I am reading up on the Schrödinger equation and I quote:

Because the potential is symmetric under $x\to-x$, we expect that there will be solutions of definite parity.

Could someone kindly explain why this is true? And perhaps also what it means physically?


Good question! First you need to know that parity refers to the behavior of a physical system, or one of the mathematical functions that describe such a system, under reflection. There are two "kinds" of parity:

  • If $f(x) = f(-x)$, we say the function $f$ has even parity
  • If $f(x) = -f(-x)$, we say the function $f$ has odd parity

Of course, for most functions, neither of those conditions are true, and in that case we would say the function $f$ has indefinite parity.

Now, have a look at the time-independent Schrödinger equation in 1D:

$$-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}\psi(x) + V(x)\psi(x) = E\psi(x)$$

and notice what happens when you reflect $x\to -x$:

$$-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}\psi(-x) + V(-x)\psi(-x) = E\psi(-x)$$

If you have a symmetric (even) potential, $V(x) = V(-x)$, this is exactly the same as the original equation except that we've transformed $\psi(x) \to \psi(-x)$. Since the two functions $\psi(x)$ and $\psi(-x)$ satisfy the same equation, you should get the same solutions for them, except for an overall multiplicative constant; in other words,

$$\psi(x) = a\psi(-x)$$

Normalizing $\psi$ requires that $|a| = 1$, which leaves two possibilities: $a = +1$ (even parity) and $a = -1$ (odd parity).

As for what this means physically, it tells you that whenever you have a symmetric potential, you should be able to find a basis of eigenstates which have definite even or odd parity (though I haven't proved that here,* only made it seem reasonable). In practice, you get linear combinations of eigenstates with different parities, so the actual state may not actually be symmetric (or antisymmetric) around the origin, but it does at least tell you that if your potential is symmetric, you could construct a symmetric (or antisymmetric) state. That's not guaranteed otherwise. You'd probably have to get input from someone else as to what exactly definite-parity states are used for, though, since that's out of my area of expertise (unless you care about parity of elementary particles, which is rather weirder).

*There is a parity operator $P$ that reverses the orientation of the space: $Pf(x) = f(-x)$. Functions of definite parity are eigenfunctions of this operator. I believe you can demonstrate the existence of a definite-parity eigenbasis by showing that $[H,P] = 0$.

  • $\begingroup$ David, thank you so much! Your explanation is superb! :) $\endgroup$ – bra-ket Aug 26 '11 at 7:16
  • $\begingroup$ States with a definite parity are eigenstates possessing certain energy $E_n$. A perturbed system becomes a superposition with no certain energy and with no certain parity $\Psi(t)=\sum_n C_n \psi_n e^\frac{-iE_n t}{\hbar}$, i.e., it is time-dependent. A tme-independent superposition (without those oscillating exponentials) arise only if you want to represent some time-independent function $\Phi$ as a spectral sum $\sum_n A_n \psi_n$. $\endgroup$ – Vladimir Kalitvianski Aug 26 '11 at 7:49
  • $\begingroup$ Ooh, btw why can't $a\in\mathbb{C}$ and $|a|=1$? $\endgroup$ – bra-ket Aug 26 '11 at 7:52
  • 2
    $\begingroup$ Doing reflection twice, you get $a^2 =1$ so it is a real number. $\endgroup$ – Vladimir Kalitvianski Aug 26 '11 at 7:54
  • $\begingroup$ @VladimirKalitvianski Since the two equations are second order ordinary differential equations why must $\psi(x) = a\psi(-x)$. In principle, couldn't $\psi(x)$ and $\psi(-x)$ be two different particular solutions out of which you could build a general solution to the homogeneous differential equation? $\endgroup$ – IanDsouza May 7 '16 at 10:12

Sorry I found David Z' answer a bit confused just when discussing the crucial point.

Since the two functions ψ(x) and ψ(−x) satisfy the same equation, you should get the same solutions for them, except for an overall multiplicative constant; in other words,


Normalizing ψ requires that |a|=1, which leaves two possibilities: a=+1 (even >parity) and a=−1 (odd parity).

The first part "Since the two functions... multiplicative constant" is generally false without an important further requirement that is not garanted here. It is indeed true under the hypothesis that the eigenspace of the Hamiltonian operator with eigenvalue $E$ we are considering is one-dimensional. However this is not the case in general. Finally the remaining part of the statement above "Normalizing ... parity)." is incorrect anyway as it stands: normalization just requires $|a|=1$.

Let me propose an alternative answer.

First of all, one introduces the parity transformation, $P: {\cal H} \to {\cal H}$, where ${\cal H} = L^2(R)$, defined as follows without referring to any Hamiltonian operator:

$$(P\psi)(x):= \eta_\psi \psi(-x)\:.$$

Above $\eta_\psi$ is a complex number with $|\eta_\psi|=1$. It is necessary to leave this possibility because, as is well known in QM, states are wavefunctions up to a phase so that $\phi$ and $e^{i\alpha} \phi$ are indistinguishable as states and, physically, we can only handle states. As the map $P$ is (1) bijective and (2) it preserves the probabilities of transition between states, it is a so-called quantum symmetry. A celebrated theorem by Wigner guarantees that every quantum symmetry can be represented by either an unitary or an antiunitary operator (depending on the nature of the symmetry itself). In the present case all that means that it must be possible to fix the map $\psi \mapsto \eta_\psi$ in order that $P$ becomes linear (or anti-linear) and unitary (or anti-unitary). As a matter of fact $P$ becomes unitary if $\eta$ is assumed to be independent form $\psi$. So we end up with the unitary parity operator:

$$(P\psi)(x):= \eta \psi(-x) \quad \psi \in L^2(R)$$

where $\eta \in C$ with $|\eta|=1$ is any fixed number. We can make more precise our choice of $\eta$ requiring that $P$ is also an observable, that is $P=P^\dagger$. It is immediate to verify that it happens only for $\eta = \pm 1$. It is matter of convenience to fix the sign. We henceforth assume $\eta=1$ (nothing follows would change with the other choice). We have our parity observable/symmetry given by:

$$(P\psi)(x):= \psi(-x) \quad \psi \in L^2(R)$$

What is the spectrum of $P$? As $P$ is unitary, the elements $\lambda$ of the spectrum must verify $|\lambda|=1$. As $P$ is self-adjoint the spectrum has to belong in the real line. We conclude that the spectrum of $P$ contains $\{-1,1\}$ at most. Since these are discrete points they must be proper eigenvalues with associated proper eigenvectors (I mean :Things like Dirac's delta are excluded).

It is impossible that the spectrum contains $1$ only or $-1$ only, otherwise we would have $P=I$ or $P=-I$ respectively, that is evidently false. We have found that $P$ has exactly two eigenvalues $-1$ and $1$.

At this point we can define a state, represented by $\psi$, to have even parity if $P\psi = \psi$ or odd parity if $P\psi = -\psi$.

Let us come to the problem with our Hamiltonian. If $V(x) = V(-x)$, by direct inspection one immediately sees that:

$$[P, H] =0\:.$$

Assuming that the spectrum of $H$ is a pure point spectrum (otherwise we can restrict ourselves to deal with the Hilbert space associated with the point spectrum of $H$ disregarding that associated with the continuous one), a known theorem assures that there is a Hilbert basis of eigenvectors of $H$ and $P$ simultaneously.

If $\psi_E$ is such a common eigenvector (associated with the eigenvalue $E$ of $H$), it must verify either $P\psi_E= \psi_E$ or $P\psi_E= -\psi_E$, namely:

$\qquad\qquad \qquad \psi_E(-x) = \psi_E(x)$ or, respectively, $\psi_E(-x)= \psi_E(x)$.

To conclude, I stress that it is generally false that an eigenvector of $H$ has defined parity. If the eigenspace of the given eigenvalue has dimension $\geq 2$, it is easy to construct counterexamples. It is necessarily true, however, if the considered eigenspace of $H$ has dimension $1$.

  • $\begingroup$ Why it is not generally true for higher dimensions? Which part of your argument requires 1-dimension restriction? $\endgroup$ – Shadumu Apr 5 '15 at 0:40
  • $\begingroup$ Suppose that the eigenspace ${\cal H}_E$ with eigenvalue $E$ of $H$ is two-dimensional (this argument more gemerally works for dimension $\geq 2$). Since $H$ commutes with $P$ there must be a Hilbertian basis of eigenvectors of $P$ of ${\cal H}_E$, say $\psi_+, \psi_-$ with $P\psi_\pm = \pm \psi_\pm$. With this situation, $\psi := \frac{1}{\sqrt{2}}\psi_+ + \frac{1}{\sqrt{2}} \psi_-$ satisfies $H\psi=E\psi$ but it is not an eigevector of $P$: It has not defined parity. Both $\psi(-x) = \psi(x)$ and $\psi(-x)= -\psi(x)$ are false. $\endgroup$ – Valter Moretti Apr 5 '15 at 11:24
  • $\begingroup$ But the question is to show 'we expect that there will be solutions of definite parity.' We can certainly construct definite-parity solutions in the example you give, it is just that not every solution satisfies. $\endgroup$ – Shadumu Apr 5 '15 at 16:01
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    $\begingroup$ Indeed, you are right. My point, at the end of my answer was different: that not all eigenvectors of the Hamiltonian operator are eigenvectors of the parity operator. $\endgroup$ – Valter Moretti Apr 5 '15 at 17:29

I think this should be an alternative answer from first principles.

Assume the Hamiltonian is one dimensional, and $V(x)$ is an even function. Given a general solution $\psi (x)$, we also have that $\psi (-x)$ is a solution. If the eigenspace of the Hamiltonian is one dimensional, then we must have $$ \psi (x) = \alpha \psi ( -x) $$ where $|\alpha | = 1$. Multiply both sides by $\overline{\psi (x) }$ and integrate over $\mathbb{R}$: $$ \int _{- \infty} ^{\infty} \mathrm{d} x \ \overline{\psi (x) }\psi (x) = \alpha \int _{- \infty} ^{\infty} \mathrm{d} x \ \overline{\psi (x) }\psi (-x) $$ It is straightforward to show by changing variables $ x=-z$ that the integral on the right hand side is equal to its complex conjugate, and hence it's real. Because the left hand side is equal to $1$, then we must have that $ \alpha = \pm 1$.

If $ \psi (x) $ and $ \psi (-x) $ are linearly independent, then I don't know how to deal with the situation.

Please let me know if you have found a flaw in my argument. Cheers.

  • $\begingroup$ In 1D, there are no degenerate eigenstates. So there can't be a linearly independent $\psi(x)$ and $\psi(-x)$. Since they have the same energy, they must be the same state! $\endgroup$ – Jahan Claes Sep 6 '18 at 19:45

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