we show that if two answers are the same at one point, then their difference is constant $0$.

The time-dependent Schrödinger equation is linear; that means that if $\Psi_1$ and $\Psi_2$ are solutions to the time-dependent Schrödinger equation, then so is $\alpha_1 \Psi_1 + \alpha_2 \Psi_2$ for any $\alpha_1, \alpha_2$. In particular, $\Psi=\Psi_1-\Psi_2$ is a solution. So we have $$ -i\hbar \partial_t \Psi = \hat{H} \Psi,\quad\Psi(0)=0 $$ Now we get $$ \begin{align*} \partial_t \langle\Psi, \Psi\rangle &= \langle\partial_t \Psi, \Psi\rangle + \langle\Psi, \partial_t \Psi\rangle \\ &= \langle \frac{i}{\hbar} \hat{H} \Psi, \Psi \rangle + \langle \Psi, \frac{i}{\hbar} \hat{H} \Psi\rangle \\ = & -\frac{i}{\hbar}\langle\hat{H}\Psi,\Psi\rangle + \frac{i}{\hbar} \langle \Psi, \hat{H} \Psi\rangle \\ &= -\frac{i}{\hbar}\langle\Psi,\hat{H}\Psi\rangle + \frac{i}{\hbar} \langle \Psi, \hat{H} \Psi\rangle \\ &= 0, \end{align*} $$ where the next-to-last equation is due to the fact that the Hamiltonian is Hermitian. So now we have a differential equation for the real function $f$ defined by $f(t) \equiv \langle\Psi(t),\Psi(t)\rangle$, $$ \partial_t f = 0 \, . $$ Zero derivative means that $f(t)$ is a constant. Therefore, $$f(t)=f(0)=\left<\Psi(0),\Psi(0)\right>=0 \, .$$ Therefore, $\Psi=0$, so $\Psi_1=\Psi_2$.


The form of the solution shown by Griffiths is not unique. That means that there exist cases where a basis $\{\psi_n(x)\}$ will reproduce $\Psi$ as $$ \Psi(x,t)=\sum_{n=1}^\infty c_n\psi_n(x) e^{-iE_nt/\hbar}, $$ but there exists a second, different basis $\{\varphi_n(x)\}$ which (with different coefficients) also reconstructs $\Psi$: $$ \Psi(x,t)=\sum_{n=1}^\infty c_n'\varphi_n(x) e^{-iE_n't/\hbar}. $$ Some notes:

  • In all such cases, the solution itself is still unique. It's the representation that changes, in a way that is exactly analogous to how a vector's coordinates will change in a different frame of reference, but the vector stays the same. The unicity of the solution is not due to way we solved the equation - it's a property of the Schrödinger equation itself.

  • If $\psi_n$ and $\varphi_m$ are eigenfunctions of $H$, and they have different eigenvalues, then they must be orthogonal and therefore cannot be equal. This follows from the fact that $$E_n⟨\varphi_m,\psi_n⟩=⟨\varphi_m,H\psi_n⟩=⟨\varphi_mH,\psi_n⟩=E_m⟨\varphi_m,\psi_n⟩$$ so either $E_n=E_m$ or $⟨\varphi_m,H\psi_n⟩=0$.

  • The non-unicity of representation is therefore combined to subspaces where $H$ is degenerate: that is, when it has two of more linearly independent eigenfunctions with the same eigenvalue. In this case, any basis for this subspace will work, but all the bases are equivalent. You choose the one that's most convenient for your purposes but you keep in mind that you have made an arbitrary choice of basis.

    The easiest example is the hydrogen atom, where eigenfunctions are degenerate if they share the $n$ and $l$ quantum numbers.* You then need to choose a basis for this subspace, and to do this you need to break the isotropy - you need to choose a particular direction for your basis. This is usually taken as the $z$ axis (wherever that is pointing towards), in the understanding that any other direction is a short basis change away.

    * There's an additional degeneracy in $l$ but it's irrelevant for these purposes.

  • 1
    $\begingroup$ "In short, if $i\hbar\partial_t \Psi=H\Psi$ and $i\hbar\partial_t \Phi=H\Phi$, then $i\hbar\partial_t(\Psi-\Phi)=0$" This skips some steps. What you get directly is that $i\hbar \partial_t(\Psi-\Phi) = H (\Psi-\Phi)$, which doesn't directly imply that $\Psi-\Phi$ is constant in time. In case of the time-dependent Schrödinger equation does, but there are many differential equations where $\partial_t f = F[f]$, $f(0)=0$ doesn't imply $f=0$ constant even if $F[0]=0$. $\endgroup$
    – JiK
    Jul 3, 2015 at 20:45
  • $\begingroup$ Hah, good catch. That's me being sloppy right there. $\endgroup$ Jul 3, 2015 at 21:34

So I see your whole question as this:

How can we guarantee that two solutions $\boldsymbol {\psi_1}$ and $\boldsymbol {\psi_2}$ to the time-dependent equation don't have $\boldsymbol {\psi_1(x,0)} = \boldsymbol {\psi_2(x,0)}$. If we can't guarantee this, then how do we know that the solution found by Griffith's method is unique?

This is a really basic property of vectors and matrices; if you have two $n\times 1$ "column vectors" $u$, $v$ in $\mathbb C^n$ which are eigenvectors of an $n\times n$ Hermitian matrix $H = H^\dagger$ then either they are orthogonal in the sense that $u^\dagger v = v^\dagger u = 0$ or else they have the same eigenvalue.

