On factorisation of Schrodinger operators

Lecture 6

<latexit sha1_base64="lugo5Y0kNnTrTemCbJadF2hVKmg=">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</latexit>

Consider a 1D Schrodinger operator

H = �d2

dx2� V (x), inL2(R),

where V ! 0 as |x| ! 1 and V � 0.

Let {��j} be negative eigenvalues of H.

<latexit sha1_base64="q9Yc3bxYFrsvqRsM6UjAe3Ko+N0=">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</latexit>

Let (��1, 1) be the lowest eigenvalue and its respective eigenfunction. It isknown that 1 6= 0 and we can choose 1 > 0.

Denote

f1 = 01

1, f 0

1 = 001

1�

✓ 01

1

◆2

.

Therefore

f 01 + f2

1 = 001

1= �1 � V.

Let us introduce

Q1 =d

dx� f1 & Q⇤

1 = � d

dx� f1.

<latexit sha1_base64="/EhuPN2qeFykpyDmy+6jva6IHpE=">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</latexit>

Then

Q⇤1Q1 =

✓�

d

dx� f1

◆✓d

dx� f1

◆= �

d2

dx2+ f 0

1 + f21

= �d2

dx2� V + �1 = H+ �1.

The discrete spectrum �d(Q⇤1Q1) of the operator Q⇤

1Q1 coincides with

�d(Q⇤1Q1) = {0,��2 + �1, ��3 + �1, . . . }.

In particular,Q⇤

1Q1 1 = 0,

where

1(x) ⇠

(e�

p�1x, x ! +1,

ep�1x, x ! �1.

and also

f1(x) = 01(x)

1(x)⇠

(�p�1, x ! +1,p�1, x ! �1.

<latexit sha1_base64="tQ6Vv/eDx355b5ADB83XrIi0qlc=">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</latexit>

Commuting Q⇤1 and Q1 we obtain

Q1Q⇤1 =

✓d

dx� f1

◆✓�

d

dx� f1

◆= �

d2

dx2� f 0

1 + f21

= �d2

dx2� 2f 0

1 � V + �1 = H� 2f 01 + �1.

The operators Q⇤1Q1 and Q1Q⇤

1 have the same non-zero spectrum.

Moreover, 0 62 �(Q1Q⇤1), indeed, assume that there is 2 L2(R) s.t.

Q1Q⇤1 = 0 =) kQ⇤

1 k = 0 =)

� 0� f1 = 0 =) (f1 ⇠ �

p�1, x ! +1) =)

⇠ ep�1 x, x ! +1.

Therefore 62 L2(R).

<latexit sha1_base64="E4wA1k2zj2WDpvdbGJiTGt9UMGQ=">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</latexit>

Conclusion:�d(H) = {��1,��2,��3, . . . }

and�d(H� 2f 0

1) = {��2,��3, . . . }.

Denote now V1 = V + 2f 01, H1 = H� 2f 0

1.

Considering the class of potentials with the finite number of eigenvalues andrepeating this process, we obtain a non-negative Schrodinger operator with thepotential

�Vn = �V � 2f 01 � 2f 0

2 � · · ·� 2f 0n,

wheref 0n + f2

n = �n � Vn�1

and�d(H� 2f 0

1 � 2f 02 � · · ·� 2f 0

n) = ;.

<latexit sha1_base64="d3hNEuP/ZqNH0CErv5be23Ih2ys=">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</latexit>

Finally we have

0 Z

R(Vn�1 + 2f 0

n)2 dx =

Z

R

⇥V 2n�1 + 4f 0

n(Vn�1 + f 0n)⇤dx

=

Z

R

⇥V 2n�1 + 4f 0

n(�n � f2n)⇤dx =

Z

RV 2n�1 dx+ 4�nfn

���1

�1� 4

3f3n

���1

�1

=

Z

RV 2n�1 dx� 8�3/2

n +8

3�3/2n =

Z

RV 2n�1 dx� 16

3�3/2n

= · · · =Z

RV 2 dx� 16

3

nX

j=1

�3/2j .

