# What is the quantum mechanical turning point of the $n^{th}$ energy eigenstate of an oscillator? [closed]

I am looking for an analytical expression for the most likely position for a quantum harmonic oscillator (which I refer to as the quantum mechanical "turning points"), in terms of $$n$$. For the quantum mechanical simple harmonic oscillator, these points lie closer to the center than the corresponding classical turning points. I have been unable to find an analytical expression for these points in terms of $$n$$ in the literature which generally refer to the classical amplitude for the same energy.

• Do you mean the classical turning point? – jacob1729 Aug 29 '19 at 22:56
• @jacob1729 No, I was referring to the positions where the probability amplitude is the greatest. This occurs at positions which is closer to the mean central value of the position than the corresponding classical turning points. – Saurabh Uday Shringarpure Aug 30 '19 at 18:49

If by "most likely position" you mean the position(s) where the probability density is greatest, then there is no analytical expression for arbitrary $$n$$ because the positions are determined by roots of a polynomial of degree $$n/2$$ (for even $$n$$) or $$(n+1)/2$$ (for odd $$n$$).

To see this, take units where $$m\omega/\hbar=1$$. Then the $$n$$-th wavefunction is

$$\psi_n(x)=N_ne^{-x^2/2}H_n(x),$$

where $$N_n$$ is a normalization constant and $$H_n(x)$$ is the Hermite polynomial of order $$n$$.

The probability density,

$$P_n(x)=|\psi_n(x)|^2=N_n^2e^{-x^2}H_n(x)^2,$$

has maxima and minima when

$$0=\frac{dP_n(x)}{dx}=-2N_n^2e^{-x^2}H_n(x)\left(xH_n(x)-\frac{dH_n(x)}{dx}\right).$$

Using the identity

$$\frac{dH_n(x)}{dx}=2nH_{n-1}(x),$$

the polynomial equation for the maxima/minima becomes

$$H_n(x)[xH_n(x)-2nH_{n-1}(x)]=0.$$

The roots of the first factor, $$H_n(x)$$, are the minima where the wave function and the probability density are zero. The roots of the second factor,

$$xH_n(x)-2nH_{n-1}(x),$$

are the maxima; from looking at graphs of the wave functions, one sees that the most positive and most negative roots give the position of the greatest probability density.

The first few such polynomials for the maxima are

$$x$$

$$2x^2-2$$

$$4x^3-10x$$

$$8x^4-36x^2+12$$

$$16x^5-112x^3+108x$$

Note that for the odd-order polynomials we can factor out $$x$$. Then each remaining polynomial is a polynomial of degree $$n/2$$ or $$(n+1)/2$$ in $$x^2$$.

The roots of polynomials up to degree $$4$$ can always be expressed in radicals, but for degree greater than $$4$$, the roots do not have analytical expressions in general. (Sometimes they do, but not in general.) So for $$n$$ from $$0$$ to $$8$$, the roots can be found as radicals. But beginning at $$n=9$$, the fifth-degree polynomial in $$x^2$$ does not (at least, according to Mathematica) have roots expressible as radicals.

As for the “quantum-mechanical turning point”, I have no idea what that means. The turning point is a classical concept, and quantum-mechanically the particle can be found beyond the turning point.

• The classical turning points are also the highest points in the classical PDF (the distribution you get by sampling position randomly in time), so the interpretation you have chosen here is reasonable. Elsewhere on the site I posted some plots comparing classical and quantum PDFs for selected values of $n$. You can see that in the limit of high $n$ the trend is approaching the classical turning points. – dmckee --- ex-moderator kitten Aug 30 '19 at 14:24
• Hmmmmm. That linear combination is very close to the standard recurrence relations, but it does look like there's no dice reducing them to something simpler, at least using the top layer of standard tools. – Emilio Pisanty Aug 30 '19 at 19:59