Why? Well it's simple interplay of the pieces involved. Hermitian matrices have to have real eigenvalues because $u^\dagger ~ H ~ u = \lambda_u ~ u^\dagger ~ u$ needs to be its own complex conjugate -- because $(A B)^\dagger = B^\dagger A^\dagger$ and the fact that the $1\times 1$ adjoint operation is a complex conjugation you find $(u^\dagger ~ H ~ u)^* = \lambda_u^* ~ u^\dagger ~ u$ is, by the very definition of Hermitian ($H = H^\dagger$), also $ u^\dagger ~H^\dagger~u = u^\dagger ~ H ~ u = \lambda_u u$, hence we either trivially have $u^\dagger ~ u = 0$ and $u = 0$, or else $\lambda_u = \lambda_u^*$ and hence $\lambda_u$ is a real number.

Now look at $u^\dagger \cdot H \cdot v = \lambda_v u^\dagger v$ but its complex conjugate must be $v^\dagger \cdot H \cdot u = \lambda_u v^\dagger u$; taking a second complex conjugate we find that $\lambda_v u^\dagger v = \lambda_u u^\dagger v$. We only have two options here: either $\lambda_u = \lambda v$ or else $u^\dagger v = 0$.

Essentially the problem is: matrices always have the same "left" and "right" eigenvalues, since the expression $\det (A - \lambda I) = 0$ that you use to find them uses a determinant, and $\det A = \det A^T$ because determinants don't care whether you iterate on minors by-row or by-column. When you say that a matrix is its own transpose, as Hermitian ones basically are, then you're saying that also the left- and right-eigenvectors are, respectively, transposes of each other. And this fact forces those eigenspaces to become orthogonal to each other.

Now I claim that the above points do not depend on the discrete indices of the row vector or column vector, but only on the relationship of the objects, and hence when we go from a discrete-indexed vector to a continuous-indexed wavefunction, we get the same conclusion. Let's follow it through. Let $\mathcal A$ map wavefunctions to wavefunctions, so that we have$$\langle \phi | \hat A | \psi \rangle = \int_{\mathbb R^n} d^n r ~ \phi^*(\vec r) ~ \mathcal A[\psi](\vec r).$$ Furthermore let's define the shorthand notation that $$ \langle \phi | \psi \rangle = \langle \phi | 1 | \psi \rangle = \int_{\mathbb R^n} d^n r ~ \phi^*(\vec r) ~ \psi(\vec r).$$We assert that $\mathcal A$ is Hermitian, which means that: $$ \langle \phi | \hat A | \psi \rangle = (\langle \psi | \hat A | \phi \rangle)^* = \left[ \int d^n r ~ \psi^*(\vec r) ~ \mathcal A[\phi](\vec r) \right]^*.$$ It follows that any eigenvalue $\hat A | a \rangle = a | a \rangle$ is real, because we find similarly to the above that $$(\langle a | \hat A | a \rangle)^* = a^* \langle a | a \rangle = \langle a | \hat A | a \rangle = a \langle a | a \rangle, ~~\text{ so }a = a^*.$$ But similarly we have that if $\hat A | a \rangle = a | a \rangle$ and $\hat A | b \rangle = b | b \rangle$ then $$\langle a | \hat A | b \rangle = b \langle a | b \rangle = (\langle a | \hat A | b \rangle)^{**} = (\langle b | \hat A | a \rangle)^* = (a \langle b | a \rangle)^* = a^* \langle a | b \rangle.$$This again forces us to conclude that either $b = a$ or else $\langle a | b \rangle = 0$.

Now consider the very special 1D case where $$\mathcal A[\psi](x) = i~\psi'(x)$$. This imaginary-single-derivative operator is Hermitian, I claim. To see why, let's look at the definition; it's Hermitian if and only if $$\int_{-\infty}^\infty dx ~ \phi^*(x) ~ i~\psi'(x) = \left[ \int_{-\infty}^\infty dx ~ \psi^*(x) ~ i ~ \phi'(x) \right]^*.$$ Is this true? Yes, we can integrate the left hand side by parts, raising $\psi'$ while we lower $\phi^*$, to find: $$\int_{-\infty}^\infty dx ~ \phi^*(x) ~ i~\psi'(x) = \left[i ~ \phi^*(x) ~\psi(x)\right]_{-\infty}^\infty - \int_{-\infty}^\infty dx ~ [\phi^*]'(x) ~ i~\psi(x), $$ the first term of which can be discarded as 0 since both wavefunctions tend to 0 at infinity. This leaves only $$ \int_{-\infty}^\infty dx ~ [\phi'(x)]^* ~ (-i) ~\psi(x) = \int_{-\infty}^\infty dx ~ [\phi'(x) ~ i ~\psi^*(x)]^* = \left[\int_{-\infty}^\infty dx ~ \psi^*(x)~ i ~ \phi'(x)\right],$$which is what we set out to prove.

It follows that $- \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}$, the composition of two such operators, which has been multiplied by a constant, is also Hermitian, as is any constant real function -- thus so is $- \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x)$.

Since it's Hermitian, when we solve $E \psi(x) = - \frac{\hbar^2}{2m} \frac{\partial^2\psi}{\partial x^2} + V(x)~\psi(x)$ we find that any two distinct solutions $\phi, \psi$ either have the same energy or else $\int dx ~ \phi^*(x) ~ \psi(x) = 0$, precluding them from ever being the same function.

Furthermore, if we ever find the complete spectrum $\{\phi_i\}$, this gives us an easy way to find the components:$$\psi(x) = \sum_i c_i \phi_i(x) \text{ where } c_i = \int dx~\phi_i^*(x) \psi(x).$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.