<latexit sha1_base64="LZ4YSbVPkE+d/NynfGtGRhkyxOM=">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</latexit>

Theorem. (Benguria and Loss)

Let H = �d2

dx2 � V , in L2(R), where V 2 L2(R), V � 0. Then for the negativeeigenvalues {�j} of the operator H we have

X

j

�3/2j

3

16

Z

RV 2 dx.

Note that the constant 3/16 is semiclassical, namely

1

2⇡

Z

R

Z

R(⇠2 � V (x))3/2� dx d⇠ =

1

2⇡

Z

R(1� ⇠2)3/2+ d⇠

Z

RV 2 dx

=1

2⇡

Z 1

0(1� t)3/2t�1/2

Z

RV 2 dx =

1

2⇡B

✓1

2,5

2

◆ Z

RV 2 dx

=1

2⇡

�(1/2)�(5/2)

�(3)

Z

RV 2 dx =

3

16

Z

RV 2 dx.

<latexit sha1_base64="ZFvp7B/xG9tNYP8YDHKMLJZj70U=">AAAGnHicjVTbbttGEKVjV02ZXuLmsRcMahoQW0khpTqXhwCBU6AB6hZpESsBvJawJIfkxuSS4C4dGwRf84/5iH5Cgc5SimPJblI+UIOzZw7PXLRBmQmlPe/txo3NrU96n978zL71+RdffnV7++upKuoqxMOwyIrqZcAVZkLioRY6w5dlhTwPMnwRnDwx5y9OsVKikM/1eYnHOU+kiEXINUHz7c03TBZCRii13bAghucpFhXmo9aG/j7KpK4EBy4jOCiUcm2bBSJRJ6K03+cdoAaH5VynIc/gKTyCIYsrHjbRbNw20Rm9YQhTZwBCgnMwG/fZfhDAXy4hr1OsEJwpo6PVE8IS9JyRcSQhLirQKYLEhJyfItgoEpSnPKtR0dcbllHREZ+/Yq0DRdyRixIrrinzkjsHXiOk/BRtx7GZqvP5K7jInTWTu2SWZQiLCiZt498jYABkUM+bhb0WprMxYdHZyKhc15M/Co3kgevOSFhIpbmkNk3u+vccEAoU5iLMuFI0imwAkueYndu7pIWJkE1eZ9rM9MfWXjjx22bMStGuW1n1BX12JmZj0+7+mevOh4uKOq9sENEhTecjiiTiD6HTcec/vRcwyf/ZCGDMfgSrwhdsb+YbTe0u+6tnzdA3wQf0rld7N8Z9GlGs+x3DHw+6370xq0SSavdjLu3rlDuE/crznPfJm7sM9yhs3+ET9wOrYBzb/3drGMro8ox357d3vJHXPXA18JfBjrV8ns23N/5hURHWOW1bt0dHvlfq44ZXmvYKaW1qhSUPT3iCRxSaBVPHTXdjtLBLSNT9qeKC1rJDL2c0PFfqPA+IaVqu1s8MeN3ZUa3jB8eNkGWtUYaLD8V1BroAc/1AJCoMdXZOAQ8rQV4hTDk1TdMltf4VneYrdTRYqlgkq9iiojUoN21p7XVBU6waLGsbGPeVijsgjEgiwpjG0QkSm1xJqrKpkqBtvNHDAc3Dv2/eE3+dXFzQPEOgF43sCqlP9ynd6am7yt7rUtbYEWJJdzWVwS/YD5baFLW0MP76elwNpuOR//Po4Z/jncf7y9W5aX1j/WD1Ld+6bz22nlrPrEMr3Px7687Wt1vf9b7v/dL7rff7gnpjY5lzx1p5etN/AXn7Gyg=</latexit>

• Schrodinger operators on [0,1) with Robin boundary conditions.

Consider the equation with V � 0

� 00 � V = �� , 0(0) = µ (0).

As before let (��1, 1) be the lowest eigenvalue and its respective eigenfunctionand as before let

f1 = 01

1s.t. 1 ⇠ e�

p�1 x, f1 ⇠ �

p�1 x ! 1.

Moreover

f1(0) = 01(0)

1(0)= µ.

<latexit sha1_base64="UYR6d6GIWo5MGtAotGKTXNXHd2k=">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</latexit>

We introduce

Q1 =d

dx� f1

and

Q⇤1Q1 = � d2

dx2� V + �1 & Q1Q

⇤1 = � d2

dx2� V � 2f 0

1 + �1.

Besides,�d(Q

⇤1Q1) = {0,��2 + �1,��3 + �1, . . . },

and�d(Q1Q

⇤1) = {��2 + �1,��3 + �1, . . . }.

<latexit sha1_base64="UJDSrE2Stul9a2+3Y3IL/AZGe5I=">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</latexit>

The eigenfunctions of the operator Q1Q⇤1 satisfy

'j = Q1 j = 0j � f1 j

and'j(0) = 0

j(0)� µ j(0) = 0.

Therefore the problem is reduced to the Dirichlet boundary value problem andthus

16

3

nX

j=2

�3/2j Z 1

0(V + 2f 0

1)2 dx =

Z 1

0

�V 2 + 4f 0

1(V + f 01)�dx

=

Z 1

0

�V 2 + 4f 0

1(�1 � f21 )�dx =

Z 1

0V 2 dx+ 4�1f1

���1

0� 4

3f31

���1

0

=

Z 1

0V 2 dx� 4�3/21 � 4�1µ+

4

3�3/21 +

4

3µ3

=

Z 1

0V 2 dx� 8

3�3/21 � 4�1µ+

4

3µ3.

<latexit sha1_base64="nXFamfnY4lHXtbOBnHeRXS8SKIg=">AAAFHHiclVTLbtNAFHWbAMW8WliyuaKpmj4SbCeCsohUFRYsW6lNK3USazIeJ9PaY2tmXBGZfAgbfoUNCxBiwwKJv2H86CNJJcRIVq7vPefcOyczHsQBk8qy/iwsVqp37t5bum8+ePjo8ZPlladdGSWC0CMSBZE4GWBJA8bpkWIqoCexoDgcBPR4cP42qx9fUCFZxA/VOKa9EA858xnBSqfclYqDeMS4R7kyD0cUKBtS7iecZGUJkQ9KZ6OYCqwiAbUD1wb99DdrILWE9Mdg1momusAiHjH3DDpZGVAsi5ciWIcG+FfpjIC5N0OsWxs38NlbA4UJoO0yV9StZsbSkwrqR4Lm08Ui0tsNgUkQ1EsI9UBFeeUdE4yMAqpgECXcw2IMFzhIrinZGGqUSDDXTDSgQ8bTMAlU5ubmxES+wCS1X03S1iQfRCahm551nEmfAwq0yx52z/pp66UzAZ2ggBhXrtXXP74aQ70LW+Doja9v9J1MwPsAZsecBmmar+pdDdiCdobVLB3mLCTYcKQ2SipCZgf+yb6cyy487zvTKvMSmlzoawW4Zmsu2mPDj9fABuSGtFu5bmumett0V9KNdu7fpXhpWWOqYfZvb121mMWi7bIGuphh9QBozszrhjl2p/XfbbcL7aaJKPduHIbshJTLXV61mla+YD6wy2DVKNe+u/wLeRFJQn3JSIClPLWtWPVSLBQjAdXnLJE0xuQcD+mpDjkOqeyl+eWewJrOeKAPu364gjx7k5HiUMpxONDIEKuRnK1lydtqp4nyd3op43GiKCdFIz8J8rujvxTgMUGJCsY6wEQwPSuQEdY2Kf09yUywZ7c8H3Sdpt1uvjlwVnf3SjuWjOfGC6Nu2MZrY9d4b+wbRwapfKp8qXyrfK9+rn6t/qj+LKCLCyXnmTG1qr//AjZ+jfA=</latexit>

Theorem. (P.Exner, AL, M.Usman)

We obtain the inequality

3

4µ�1 +

1

2�3/21 +

nX

j=2

�3/2j 3

16

Z 1

0V 2 dx+

1

4µ3. (⇤)

Remarks.1. Let µ = 0 (Neumann problem). Then

1

2�3/21 +

nX

j=2

�3/2j 3

16

Z 1

0V 2 dx.

<latexit sha1_base64="IzBPLuI1LLzxbdWgNskme5Ze8x8=">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</latexit>

2. Let V ⌘ 0 and µ < 0. Then

� 00 = �� , 0(0) = µ (0).

Then (x) = e�

p�x, where �

p� = µ

and the inequality (⇤) becomes equality

�3

4�3/2 +

1

2�3/2 = �1

4�3/2.

<latexit sha1_base64="1UDFTZjucIjnpzktwvnjm9jO2HM=">AAADP3icbVJNb9NAEF2brxI+msKRy4gaJYXG2GklQAKpggsHDkVq0krZNFqv182q9trZXZdGlv8ZF/4CN65cOIAQV26sHRfRhJFWmnnvzczO7AZZzJX2vC+WfeXqtes31m62bt2+c3e9vXFvqNJcUjagaZzKo4AoFnPBBprrmB1lkpEkiNlhcPqm4g/PmFQ8FQd6nrFxQk4Ejzgl2kCTDWuARcpFyIRuFTiIoO+W8I5pcIaYzXJ+Bp4DRITg4CR/6TkuHEyZgJbjtHo4U7zTgVfQAxybliGBCtoGPJvlJKyDTtfbMgqTXIcmcqvcv0Vq8LySsOOih9VM6qIpVsJ5WdWqShVYJvBhyiQrG2RZW/eoKlaX1VMGZiFGGHM9B6f7eMuBgNE0YQou4MUMkSR0ZxdvX4xwXOw87ZfwBGrG768w9bg1Z7IuUfVkC5u0Nz3Xqw1WHb9xNlFj+5P2ZxymNE/MO9CYKDXyvUyPCyI1pzErWzhXLCP0lJywkXEFMZOMi/r9S3hkkBCiVJojNNTovxkFSZSaJ4FRJkRP1TJXgf/jRrmOno8LLrJcM0EXjaI8Bp1C9Zkg5JJRHc+NQ6jk5q5Ap8TsRpsvVy3BXx551Rn2XX/XffG+v7n3ulnHGnqAHqIu8tEztIfeon00QNT6aH21vls/7E/2N/un/Wshta0m5z66ZPbvP4wu+LI=</latexit>

Recently Lukas Schimmer has obtained a di↵erent inequality by using the so-called the double commutation method:

Theorem. For any V 2 L2(R+), V � 0, the negative eigenvalues ��j of�d2/dx2 � V with Robin boundary condition '0(0)� �0'(0) = 0 satisfy

X

j

�3/2j 3

16

Z 1

0V 2 dx� 3

4

X

j

�j(�j�1 � �j) +1

4(�3

j�1 � �3j ),

where

�j = �j�1 +|'0

j(0)|2

k'jk2

with 'j denoting the eigenfunction to �j .

<latexit sha1_base64="bZpr/GOq3RFI466Ow0KMqbJLJbI=">AAAEb3icbVNLb9NAEHaTAMW82nLgUIRG1IhEbVInqXhJSFWREIceSmnTSnVirdfreBN713jXaSPXR/4gN/4DF/4Bu07ok7ns7DffzmtnvCSiQtr2r4VKtXbn7r3F++aDh48eP1laXukJnqWYHGIe8fTYQ4JElJFDSWVEjpOUoNiLyJE3/qTtRxOSCsrZgZwmpB+jIaMBxUgqyF2u/HAYp8wnTJr7BKsjmsJuNkYCvuGQxjFJIVQX7kmkQviAwKdBQFLFBAV8z1BE5RS8KWSCsiHIkIDgTYyiSLH1zeeZysbEPI4zWYaFmMiQ+x9M0/HoUIxpYl5mkTteAAch4SmJWwV85ikgNgWr51AGu4NO3dnxPNh31xvWhkaHxFaKDsTIULmfECB0SNgERRkRYDWdSLXDR+7IAh6AaTX9QWfTPxt0oNmz4JTKEPa5p5x7PGM+SqeAOfNpmajlTFCahPR13W5AExxBhzFybZjDGv0ItgVCBRbB1LQs0xFZ7I7gIuog7252ClAAASdIEc67Rd5+U6hypGsP1BGo/vVUOs4G+Gc6imZ1t+CmJ6jP4+ejZru4TGfUgHWzfNTeus4ZdK+wBt3GBugMT0P1fTDLdWZSRVxzvT7P9HxepzvSDTgfdIrcucQcDZQOdQ+tC9wC9ZFc/puG8jeCjOGyo5Ir5sWPtMyZuEtrdssuBW4r7bmyZsxlz1366fgcZ7GaGBwhIU7adiL7OUolxREpTCcTJEF4jIbkRKkMxUT083JfCnilEB8CNVkBV2Ncoldf5CgWYhp7ihkjGYqbNg3+z3aSyeBdP6csySRheBYoyCJdtF4+tTopwXrDfIpwqkYMAw6R6rRUK6qb0L5Z8m2l12m1t1rvv3bWtnfm7Vg0Vo2XRt1oG2+NbeOLsWccGrjyu7pSXa0+r/6pPau9qMGMWlmYv3lqXJNa4y81b2J6</latexit>

Corollary.

For any V 2 L2(R+), V � 0, the negative eigenvalues ��j of �d2/dx2�V withDirichlet boundary condition '(0) = 0 satisfy

X

j

�3/2j 3

16

Z 1

0V 2 dx� 3

4

X

j

|'0j(0)|2

k'jk2.

<latexit sha1_base64="uRO79o5g7VakmVvwJXZG4mVfNuM=">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</latexit>

• Schrodinger operators with Hardy’s terms.

Let ↵ � 0, V � 0, V 2 C10 (0,1). Consider the operator H↵

H↵ = �d2

dx2+

↵(↵� 1)

x2� V,

acting in L2(0,1) with the Dirichlet boundary condition at zero.

Let

D↵ =d

dx�

↵

x.

Then

D⇤↵D↵ =

⇣�

d

dx�

↵

x

⌘⇣ d

dx�

↵

x

⌘= �

d2

dx2+

↵(↵� 1)

x2.

<latexit sha1_base64="/OKDnFTMgwxZHe+sEWmWeFi80UY=">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</latexit>

Consider the lowest eigenvalue ��1 and the corresponding eigenfunction 1

H↵ 1 = D⇤↵D↵ 1 � V 1 = ��1 1.

Since V 2 C10 (R+) we have

1(x) =pxI⌫(

p�1x), as x ! 0,

and

1(x) =pxK⌫(

p�1x) as x ! 1.

Therefore

1(x) =

8>><

>>:

�⌫/21

�(1+⌫) x⌫+1/2

⇣1 +

�1x2

4(1+⌫) +O(x4)

⌘, as x ! 0,

q⇡

2p�1

e�p�1x(1 +O(x

�1), as x ! 1.

Here ⌫ + 1/2 = ↵.

<latexit sha1_base64="UXKw0MRwmsY//vdckPV95/q3n0I=">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</latexit>

Let us introduce

f1 =D↵ 1

1= 01

1� ↵

x.

We find

f1(x) =

8><

>:

�1(⌫+5/2)4(1+⌫) x+O(x3), as x ! 0.

�p�1 +O(1/x), as x ! 1.

<latexit sha1_base64="lfCyWqOP1GMAh+snvGE2r4g+ZAE=">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</latexit>

There is a useful identity

f 01 =

✓ 01

1� ↵

x

◆0

= 001

1�

✓ 01

1

◆2

+↵

x2

=� V + �1 +↵2

x2�✓ 01

1

◆2

=� V + �1 �✓ 01

1� ↵

x

◆2

� 2↵

x

✓ 01

1� ↵

x

◆

=� V + �1 � f21 � 2

↵

xf1.

<latexit sha1_base64="nM48DIotx8lubYcqkkTtrAyJjto=">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</latexit>

LetQ1(↵) = (D↵ � f1).

Elementary computations imply

Q⇤1(↵)Q1(↵) := (D⇤

↵ � f1)(D↵ � f1) = H↵ + �1.

Indeed,

Q⇤1(↵)Q1(↵) = (D⇤

↵ � f1) (D↵ � f1) =

✓D⇤

↵ �D↵ 1

1

◆✓D↵ �

D↵ 1

1

◆

= D⇤↵D↵ �D⇤

↵D↵ 1

1�

D↵ 1

1D↵ +

✓D↵ 1

1

◆2

. (1)

<latexit sha1_base64="Gpn61j77+JbFpqf4sfHJsCIqGpM=">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</latexit>

We have

�D⇤↵D↵ 1

1= �

✓� d

dx� ↵

x

◆✓ 01

1� ↵

x

◆

= 001

1�

✓ 01

1

◆2

+↵

x2+ 01

1

d

dx� ↵

x

d

dx� ↵2

x2+↵

x

01

1. (2)

Similarly we find

� D↵ 1

1D↵ = �

✓ 01

1� ↵

x

◆✓d

dx� ↵

x

◆

= � 01

1

d

dx+↵

x

d

dx+↵

x

01

1� ↵2

x2, (3)

and

✓D↵ 1

1

◆2

=

✓ 01

1� ↵

x

◆2

=

✓ 01

1

◆2

� 2↵

x

01

1+↵2

x2. (4)

<latexit sha1_base64="brpdBh0SI6TUNZOPxLTW8gJbyOw=">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</latexit>

Therefore (1) - (4) imply

Q⇤1(↵)Q1(↵) = �

d2

dx2+ 001

1.

After using the equation

� 001 +

↵(↵� 1)

x2 1 � V 1 = ��1 1

we finally arrive at

Q⇤1(↵)Q1(↵) = �

d2

dx2+↵(↵� 1)

x2� V + �1 = H↵ + �1.

<latexit sha1_base64="Yxzxji0oKPvsD5KAXrFkizrhbSI=">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</latexit>

Consider now

Q1(↵)Q⇤1(↵) =

✓D↵ � D↵u1

u1

◆✓D⇤

↵ � D↵u1

u1

◆

= D↵D⇤↵ �D↵

D↵u1

u1� D↵u1

u1D⇤

↵ +

✓D↵u1

u1

◆2

. (5)

Then

D↵D⇤↵ =

✓d

dx� ↵

x

◆✓� d

dx� ↵

x

◆= � d2

dx2+↵(↵+ 1)

x2. (6)

�D↵D↵ 1

1= �

✓d

dx� ↵

x

◆✓ 01

1� ↵

x

◆

= � 001

1+

✓ 1

1

◆2

� ↵

x2� 0

1

1

d

dx+↵

x

d

dx� ↵2

x2+↵

x

01

1. (7)

<latexit sha1_base64="WyZf5Gi3OUM1b7n/01p2nQPdLoA=">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</latexit>

Moreover,

� D↵ 1

1D⇤

↵ = �✓ 01

1� ↵

x

◆✓� d

dx� ↵

x

◆

= 01

1

d

dx� ↵

x

d

dx+↵

x

01

1� ↵2

x2, (8)

and as in (4)

✓D↵ 1

1

◆2

=

✓ 01

1� ↵

x

◆2

=

✓ 01

1

◆2

� 2↵

x

01

1+↵2

x2. (9)

<latexit sha1_base64="wG/Ib+KjDS688QNzgsMRvpomzws=">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</latexit>

Adding together (6) - (9) we arrive at

Q1(↵)Q⇤1(↵) =�

d2

dx2� 001

1+ 2

✓ 01

1

◆2

=�d2

dx2+ 001

1� 2

001

1�

✓ 01

1

◆2!

=�d2

dx2+↵(↵� 1)

x2� V + �1 � 2

✓ 01

1�↵

x+↵

x

◆0

=�d2

dx2+↵(↵+ 1)

x2� V + �1 � 2f 0

1 = H↵+1 + �1 � 2f 01.

<latexit sha1_base64="Cj6nKCsbzjKwVsyLDw4RF+135Xw=">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</latexit>

We use now thatf 01 = �V + �1 � f2

1 � 2↵

xf1

and obtain

Z 1

0(V + 2f 0

1)2 dx =

Z 1

0V 2 dx+ 4

Z 1

0(V + f 0

1)f01 dx

=

Z 1

0V 2 dx+ 4

Z 1

0

⇣�1 � f2

1 � 2↵

xf1⌘f 01 dx.

<latexit sha1_base64="0sZGzV9/qxNUgO3HDUZHoBnUXmk=">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</latexit>

Using asymptotic properties of f 01 and f1 at zero and infinity we find

�1

Z 1

0f 01(x) dx = ��3/2

1 and �Z 1

0f21 f 0

1 dx = �1

3f31

���1

0=

1

3�3/21 ,

Integrating by parts we also obtain

� 2

Z 1

0

↵

xf1(x) f

01 dx = �

Z 1

0

↵

x(f2

1 (x))0 dx

= �↵

xf21 (x)

���1

0�

Z 1

0

↵

x2f21 (x) dx = �

Z 1

0

↵

x2f21 (x) dx 0.

This finally implies

0 Z 1

0(V + 2f 0

1)2 dx =

Z 1

0V 2 dx� 8

3�3/21 � 4

Z 1

0

↵

x2f21 (x) dx

<latexit sha1_base64="ESD9wDa2UQJEq7FWIBdOri7iANc=">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</latexit>

Note that after the first swap of Q1 and Q⇤1 the original operator

H↵ = �d2

dx2+

↵(↵� 1)

x2� V,

becomes

H↵+1 = �d2

dx2+

↵(↵+ 1)

x2� V � 2f

01.

Since ↵ � 0, the variational principle allows us to omit the positive Hardy term

and reduce the problem to the operator

H = �d2

dx2� V � 2f

01, in L

2(R),

where V + 2f01 is extended to the negative semi-axis by zero.

<latexit sha1_base64="1EVcfWc97HlUrZfGl3AXV0CRqvI=">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</latexit>

Theorem.

If ↵ � 1/2 and V � 0. Then for the negative eigenvalues of the Schrodingeroperator

H↵ = �d2

dx2+

↵(↵� 1)

x2� V in L2(R+)

we have

8

3�3/21 (H↵) +

16

3

1X

k=2

�3/2k (H↵)

Z 1

0V 2(x) dx� 4↵

Z 1

0

1

x2f21 (x) dx.

<latexit sha1_base64="cvZqx+rI3frPxFuZNxPaVhaJj5U=">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</latexit>

Remark. In particular, assuming that ↵ = 1/2 we immediately obtain an in-

equality for the Schrodinger operatorH1/2 = �d2

dx2 �1

4x2 �V with the subtracted

sharp Hardy term

8

3�3/21 (H1/2) +

16

3

1X

k=2

�3/2k (H1/2)

Z 1

0V 2

(x) dx� 2

Z 1

0

1

x2f21 (x) dx.

<latexit sha1_base64="YDHGiYXd/+hnWN8HC/Mo7VmMLVo=">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</latexit>

• Bounds on the lowest eigenvalue.

Consider

H = ��� V, in L2(Rd

).

Proposition. Let d � 3 . If kV+kd/2 Sd , where Sd is the constant in

the Sobolev inequality, then the Schrodinger operator ��� V has no negative

eigenvalue.

Proof. By Holder’s and Sobolev’s inequalities we have

Z

Rd

(|ru|2 � V |u|2) dx � kruk2 � kV+kd/2 kuk22d/(d�2)

� kruk2�1� S�1

d kV+kd/2

�� 0.

<latexit sha1_base64="mXoTcabcachfOgBqKh8ULTt4cWs=">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</latexit>

Conversely, if there is a constant C such that ���V has no negative eigenvaluesfor kV+kd/2 C, then for such V we have

Z

Rd

(|ru|2 � V |u|2) dx � 0

and therefore Z

Rd

|ru|2 dx � supkV kd/2C

Z

Rd

V kuk2 dx.

<latexit sha1_base64="fTKiRrk2mnb41yFqbSZj+RwRuMQ=">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</latexit>